This is a simplified approach, just to explain how it works.
The principles and calculations are basically the same for a push pull stage as it is for an SE stage, so everything described in the Impedance matching of loudspeakers to output valves article are valid for any transformer, but there are other concerns to be taken when calculating a push-pull transformer. However, let me take a little look at the calculations and the transformer, before we look at the output stage itself.
To use the formulas to calculate transformers it is important to remember that an output transformer has no impedance in itself; it reflects the impedance and any resistances on the secondary side over to the primary side and thereby present an impedance as a load to the output valves.
The load to valves connected as a push-pull stage is always specified as “anode to anode” (Zaa) illustrated in the drawing below, and since B+ is AC ground, each valve actually sees a load impedance that is Zaa/2.
Now that we have established how to specify load, let us look at the formulas for calculating the transformer. The key factor is, as always, the turnover ratio of the transformer. That is the ratio between the number of turns on the primary side divided by the number of turns on the secondary, which equals the voltage on primary divided on the voltage on secondary, and finally the square root of the reflected impedance on the primary side divided by the impedance on the secondary side, and we could off course us the current too, so in a formula it looks like this:
n = N1/N2 = V1/V2=√(R1/R2)=Is/Ip
Moreover, as before:
N1 = number of turns on primary (also Np)
N2 = Number of turns on secondary (also Ns)
V1 = Voltage on primary side
V2 = Voltage on secondary side
R2 = Load impedance on secondary side (loudspeaker)
R1 = Impact of R2 reflected over to the secondary side (here: anode-to-anode load)
Ip = primary ac current
Is = Secondary ac current
n = the transformer turnover ratio
All values illustrated below:
One of the common situations is that we want to calculate the load a given transformer presents to the valves. So let us look at these calculations.
The specifications for the Toroidy transformer, without the primary impedance is:
Technical data
Intended for | Push-Pull |
Core type | Toroidal |
Ultralinear tap | 43% |
Nominal Power | 50W |
Nominal anode current | 200 mA |
Frequency bandwidth (-3dB) | 8 Hz – 54 kHz |
Secondary Impedance | 4 and 8 Ω |
Primary Impedance | |
Turns Ratio (Np:Ns) | 40,62:1(4Ω) , 28,72:1 (8Ω) |
Primary Inductance Lp | 591 H |
Primary Leakage Inductance Lsp | 3,5 mH |
Total Primary DC Resistance | 96,1 Ω |
Effective Primary Capacitance | 11,2 nF |
Dimensions (standard) | 115 mm (OD) x 65 mm (h), weight: 2,7 kg |
Dimensions EXPO | 120 mm (OD) x 80 mm (h), weight: 4,4 kg |
Let us now calculate the primary impedance of this transformer, we can choose 8 or 4 ohms secondary output for the calculations, the primary should be the same anyway:
The formula solved for R1 (primary impedance) is:
R1 = (n) ² x R2 => (28.72) ² x 8 = 6598.7 ohm anode to anode. (If solved for R2 = 4 ohm, we get 6599.9)
However, we could, if wanted use the 4-ohm tap connected to an 8-ohm speaker, and what would the load on the valves then be? The 4-ohm tap has a turnover ratio of 40.62 so the calculated value will be:
R1 = (n) ² x R2 => (40.62) ² x 8 = 13 199.9 ohm.
That means, if we use the 4-ohm tapping for an 8-ohm speaker, the same transformer would reflect a load of 13.2 Kohm to the valves. This shows that the transformer has no impedance of its own; it is all due to the turnover ratio.
The spec from the manufacturer is 6.6 Kohm, under the usual conditions, so the calculation is correct. This solves the calculation of the turns ratio as far as the load on the secondary side is concerned, but we also have some losses in the transformer to take into consideration. Let us look at the loss due to the resistance in the copper wire of the transformer itself.
Although this is the usual method used to calculate a push-pull transformer, the a-a impedance is not the impedance each valve sees during signal conditions, more about that a little later.
Now let us take another example, we are going to make a power amplifier with two EL34 Pentodes in push pull configuration. We are going to use cathode bias and 450 volts on the anode. According to the Mullard datasheet, we need the load to the valves to be 6.5 k (Zaa), and the quiescent current should be 60 mA in each valve.
The maximum signal current should be 71.5 mA. The power output will be 40 W. The loudspeakers are 8 ohms.
Now let us calculate the turnover ratio for the needed transformer:
n = √(Rp/Rs) = √(6500/8) = 28.5
We also need to check that the transformer can take the voltage and maximum current. If we look at the transformer specified above, it is within the margins for the wanted specifications.
Primary DC resistance:
Let us introduce the dc resistance to the case:
When no signal is present, the valve sees only the dc resistance of the primary winding ( actually half of the total primary winding in a push-pull stage, at least in a class B stage). However, as soon as ac signal is present, the valve will see the dc resistance of the primary turns of the transformer in series with the reflected impedance from the load. Let me take an example, and for this example II will go back to the single ended stage, it works in the same way, just imagine that this is half of the push-pull stage seeing half of the transformer:
Let us say that the DC resistance of the primary winding is 200 ohm (R1)( it is a very poor transformer,but it only to show how it works); drawing “a” above show the no signal condition, the valve will only see the resistance of the primary copper winding, presented by R1 in the drawing. This is the resistance used to draw the DC load line, quiescent condition, with no signal present.
Once the ac signal is present there will flow AC currents in the transformer and the valve will not see only the 200 ohm, but also the reflected secondary impedance n² x R2 in series with it. If we set “n” to be 14 and the loudspeaker to be 8 ohms n² x R2 = (14) ² x 8 the reflected impedance is 1568 ohms. The total AC load the valve see is then 1568 + 200 = 1768 ohms.
Therefore, the DC load presented to the valve is 200 ohm, and the AC load is 1768 ohm.
We have not taken the winding resistance of the secondary into consideration in the formula, but it is reflected the same way as the loudspeaker impedance.
If we look at the data for a real transformer above, the DC resistance of the primary winding is only 96 ohm, and the resistance for the secondary will always be less than this, the normal tolerances among different valves (ri variation) is much larger than this. Therefore, in the real world it will not make any noticeable difference whether it is included in the calculations or not, except when we want to draw the DC load line, or off course if we are designing a transformer, but that is not what this article is dealing with.
We also have other losses in an output transformer, among them are:
Power losses because of setting up the magnetic field, these are hysteresis loss and eddy current loss, these are among the main reasons that a transformer for a SE amp must be a lot larger than for a P-P amp, with similar specifications. There is some more, losses for the low and high frequency range, these depends on the physical construction and design of the transformer, but none of them will be dealt with in this article.
Triode Push-Pull stage:
Figure 1
The push-pull output stage will act different in class A than it would in class B, when it comes to how the transformer is used, with class AB somewhere between these. Let us look at class A operation, but first another drawing to explain things.
This drawing is a so-called equivalent circuit to the output stage. The following are new to this drawing:
ri | Internal dynamic resistance of the valve (rp in us) (AC value) |
Rp | Primary copper resistance |
Rs | Secondary copper resistance |
Ns | Secondary turns number |
Np | Primary turns number |
Zs | Secondary load impedance |
0.5 x Np is the turns for half of the primary (a to a), of the transformer and, 0.5 x Rp is half the total copper resistance of the primary, it will be important later. Everything to the left of the black dots are internal to the valve itself.
Let us first look at what happens in a class A output stage:
In a class A stage both of the primary turns (0.5 x Np) will conduct at the same time, but in opposite phase. This will cause a secondary current since the primary halves are wound in opposite direction. Now the impedance of the transformer presented to the valves will be:
Remember the valves uses both primary halves at the same time:
We can make it a little simpler:
In addition, even simpler if we do not take the copper loss in to the calculation and this is the formula you will use in most cases:
The load (anode impedance) to each triode will be, without copper losses:
Now to class B triode push-pull amplifier:
Here the primary transformer halves conduct the anode currents alternately. Each triode is conducting or blocking alternately, so here we have only one half primary conducting at the time.
The load to each valve, without the math leading to it, is:
This is for the triode that is conducting, the other triode is blocking and has Za = ꝏ, but the anode voltage is still changing. The load impedances pr. Triode in class B varies between 0.25 times Zaa through Za, to infinity depending on the signal, or put in another way:
0.25 x Zaa < Za < ꝏ
Class AB lies somewhere between A and B. and the load impedance pr. Triode varies between:
0.25 x Zaa < Za < 0.5 x Zaa
We may say that the load that the transformer presents to the valves does not have a constant value for all values of current through the transformer windings. In addition, that goes for the internal AC resistance of the valve (ri) too.
This shows that the dynamic (AC) load line for an output stage actually never will be a straight line, it is in fact an ellipse, but we cannot use that for calculations, so we assume that the load is resistive and that gives us a satisfactory result, but that is not the subject here.
When it comes to selecting the load for a given set of valves, many different compromises have to be made, and selected. I recommend checking the manufacturer’s data.
Although I have presented the formulas above for a triode stage, they are also valid for a pentode output stage.
When specifying an output transformer for an amplifier we usually specify the load impedance wanted a-a (or you could use the turnover ratio), but we also must specify the maximum anode current you expect the output valves to draw.
There are a lot more to say about the design of push-pull power amplifiers, and transformers, but it is not the subject of this article. Among the things not covered are Composite load lines, B.J Thomson method of load lines, most favourable load and a lot of other stuff. Some of which will be the subject of a later article.
]]>Triode.
The triode consists of three electrodes, cathode, grid and anode and these are different electrical conductors with insulator between, so there has to be capacitance between them too. We normally calculates with three capacitances in a triode; the grid to anode capacitance, Cga, the grid to cathode capacitance, Cgk and the anode to cathode capacitance, Cak. We visualize them like this:
These capacitances are in fact unwanted; we actually call them parasitic capacities, so the manufacturers did as much as they could to minimize them. Another way to draw them, is like this:
In this drawing, we can see that Cgk is at the input circuit and that Cak is at the output circuit. Cga is between the input and output circuit, so it must influence both the input and output capacitance of the triode circuit.
Cak is AC wise in parallel with the anode load. Moreover, it will start to influence the circuit as soon as the frequency gets high enough so its reactance approaches the same value as the anode load. It will then gradually reduce the amplification of the circuit as the frequency increases.
Let me show you how this looks, I have altered a spice model for 6N1P-EV, in the original model Cak is 1.6 pF and the frequency plot from LtSpice looks like this:
We see that the stage amplification is constant above 100 kHz.
Now let us try the changed model, with Cak set to 150 pF (I could just have put a capacitor in parallel with the tube, anode to cathode, but I also wanted to test the spice model.):
Here we can see that the amplification starts to decrease slightly at about 30 kHz, and falls off due to the high value of Cak. The total load presented to the stage was 6875 Ohms. (Ra 22k and load 10k) The 150 pF value used for Cak here is not very realistic, but it is just to illustrate how it works. Cak will rarely be problematic at audio frequencies, but should be taken into account at frequencies in the MHz range.
However, the output capacitance of a triode stage does not consist of Cak alone, Cga is connected to the anode, and it has some influence. Moreover, without writing the entire mathematical derivation, the formula for output capacitance is:
Where A is Dynamic amplification =
We can see that the output capacitance is not particularly affected by the amplification because:
Cak in the 6N1P-EV is 1.6 pF and at 10 MHz has a reactance of approx. 9950 ohms.
The formula for Capacitor reactance is:
Now let us look at Cgk, this capacitor will act as a load to the signal source. It will be in series with the signal source internal resistance. As in the figure below:
The cathode capacitor Ck has very little reactance in the audio frequencies, so we consider it as a short. Ri (signal source internal resistance) and Cgk forms a voltage divider for the signal voltage. As the frequency increases, and the reactance of Cgk decreases, less of the signal voltage will be presented over Cgk, and thereby to the valve. A high value of Ri will make the frequency where this happens to decrease, and the signal voltage to the tube decreases at a lower frequency. The typical values of Cgk, in the order of a couple of pF will probably not make any problems at audio frequencies, but can be worth considering when designing a filter circuit, and off course at radio frequencies.
Now let us look at Cga, which is the capacitor that has the largest influence on the amplifier stage. This is mainly because it is presented between the anode circuit (output circuit) and the input circuit. This causes it to have a major impact on the stage input capacitance. Let us take a closer look at this, with a restive load to the valve, as in the figure below:
The following will be used in this explanation:
A= Dynamic amplification
Xcgk = Cgk reactance
Xcga = Cga reactance
The anode AC voltage Ua are in the opposite phase of the grid input AC voltage Ug, and is presented over serial connection of Cgk and Cga, as in the figure below.
We then have:
This means that the voltages over Cgk and Cga are in opposite phase and that the voltage over Cga is (1+A) times larger than the voltage over Cgk. The current through Cgk (I1) is 90 degrees ahead of Ug, and the current through Cga (I2) is 90 degrees ahead of Ucga. I1 and I2 are then in opposite phase so when I1 floats down I2 floats up, as shown by the arrows in the drawing above.
By using Kirchhoff’s Law’s we can find that the total current ig, in the grid circuit is:
( Kirchoff’s circuit laws can be found here)
And this can be written as:
And here I could write a lot of math, which will eventually lead us to a formula for the input capacitance of the valve when signal is connected between grid and cathode:
Alternatively
which actually is the same formula.
The conclusion is that the valves input capacitance is a lot larger than Cgk and Cga alone. It is dependent of the amplification (A) which depends on μ and Ra. More amplification in a stage means higher input capacitance, and the input capacitance is mostly dependent on Cga and the amplification. The input capacitance is increased by the amplification by the factor:. This is known as the “Miller effect”.
Let us calculate an example:
Valve: 6N1P-EV
Cgk=3.2 pF
Cga=2.7pF
Ug= – 4.2 v
Ri= 7.5 k
Ua= 300 v
μ (A)= 35.7
Cin = 83.63 pF
Which is a considerably larger value than both Cga and Cgk and should be taken into consideration when designing a valve stage The input capacitance can also be somewhat problematic if the anode load is reactive, the keyword here is resonance.
All of the above calculations is for a basic cathode circuit, like this:
The Cathode follower also has some input capacitance, but when using a cathode follower the output capacity, Cak, has no influence. The cathode follower (or anode circuit) looks like this:
The formula for the input capacitance is (without any math leading to it):
At last, if you want to use a “common grid” coupling, the input capacitance equals Cgk.
and output:
Now let us look at the Tetrode and Pentode:
One of the reasons for the development of the tetrode and pentode was the large input capacitance of the triode, and the problems it caused, especially at radio frequencies.
This is the interelectrode capacities of a tetrode:
If the screen grid is effectively decoupled to ground this can be simplified. The two capacities between control grid and cathode, and control grid and screen grid, are in parallel and the capacities between anode and cathode, screen and anode are in parallel, so we can draw it like this:
Input as always on the left hand side and output on the right hand side.
The grid to anode capacity of a tetrode or pentode is usually in the order of 0.005 pF, and this avoids the problems with input capacities that we have in triode stages, even if we still has the “Miller effect” in both tetrode and pentode stages.
Let us look at the capacities in a pentode:
The drawing shows that there is lot of different capacitances in a pentode, but fortunately, it is easy to simplify. The suppressor grid (g3) is connected to the cathode, and the screen grid usually has a large capacitor to ground, unless it is in ultralinear coupling. Therefore, their capacitances are not of importance. This actually leaves us with the same capacities as in the triode, Cg1k, Cg1a and Cak.
Just as in a triode the Miller effect will increase the input capacitance the same way as in the triode, but the suppressor grid is grounded (connected to cathode) so this will help, it will decrease Cga with a large factor depending on the amplification, and the net input capacitance will remain low. For all practical purposes, we can neglect the Miller effect for pentodes, at least when it comes to audio frequencies.
For the pentode, (common cathode stage) we then have:
Input capacitance:
And finally, output capacitance:
Cout = Cak
That is the end of this little primer on tube capacitance’s, without too much math involved.
]]>B + denotes the high voltage of the amplifier.
IMPORTANT:
Disconnect power and let all the capacitors discharge before you begin. Remove the output tubes from the amplifier.
There are dangerous voltages in a tube amp, so be careful, check that there is no tension on the circuit by measuring the B +.
It is important to identify the correct wiring of the transformer before beginning. The primary side will have two lines if it is a “single ended” amplifier, 3 if it is also ultralinear. One wire is connected to B + and the other is in most cases connected to the anode of the tube. The UL wire will be connected to the screen grid on the output tube.
A push pull amplifier will have a minimum of 3 wires on the primary side, and 5 if it is ultra-linear. You need to find out which wires are what. Begin by finding out what is B +, it will be the center leader at pushpull.
I take this as a starting point for a push – pull transformer.
Connect the multimeter to both wires that go to the anodes and note the resistance value (ohms). Then connect the multimeter to B + and the line that goes to one of the anodes of the valve. Note the value and measure to the second valve if it is a push pull transformer. The readings from the B + to the valves should be about half of the measurement between the anode connections at push-pull, if the difference is great, it indicates errors with the transformer.
The resistance in the winding’s from the anode to the anode will be on the order of 10 to a couple of 100 ohms, the same applies from the anode to the B +, but about half the a-a reading. Readings of values more than 1 Kohm almost always indicate that there is a break in the winding’s. Readings below about 10 ohms indicate short between two or more of the winding’s.
Do a similar test on the secondary side (speaker side), the value here will be lower than the primary side, usually from 0.5 to 10 ohms because there will generally be fewer winding’s and a thicker copper wire used.
Also measure the resistance between the winding’s and earth, on the primary side there should be hundreds of Kohm , if there is a low value here, it indicates a short to earth . On the secondary side, we has to check if one side is connected to ground before doing the test, in some cases it can, if so, disconnect and check. The readings should then be as on the primary side.
In some cases, it may be a good idea to disconnect from B + as one may be fooled by the power supply when measuring on the primary side.
Finally test between secondary and primary side, a low value here (should be many Kohm ) indicates short circuit.
If the transformer fails in any of these tests is probably faulty and should be replaced, but there may be other faults with it even if these simple tests are not showing anything. If the transformer seems to be OK after these measurements, we know that it is not completely short-circuited or has broken, but if it still does not work as it should, we could use some more advanced methods to find out if there is something wrong with it. This means connecting a signal to the transformer and making some simple measurements, using a voltmeter that can measure the frequency we are going to use, or an oscilloscope. And do some simple calculations.
This method can also be used to find out some of the transformer data if it is unknown.
We can use another transformer as a signal source, this will give a 50 Hz (60 in the US) signal we can test with, but 50 Hz can be in the lowest layer as frequency so I prefer to use a signal generator that can provide 1 kHz signal with fixed amplitude.
Let me show you an example of how to do this, I’ll check an output transformer for an EL34 push-pull output stage. The drawing shows a transformer for push pull, with outlets for Ultralinear coupling, the picture shows a similar transformer but without UL connection.
We first check if the transformer is in order by measuring the resistance in the windings:
Here we measure between the points to be connected to the anodes on the valves, and get a value of 33,5 ohms, measurement from anode to B + gives 15.6 ohms on one winding and 15.3 on the other so the primary side seems like to be ok. Secondary side measurement (speaker side) gives us 1.2 ohms, and due to fewer windings and thicker wire, this is normal.
This transformer thus seems to be fine, based on windings resistance, but we still do not know Transformer conversion ratio and thus the load impedance it presents to the valves.
I will now try to find the load impedance using a signal generator.
We should be aware that a transformer does not have its own impedance when using it in an output stage, it has only one conversion ratio, and the actual impedance it presents to the output tubes depends on the value connected on the secondary side. See my article: Impedance matching of loudspeakers to output tubes
When we want to check a transformer, we can connect the signal to either primary or secondary, it does not really matter, but we must be aware that N1 is the winding on the primary side and N2 is the winding on the secondary side in the formula, whichever side we choose to connect the signal to when measuring the voltage. I’ve connected the signal source to the primary side in the measurements I’ve made here.
I connect the signal to the primary side (anode – anode connections), check this signal with the meter and note the value 7.597 volts rms.
I do the same on the secondary side and note the value 0.322 volts rms, I round up and use 0.323 as value.
We are looking for a ratio so it does not matter if you read the value in volt rms, p-p or whatever, as long as you use the same type of reading on both the primary and secondary sides.
The readings entered in the schematic will be as follows:
Now, we can calculate the transformer’s impedance on the primary side, ie the load it will give to the output valves with a given load on the secondary side.
We know that the turnover ratio of a transformer is:
n = N1 / N2 = V1 / V2 = √ (R1 / R2)
Where:
N1 = The quantity of turns on the primary side N2 = The quantity of turns on the primary side V1 = The voltage on the primary side V2 = The secondary voltage R2 = Secondary side load impedance ( speaker ) R1 = The effect of R2 reflected over to the secondary side ( anode load ) n = Conversion (turnover) ratio of the transformer
Based on what we know, we can calculate the transformer’s turnover ratio:
n = V1 / V2 = 7.597 / 0.323 = 23.52
We assume that the speaker’s impedance (load) is 8 ohms, which is quite common, this becomes R2 in the formula.
n = √ (R1 / R2) now let us solve this formula for R1:
R1 = (n) ² x R2 => ( 23.52 ) ² x 8 = 4425 ohms.
The measured transformer will thus present a load (anode to anode) of 4425 ohm (4.4 Kohm) to the valves at a speaker load of 8 ohms.
In these measurements and calculations, I have used an output transformer for push – pull, the principle is exactly the same at a “single ended” transformer, then you measure from B + to the anode and calculate in the same way.
If the transformer has multiple taps on the secondary side, we can use this method to find the conversion ratio for one tap at the time, The multiple taps will give you a transformer that can be used to present multiple loads to the valves, or it can be designed to be used with different speaker impedance’s.
]]>The key to understand this is to look at how a valve works. To make a long story short, this is what happens: The heater heats the cathode. The cathode, when heated, makes a cloud of electrons around it, some of the electrons fall back to the cathode, but some escapes. A few of them has enough energy and the right direction to hit the grid wires, and stay there. Now the grid has more electrons than it had in the beginning, and if it were not for the grid resistor it would be negatively charged, but the grid resistor leaks this charge back to ground and the grid is maintained at the same level as before. However, if we want grid leak bias we make the grid resistor several mega ohms (5 – 20 Mohm) so there only will be a minimal current to ground. This lets the grid develop a negative charge with respect to the cathode and the grid is then negatively biased. That is the whole story……. well off course not, there is more to it. Let us take a closer look.
Grid leak bias or Contact-potential grid bias is probably the least understood of the biasing methods. In addition, in a lot of literature, and at some places on the internet, grid Leak Bias is said to be dependent on the input capacitor, and that bias develops due to charging of the capacitor. This is not quite true; the capacitor will develop a charge and maintain the bias voltage during negative alterations of the signal, but it will not alter the bias voltage developed over Rg. The bias voltage is only dependent on the grid current, and the resistor. That is unless we overdrive the stage, then some other variables also plays a role, but the grid current still is the source of the bias voltage, more about this later.
The contact potential grid current exists together with the other grid currents as described in a previous article; I will threat it here as if it were present alone. The term contact potential can have more than one meaning, dependent on who uses it. To the valve user it is a biasing mechanism inherent to the valve when we use a large resistance in the grid circuit, that is what I will discuss here, and I will call it contact potential bias (CPB). CPB is the potential that is established across a high value resistor in the external grid circuit due to the flow of electrons from the cathode, collected by the grid.
The work function of a material can be expressed as the relative ease of emitting electrons. The cathode has a low work function to ensure efficient emission, and the grid has a high work function to prevent it from emitting electrons. Work function varies between 1 and 5 volts for typical materials used in valve manufacture.
Electrons emitted from the cathode has an initial velocity related to the temperature of the cathode and its work function. Some of these electrons have enough energy to reach the grid, even though the grid is has no positive potential, and these electrons develop a retarding field. This retarding field is dependent upon the work function of the grid and any emissive material that are deposited on the grid wires, and off course on the temperature of the grid itself.
Therefore, the potential that exists as a bias between the grid and cathode does depend on the difference in work functions of the grid and cathode. The tube manufacturers did actually attempt to deposit some amount of emissive material on the grid wires during processing, to establish a work function that would reduce the contact potential value to a reasonable value. This process will continue through the valve life, and will make the contact potential bias change as the valve is used and ages. When using contact potential bias several disadvantages should be noted.
When a valve is operated with contact potential bias, the valves input resistance may no longer be in the order of thousands of Mohm but is determined by the slope of the tangent to the grid current curve at the operating bias. (The value will often be in the order of 100 – 250kohm). The result of this is that the valve is loading the input circuit more than expected.
If the valve is biased, by other means, outside the contact potential bias region, but is driven into it by positive peaks that causes grid current to flow, the signal source is also loaded, and the result could be distortion in audio amplifiers (and lowered Q when tuned circuits are involved).
When high mu triodes and sharp cutoff pentodes with limited signal handling capabilities are used, grid leak bias restricts this even more. A common mistake is to operate high mu triodes at low anode voltages, when this is done, the contact potential bias voltage may be sufficient to cut the anode current to almost zero.
When contact potential bias is used, and the only place we can use it is in low-level stages of an audio amplifier, a high value of grid resistance (5 Mohm or more) is used to hold the bias voltage. It is important to remember that the input resistance may be considerably lower than the grid resistor, and the signal handling capacity is limited to signals not much larger than 300mV if you want to make sure that you do not overload the stage, increase the grid current, and get severe distortion.
The manufacturers did try to control contact potential in the valve types commonly used in this mode, but there may anyway be considerable variations between similar valves, and during valve life span, and this must be taken into account when designing such circuits.
Now comes the question; how do we calculate the bias voltage when using contact potential bias; the answer is, it is practically impossible.
There are many variable factors that act upon this mechanism, gas current (as described in an earlier article) that drives the grid positive, and the number of electrons that have an initial velocity high enough to attract to the grid even if it does not have a positive potential are among them. How all of these factors balance out is not easy to predict, and then comes the aging of the valve as another variable. All of these factors vary even among valves of the same kind and brand.
However, if we do not take all of these factors into consideration, and if we knew the grid current it would be rather easy. Let us say that the grid current for a given valve is 36 nA (0.036 uA) and the resistor is 22 Mohm, the resulting grid bias would be -0.8 Volt, just as in the schematic below, from a Fisher amplifier. In this example, this is the resulting grid current taking all factors mentioned into consideration, but the voltage given on the schematic would probably not be the same during the valves life span. In addition, if you try to measure it, your voltmeter would load the circuit so you may not read the correct grid voltage. (Even if the voltmeter has 10Mohm input resistance).
The most reliable way to check the grid bias voltage on a grid contact bias stage is to measure the voltage drop over the anode resistor, and then use the valve characteristic curves to determine what the bias voltage is. Let us look at the schematic below, B+ is 110 volts and the anode voltage is 80, so the drop over the resistor is 30 volts, the resistor is 68k. This gives us a current of: 80/68 = 1.17 mA.
A quick look at the characteristics shows that this does not add up, the grid voltage at -0.8 volts given on the schematic has to be the voltage you will measure with a voltmeter attached to the circuit, which makes sense since this is from a service manual.
I decided to do some practical tests and measurements, and made the following setup for the tests:
I also connected a signal source to the capacitor. All measurements done using a Fluke 867B. The first try was with no signal and a 6N2P-EV valve from the Voskhod factory produced in 1989. The voltage measured over Rg (Vg) was – 2 volts, and the voltage drop over Ra was 51 volts. Let us see how this confirms with the valve characteristics. 51 volts over 100k gives an anode current of 0.51 mA. As we can see in the figure below, it confirms very good with the measured value for Vg. I also tried with and without the capacitor in the circuit, and there was, off course, no difference in the voltages.
I measured with some other valves, and the results shown in the table below:
Valve | Measured Grid voltage | Calculated Grid voltage using Ia/Va curves | Grid Current | Anode Current | Anode resistor voltage |
6N2P-EV Voskhod 89 | -2 | -2 | 0.35 uA | 0.51 mA | 51 |
6N2P-EV Voskhod 81 | -2.35 | Approx -2.2 | 0.45 uA | 0.37 mA | 37 |
6N2P-EV Voskhod 81 | -1.22 | 0.22 uA | 0.88 mA | 88 | |
6N2P-EV Voskhod 81 | -0.6 | 0.11 uA | 1.1 mA | 111 | |
6N1P-EV Voskhod 91 | -1.4 | Approx – 1.5 | 0.25 uA | 1.5 mA | 150 |
6N1P-EV Voskhod 81 | -0.7 | 0.12 uA | 1.68 mA | 168 | |
6N5P Nevz 68 | -0.7 | 0.12 uA | 2.02 mA | 202 | |
6N6P Voskhod 81 | -0.9 | 0.16 uA | 2 mA | 200 | |
ECC83 Siemens | -1.35 | 0.24 uA | 0.11 mA | 11.1 |
The parallel resistance of the grid leak resistor (13 M) and the Fluke 867B (10M) are used when calculating the grid current. The parallel resistance is 5.65 Mohm, and this is the value used in ohms law to calculate grid current.
I also observed that connecting the Fluke 867B to the grid leak resistor did not alter the voltage over the anode resistor more than a few volts, so it probably did almost nothing to the CPB voltage on the grid.
All of these voltages measured without any signal applied to the circuit.
The table show us that the CPB voltage varies between different valves, as expected, but also between different valves of the same kind. That is also as expected due to causes mentioned before. This tells us that the grid bias, when set up as contact potential bias, will vary a lot with different valves, also off the same kind. The difference between the 6N2P-EV’s is actually 1.9 volts. The difference between the tested 6N1P’s is -0.7 volts. The variations will give a different working point (bias), even with the same valve types.
Some sources tell us that the Grid Leak Bias is dependent on the input capacitor, and the charging of it due to the signal, and that is not quite correct. The charging of the capacitor by the grid current will keep the bias of the valve stable during the negative signal cycle, and the signal voltage will, if it exceeds the CPB voltage, alter the bias by increasing the grid current. This mechanism comes into play as soon as we input a signal that is close to or larger than the Contact Potential Bias Voltage (CPB Voltage), and it leads to distortion, as I will show in a moment. First, a little look at the theory behind it. The illustrations and parts of this explanation is copied from “Neets book 6, pages 1-34 to 1-36”. Witch may be found on the internet. However, the text in Neets is, in my opinion, not correct, so I have altered parts of it.
GRID-LEAK BIAS is as the name implies, bias voltage developed in the grid leg portion of the circuit. We have the grid resistor, Rg_{ , }coupling capacitor Cc, and resistance Rgk. Resistance Rgk does not exist as a physical component, but it is used to represent the internal valve resistance between the triode’s cathode and grid. Electrically, Rgk is very large, in the order of several Megaohms, but if we drive the grid positive Rgk will decrease to a quite small value, down to about 500 ohms depending on the grid current, and this is important to remember, to understand what happens when we add a signal to the circuit.
Under quiescent conditions, some conduction occurs through the valve. Some electrons will strike the wires of the grid, and a small amount of GRID CURRENT will flow through Rg to ground. There will develop a voltage over Rg as long as it has a value of 5 – 20 Mohm. This voltage is due to the contact potential difference between the cathode and the grid, this voltage will develop over Rgk even if there is no capacitor in the circuit, as long as it is not connected to any low resistance circuit, as shown in the measurements described in the text above. The grid will be kept slightly negative. This limits the number of electrons that strike the grid wires, and an equilibrium is obtained where the grid stay negative as long as there is no signal to the circuit and as long as the signal is kept below the contact potential voltage.
Now let us see what happens when we apply signal to the circuit.
In the charge figure above, the positive-going voltage at the top of Rg will be coupled to the grid causing the grid to go positive. The positively charged grid will attract more electrons from the electron stream in the tube. Grid current will flow from the grid and the negative voltage over Rg will increase, this will also affect the right-hand plate of Cc. It causes the right-hand plate to go negative. (Electrostatic repulsion from the right-hand plate of Cc will force electrons from the left-hand plate of Cc, causing it to go positive.) The electrons will flow through the signal source, to ground, from ground to the cathode, from the cathode to the grid, and finally to the right-hand plate of Cc. This charge cycle charges the Cc to the same voltage that is developed across Rg during the positive half cycle.
You may wonder why the charge current went through the valve rather than through Rg. When the grid goes positive in response to the positive-going input signal, electrostatic attraction between the grid and cathode increases. This, in turn, reduces the resistance (rgk) between the grid and cathode. Current always follows the path of least resistance. Thus, the capacitor charge path is through the valve and not through Rg.
When the first negative alternation is applied to the circuit (Discharge circuit), the left-hand plate of Cc must go negative. To do this, electrons are drawn from the right-hand plate.
The electrons travel from the right-hand plate of Cc, through Rg keeping the voltage drop over Rg negative during the negative alternation of the signal. So, Cc will discharge for the duration of the negative alternation.
BUT Cc can only discharge through Rg, which is a high-resistance path, compared to the charge path.
Remember that RC time constants and the rate of discharge increase with the size of R. Cc can therefore charge through the low resistance of rgk to its maximum negative value during the positive half-cycle. Because Cc discharges through Rg (the high resistance path), it cannot completely discharge during the duration of the negative half-cycle. As a result, at the completion of the negative alternation, Cc still retains most of the negative charge it gained during the positive alternation, and the voltage over Rg is kept nearly constant. When the next positive alternation starts, the right-hand plate of Cc will be more negative than when the first positive alternation started. During the next cycle, the same process will be repeated, with Cc charging on the positive alternation and discharging a lesser amount during the negative alternation. Therefore, at the end of the second cycle, Cc will have an even larger negative charge than it did after the first cycle.
You might think that the voltage over Rg, and thereby the charge on Cc will continue to increase until the valve is forced into cut-off. Testing show that this only will happen if the input signal is very large, but not under normal conditions. As the negative voltage over Rg and the charge on the right-hand plate of Cc keeps the grid more negative, electrostatic attraction between the grid and cathode decreases, and so does the grid current that is the source of this voltage. This, in effect, increases the resistance (Rgk) between the cathode and the grid, until Rgk becomes, in effect, the same size as Rg. At this point, the grid current decreases, charge and discharge of Cc will equal one another and the grid will remain at some negative, relatively constant voltage.
What has happened in this circuit is that Cc and Rg, with unequal charge and discharge RC time constants (paths), have acted to change the ac input to a negative dc voltage by use of the grid current. The extent of the bias on the grid will depend on several things: the amplitude of the input, the frequency of the input, the valve (grid current, contact potential), and the time constant of Rg and Cc.
This type of biasing has the advantage of being directly related to the amplitude of the input signal. If the amplitude increases, biasing increases in step with it, this in turn reduces the anode current, and thereby the amplification of the valve. We have a sort of dynamic bias that reduces the amplification if the signal get large, and increases it if the signal decreases. However, remember, the signal must be of a large enough voltage to drive the circuit into overload condition to make all of this happen, so it is not very useful, at least not in a HiFi amplifier. The main limiting factor is the amount of distortion that you may be willing to tolerate, and my testing show that the distortion rises rapidly with increasing signal voltage.
Distortion occurs during the positive alternation when the grid draws current. Current drawn from the electron stream by the grid never reaches the plate; therefore, the negative-going output is not a faithful reproduction of the input, while the positive-going output (during the negative input cycle) will be a faithful reproduction of the input, unless the signal is large enough to drive the valve into cut-off, which is unlikely during normal use.
The picture above show input and output signal to a 6N2P-ev with 37 % distortion. The output signal is at the bottom of the picture, and we clearly see that the output’s negative going alteration is the most distorted. This is also when there is grid current.
If we are going to use this type of biasing in a HiFi amplifier, we must ensure that the signal never exceeds the CPB voltage, and this voltage is dependent on the actual tube in use. As mentioned before, to use this biasing method in a HiFi amplifier, the signal should stay below 300 mV to ensure that we don’t get severe distortion, and because you actually don’t know the value of the CPB voltage due to the variations among valves.
Let us have a look at some practical tests and measurements I have done:
The table below show measurements done with some different valves. The circuit used is the same as the one used before. Distortion measured with a LEADER LDM-170 distortion meter. The input signal is 1000 Hz sine wave. < = less than.
Input | JJ ECC83S | Mullard ECC82 | Siemens ECC83 | Siemens ECC81 | Voskhod 6N1P-EV 81 | Voskhod 6N2P-EV 81 | ||||||
Signal
Voltage (p-p) |
-Vg | THD
% |
-Vg | THD
% |
-Vg | THD
% |
-Vg | THD
% |
-Vg | THD
% |
-Vg | THD
% |
0 | 0.97 | 0.86 | 0.9 | 0.7 | 0.56 | 0.69 | ||||||
0.4 | 1 | 0.25 | 0.87 | <0.2 | 0.9 | 0.4 | 0.8 | 0.5 | 0.57 | 0.4 | 0.78 | 0.72 |
0.7 | 1.1 | 1 | 1 | 0.4 | 0.95 | 1.2 | 0.88 | 1.2 | 0.68 | 0.6 | 0.87 | 1.9 |
0.8 | 1.17 | 2 | 1.1 | 0.8 | 1 | 2.2 | 0.9 | 1.4 | 0.71 | 0.7 | 0.9 | 2.5 |
1 | 1.22 | 2.6 | 1.25 | 0.9 | 1.1 | 4 | 0.97 | 2.1 | 0.8 | 0.95 | 0.97 | 2.75 |
1.4 | 1.4 | 6.2 | 1.3 | 1.4 | 1.2 | 7 | 1.15 | 4.1 | 0.9 | 1.25 | 1.13 | 7.2 |
1.8 | 1.5 | 11 | 1.4 | 2 | 1.35 | 12.8 | 1.3 | 6.3 | 1 | 2.35 | 1.2 | 9.5 |
2 | 1.6 | 15 | 1.5 | 2.2 | 1.4 | 17.5 | 1.4 | 8 | 1.18 | 2.6 | 1.4 | 16.5 |
2.5 | 1.8 | 26 | 1.7 | 3.2 | 1.65 | 34 | 1.5 | 11 | 1.4 | 4.8 | 1.65 | 27.5 |
3 | 2.0 | – | 1.9 | 3.8 | 1.8 | – | 1.8 | 15 | 1.6 | 6.4 | 1.8 | 37 |
If we look at –Vg in the table, we clearly see that the grid voltage increases (gets more negative) with increasing signal voltage.
What we also see is that the valves with the highest amplification factor (u) gives much higher distortion than valves with lower amplification factor, and this is normal. The output signal from the ECC83/6N2P-EV is more distorted, but is also much larger than the output signal from the valves with lower amplification.
JJ ECC 83S is the only current production valve measured, and it measures very similar to the other ECC83/12AX7 types tested (Siemens ECC83 and 6N2P-EV).
The valve with the highest distortion measured is in fact Siemens ECC83. This is (I believe) an Ei produced valve labelled Siemens, in the other end of the scale is Mullard ECC82, which has the lowest distortion, followed by 6N1P-EV as the second best regarding distortion.
JJ ECC83S measured surprisingly low distortion with 0.4 v signal, but rises rapidly as the signal gets larger.
This confirms what I said earlier considering high “mu” valves.
The voltage at 0 signal is the CPB voltage. The CPB voltage develops as soon as the heater is starting and is not dependent on the anode voltage.
The signal voltages are peak-to-peak voltage; to convert this to rms we use the formula:
V_{rms} = 0.3535 * V_{pp}
0.4 volts p-p = 0.14 Vrms, 0.7 Vp-p = 0.247 Vrms and 0.8 Vp-p = 0.282 Vrms.
To summarize:
The Grid Leak Bias voltage is only dependent on the voltage drop the grid current makes over Rg. It is not dependent on the capacitor and the charge and discharge of it. The capacitor charge only helps to keep the voltage almost constant over Rg during the negative alteration of the signal voltage. If the signal exceeds the contact potential voltage the grid current increases, and so does the voltage drop over Rg, but when this happens the distortion rises rapidly and makes this type of biasing useless for signal levels that is larger than or equal to the Contact Potential Voltage. At least for HiFi amplifiers where the level of distortion that occurs is unacceptable.
One last comment:
If we drive, a normal biased stage into overdrive by allowing a signal that is larger than the grid bias voltage the stage will off course start to draw grid current, in the same manner as a CPB stage, and distortion will occur. However, as long as we stick to the manufacturers recommendations regarding maximum value for Rg, it will not develop a large negative voltage on the grid for a longer period, this because the grid current will “leak” to ground through Rg.
Although it has been used in some rather successful RIAA amplifiers, and as input stages for low-level signals from microphones. Grid leak bias, or contact potential bias gives a rather unpredictable bias point for the valve, and is, in my opinion, not the biasing method to choose for a HiFi amplifier.
]]>Domestic receiving tubes and their foreign counterparts by: В.Каинельсон, А. С. Ларионов (V. Kainel’son, A. S. Larionov 1981).
I have put the data together in a document over Russian tubes and their western equivalent and possible substitutes, as seen from the Russian side.
The first table in the document has 4 headers, first the russian tube model in cryllic, then the same in western letters, next is called “foreign counterparts, which is the equivalents, the next header is called: some European and American analogues, this is tubes that are near equivalents, but may have different socket, pin layout or other differences.
There also is a table called: possible substitutes, and that is exactly what it is, not equivalents but tubes that can be used as substitutes, if you can’t get the original tube.
All of the data comes from the russian book mentioned above, and I have not checked it in any other ways.
The document can be downloaded from the following link:
]]>A loudspeaker usually has an impedance of 4 – 8 ohms, and an output valve will usually demand a load of several k.ohms. To make this work we need a transformer to match the low impedance loudspeaker to the high impedance tube.
I will assume an ideal transformer, with no loss or capacitance; it will be close enough to the real thing.
If we see this as an ideal transformer (no loss), we can use the following expression:
V1/N1 = V2/N2
By adding that the primary power equals the secondary power, we can use the following:
n = N1/N2 = √(R1/R2)
Where:
N1 = number of turns on primary
N2 = Number of turns on secondary
R2 = Load impedance on secondary side (loudspeaker)
R1 = Impact of R2 reflected over to the primary side (plate load)
n = the transformer ratio
Let us look at an example to clarify:
(Example taken from an old textbook, I used when I was at school (a long time ago..))
We will use a power pentode as an output stage in a SE amp.
Plate voltage is 250V, Grid has a signal level of 5 V, the load on the tube (Rl) should be 5.2 K.
The available transformer has a ratio: N1/N2 = 36. We see the transformer as ideal. (At this moment)
Loudspeaker impedance needed with the available transformer:
N1/N2 = √(R1/R2)
R2 is the loadspeaker impedance, R1 (Rl) is the optimum load impedance, by solving for R2:
R2= R1/(N1/N2)²=5200/(36) ² =5200/1296 = 4.02 ohm
If the manufacturer gives the impedance ratio (in this case 5200 to 4 ohms, we would determine the ratio as √(5200/4)=36, so this transformer would give us a 5.2 k load on the tube if we use a 4 ohm speaker, but what if we were to use an 8 ohm speaker with the same transformer?
The ratio will be the same, 36, but the transformer would present a different load to the valve:
√(R1/R2)=36 => √(R1/8)=36
Solve for R1 (anode load):
R1 = (n) ² x R2 => (36) ² x 8 = 10368 ohm would be the load reflected to the valve if we used an 8-ohm speaker on the same transformer.
Witch off course is twice the load reflected for a 4-ohm speaker, but here is the math to prove it.
Primary DC resistance:
Now let us see what happens if we do not see the transformer as ideal, let us introduce the dc resistance to the case:
When ac signal is present, the valve will see the dc resistance of the primary turns of the transformer in series with the reflected impedance from the load. Let me take an example:
Dc resistance of coil is 300 ohm (Rpri); with no signal present, the valve will see a 300-ohm resistance. This resistance is the one we will use for the DC load line in the curves. But, as soon as the signal is present this resistance will appear in series with the reflected impedance, using the same transformer as above this means (4 ohm load) that the total impedance that the tube sees is 300 + 5200 ohm = 5500 0hms. The difference by taking primary dc resistance into the equations is less than 10% so it is not going to make any big difference, the tolerances between valves is in most cases larger than this, but it might be important if we want to plot the DC load line. That is not the whole story; let us look at:
Power loss and plate load impedance with a non-ideal transformer:
The secondary side does also have a dc resistance, and we must to consider this to if we want to find the power loss in the transformer, and the total impedance the anode sees. The DC resistance of the secondary will be reflected over to the primary side by the transformer. The DC resistance of the primary side is usually rather small, about 0.2 – 0.5 ohms is a normal value, but it will be reflected over by the transformer ratio. Let’s take a look at the math:
Rpri is the primary DC resistance
Rsec is the secondary DC resistance
Rpri= 300 ohm (yes it is a bit high)
Rsec= 0.2 ohm
The secondary Dc resistance reflected to the primary is then:
n ² x Rpri => 36 ² x 0.2 = 259 ohm
The total copper resistance the tube sees is:
Rtot= Rpri + (n ² x Rsec) = > 300 + (36 ² x 0.2) = 559 Ohm. This is important if we are going to find the power loss in a non-ideal transformer.
The primary losses (and that includes the reflected Rsec):
(Rtot/Rtot+R1) x 100%= (559/559+5200) x 100 = 9.7 % (not a very good transformer)
The iron loss typically is about 5% so this transformer gives a total loss of 14.7 %, then the power efficiency is about 85%. That is not a very good figure; a good output transformer should have power efficiencies between 90 and 95%.
Plate load impedance with a non-ideal transformer:
To calculate the anode load with a non-ideal transformer we have to include all of the loss resistances mentioned above, and R2 (anode load) from the first figure:
The simplified equation above, with no loss, looks like this:
R1 = (n) ² x R2 => (36) ² x 4 = 5200
The equation with a non-ideal transformer will be:
R1 = Rpri + (n ² x Rsec) + (n ² x R2) => 300 + (36 ² + 0.2) + (36 ² x 4) = 5743.2 ohm
It is less than 15% off from the simplified equation, and, in my opinion, the simplified equation is accurate enough, due to manufacturing tolerances. Unless, of course, we plot the curves for the actual valve, measures the transformer, and then use these values for calculations, but then, we have to redo the whole thing if we have to replace valves.
There are other important parameters to consider when choosing a transformer, such as; primary self-inductance and the capacitance of the transformer windings, but I am not going to write about it at this moment.
The text above is mainly written for a SE stage, but it is also valid for Push Pull. If using a PP stage remember that, the primary winding is divided into two halves and these have an opposite winding direction. Each of them has o.5xN1 windings and the loss resistance is 0.5xRpri. I will not look any deeper into it here, but maybe write a bit about it later if there is interest for it.
The text, and equations, are valid for any transformer-based stage, the impedance and resistance values will be different if we use it for an interstage transformer, but the principles are the same.
]]>We do not usually want any grid current in small signal amplifiers (voltage amplifiers), in most cases it will have an unwanted effect and will usually shorten the valve life.
As mentioned grid current is in most cases unwanted, therefore we define it as negative, unlike wanted grid current, which we define as positive. Positive grid current will, in small signal amplifiers, only be the cases where we use it to produce the negative grid bias voltage.
Negative grid current includes current because of remnants of gas in the valve (gas current), grid emission and leakage current. The vacuum in a valve is never perfect, there will always be remnants of gas in it, and more gas can be present if the valve is overheated.
Gas current
Gas current occurs because the negatively biased grid attracts positive charged gas atoms. These atoms occurs due to collisions between electrons and gas atoms, when this collision occur an electron is unleashed from the atom so the atom gets a positive charge ( it becomes an positive ion) and because of the positive charge it will be attracted to the negatively biased grid. If there is a large resistance in the external grid circuit, it will develop a positive voltage over this resistance. This will in turn cause the grid bias voltage to become less negative, and the current in the tube will increase. This can than have a cumulative effect, the reduction in the negative bias will cause the anode current to increase, and there will be more collisions. The grid voltage will become even less negative and the current increases more, which will give more gas in the valve due to heating of the elements, and so on. The result from all of this can be a thermal runaway and in the end, the result will be a damaged valve. This is most likely to occur if we use a fixed grid voltage, if we use a “self biased stage” it will be more self adjusting because the increased current causes the cathode voltage to increase, which in turn gives a more negative bias to the grid. And it will to a certain degree compensate for the increase in current through the valve.
The problems with ionization of gas molecules happens more often in valves with oxide coated filaments, or cathodes, than in valves with filaments of pure tungsten. This is because the valves with filaments of pure tungsten usually have a harder vacuum compared with the oxide coated filament valves.
Gas current is the main reason for the maximum values for grid resistors (the resistor between grid and ground) given by the valve manufacturers. Examples are: 500k for 6N1P and 6N2P, and 1 – 2 M for ECC83 ( 1Meg for fixed bias, and 2 Meg for self bias). It is not advisable to use larger resistors than recommended, though we sometimes see it done. The use of too large grid resistors can in many cases be the cause of a shorter life span for the valves.
Grid emission.
Emission current from the grid is mainly due to cathode material transferred to the grid due to high temperature on the cathode. If that has happened and because of the very short distance from grid to cathode, the grid can be enough heated to start emitting electrons. The result of grid emission will be an increase in the anode current and eventually overheating of the valve. Emission can also occur from the screen grid if this gets overheated. If the screen grid is supplied from a resistor connected to the high voltage, it could get more positively charged due to screen grid emission. Emission from screen (or control grid) could cause the valve to be overheated and the result can be a thermal runaway and a destroyed valve.
Current leak.
Current leak can occur in the supporting structures of the valve. Mica are often used, and it is a good isolator, but in certain cases, especially if the temperature in the valve gets too high, or during fabrication, there can be metallic deposits on the mica wafers. These deposits can then develop further during the use of the valve, often due to high temperature, and can cause leak current between the elements of the valve. This current can be between any of the metallic elements that supported by the structures. As the picture below shows, that generally means all of the elements.
Current leak between the grid and cathode will lead to a reduction of the input impedance of the valve to values much lower than normal. This could have a devastating effect to the circuit around the valve, and may be the cause of a malfunctioning circuit. It can also be the cause of a reduced life span for the valve.
Worst case would be a leak current between the anode, or screen grid and the control grid. Leak current between those elements would make the grid considerably less negative biased and will give an effect as described before.
Leak current could also be the source of noise in the circuit, which will be very annoying in a small signal amplifier, e.g. RIAA amplifiers.
As mentioned in the beginning, we sometimes want grid current in a small signal amplifier, this is when use the grid current to bias the grid: Grid Leak Bias or Contact Potential Bias, and this will be the subject in another article.
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