# Trying to make a more accurate bias chart

Discussion in 'Amp Tech Center' started by peteb, Jul 17, 2017.

1. ### petebFriend of Leo's

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That was interesting measuring signal voltages at the grid of the power tube on different Amps, i didn't know what to expect.

What I found

As the amps got bigger, the bias voltage got bigger, and the AC signal got bigger too.

In each case it looks like the idea is to have the AC grid signal have a voltage that can reach at max half of the bias voltage. At max the signal is opening half of the blockage caused by the negative bias voltage.

If the signal voltage equaled the bias voltage, erasing all of the blockage of the bias voltage, it would probably redplate.

Changing the bias voltage will affect the ratio of signal V / bias V quite a bit.

If the max signal voltage can reach half the bias voltage it provides a good range. The good sounds start happening much earlier, as early as 5 or 10%, or 15 or 20 %, the mild overdrive starts.

I was strictly looking at the ratio of the signal voltage divided by the bias voltage, but it could also be looked at as what is the remaining bias voltage once you subtract the signal voltage from it? Comparing across amps with different bias voltages this way would give completely different results than comparing the ratios.

Last edited: Jul 21, 2017
2. ### petebFriend of Leo's

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I think this is a good summary of what I learned.

The standard 50,60,70 rule will accurately predict the upper bias ranges with respect to over powering the tube and bringing on redplate conditions. This will always work for any amp.

There is a lot of usefull ground below the 50-60-70 rule that's harder to quantify, and this can vary from amp to amp.

An alternative I have found to evaluating bias point is to look closer at the operation of the tube and measure and look at and compare the signal voltage to the bias voltage. the bias voltage needs to be large enough to keep the tube in check, and then the signal voltage needs to be big enough to counter act the bias voltage and let the tube flow. What I have found in typical signal voltages, is that they max out at or above half of the bias voltage. If I see this, my estimate is that the tube is biased and functioning properly, and proper tube tone will ensue.

3. ### robrobPoster ExtraordinaireAd Free Member

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Adjust your amp's bias by ear as long as it's under 70%. Test the amp tone with your (or your customer's) typical playing style including overdrive. The amp's tone is what matters.

At under 70% a healthy amp and tubes won't red plate. Power tube saturation (grid clipping really) and cutoff are a major part of the amp's overdrive tone. The 5F6A's power tubes begin to overdrive with the volume at about 25%.

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4. ### petebFriend of Leo's

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Oh yea, I forgot the voltages for you technical guys

An aa764 champ has a bias voltage of -20-23 VDC

The signal on the grid maxes out at around 12 VAC

12/23=52%

A 6g2 Princeton has a bias voltage of -40 VDC

The signal on the grids max out at 22 VAC

22/40=55%

An aa165 bassman has a bias voltage of -52 VDC

The signal on the grids max out at 30 VAC

30/50 = 58%

The one of interest is the low powered Princeton. It is biased cold at around 30%. And Bendyha, I know what you mean, when expressed in the standard bias way, everybody knows where it's at. The bias voltage is set at an elavated -40 degrees over the schematic value of -35. The question is, can the AC grid voltage over come the elevated bias voltage? The answer to me is, yes it can because the AC signal, 22V, is more than half of the bias voltage, -40, so it's OK.

5. ### petebFriend of Leo's

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++++++++++++!

6. ### petebFriend of Leo's

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Another thing I learned from this

The three keys to a healthy amp are:

1. the plate voltage needs to be of good strength.

2. The bias voltage needs to high enough to keep the tube from redplating

3. The signal voltage needs to be strong enough to counter the bias voltage to open up the tube.

7. ### robrobPoster ExtraordinaireAd Free Member

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Pete, the relationship between power tube bias voltage and signal voltage on the grid is a basic design parameter. You want to be able to fully drive your power tubes for max power output.

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8. ### petebFriend of Leo's

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Thanks Rob, this is consistent with my main points, I appreciate it.

9. ### 24 trackDoctor of TeleocityAd Free Member

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great post ! i hope i can save this as a refernce!

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10. ### Old Tele manFriend of Leo's

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This post is in agreement with robrob, I say the samething, just using equations and different words:

Peteb see this paper for more info about bias & drive signal relationship:

http://thermionic.info/mccaul/McCaul_PhaseInverterDriveVoltage_2008.pdf

Last edited: Jul 21, 2017
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11. ### RLee77Friend of Leo's

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Remember that bias voltage is not a separate adjustable parameter that can be thought of as an independent power level control. The operating point of the amp is set by the designer using several fixed parameters to set an ideal range, and bias allows tweaking within that narrow range. The load that the plate sees is a major factor, as is the supply voltage. An output transformer that matches the tube impedance is chosen for the intended speaker load and the desired supply voltage to deliver the output power wanted. Bias works with these parameters to allow adjusting for ideal operation when tubes with different characteristics are swapped in -- it all hangs together.

Another method of biasing an amp is to adjust for symmetrical clipping, to give max clean head room (hifi for example). A sine wave is used to drive the amp into slight clipping, and the bias is tweaked to balance the positive and negative waveform shapes using a scope.

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12. ### SnfoilhatTele-Afflicted

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Same amp of mine, the reverb driver.

Don't know if Weber follows the very same transformer turns ratio of the original Fender reverb unit, but I always wondered why the recommended part was Weber's 5W SE OT with 7k:4R windings. Standard 8R reverb tank gives a load of 14k. Seemed to me (knowing very little) to be off the charts high. But now it looks like just the right load (shallower gradient) to deal with the very high plate voltages (near the limit of a 6K6) you'd see in a reverb unit (or in my case a small combo amp) that runs a B+ of 300-350V. Some of your choices are going to be constrained rather than every number somehow optimized.

Last edited: Jul 21, 2017
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13. ### jazzguitarTele-Afflicted

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That's good for you!

The great thing is that decades ago, some ivory tower nerds researched and developed the entire theory behind tube (and general) amplification technology, so it is not about making theories but understanding the problems.

This is not defined by the parameters given here. It would depend on the particular conditions and purpose of the circuit it operates in.

Max plate dissipation is just one parameter to observe and not exceed in the operation of the tube if you want it to work longer than say minutes.

A tube can control plate current by varying control-grid to cathode voltage. Plate current times plate-to-cathode voltage is plate dissipation.

It is entirely possible to put out vastly more power than the plate dissipation figure, depending on the waveform, distortion characteristic and other parameters, provided other maximum design ratings (plate and screen voltages, cathode and screen current, etc) are not exceeded. Modern "Class D" technology and switching power supplies use this (with no screen current there is no dissipation, with maximum design current the plate voltage is low and hence dissipation also is low).

Analog signal amplification of old uses the grid-voltage-to-cathode-current transfer curve (others posted the graphs here) which means any current value between circuit-design-maximum and minimum are used, and somewhere in between the tube is dissipating the most power. The math is a bit complex but you end up with an output power of - rule of thumb - more or less the size of the maximum dissipation power of the circuit, which may - or may not be close to what the tube's manufacturer states as the limit of acceptable.

I hope that became a little clearer. Math tells us the most efficient source of electric power delivers output power the same as internal dissipation, which means the figures are connected and probably close.

The greatest differences between max dissipation of a circuit and power output are with Class C and Class D, only the latter is used in audio designs and only with special filtering and extra adaptation circuitry.

A Champ Class A will dissipate - like all class A amps - the same amount of power with signal or no signal, in other words whether there is output or not, volume or silence. The idle current (this is a characteristic of class A and not applicable to other classes) limits the signal size, so double the idle current is maximum and zero is minimum current signal.

It depends on the tube type how much signal voltage you need to drive the tube from zero to full current. If the input signal is smaller, then the variation from the idle (half of maximum circuit design) is lower, as is the output power. The tube always dissipates the same (which is why engineers tried to come up with more efficient designs, which are Class B, Class AB, Class C and Class D).

Hope this helped ...

Last edited: Jul 21, 2017
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14. ### BendyhaFriend of Leo'sSilver Supporter

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How much drive signal you need is very much dependet on the screen-grid voltage. Here are three charts for the 6L6, with the max,dissipation of 30W shown.
Just as a random comparison point I chose 400V at 70% dissipation. All three charts are the same except one has screens at 250V, the next at 300V and then 400V.
The chosen bias point current is identical for all, but the bias voltages are vastly different, and therefore also the sensitivity and maximum drive voltages to drive the tube into cut-off.
About 18.4V, 24.5V & 37.5V.
Adjusting screen-grid voltage in relationship to plate voltage is also a form bias, that effects the bias voltage between control-grid and cathode.

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15. ### clintjFriend of Leo's

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To the OP, sitting down with this book for a couple of weeks and working through some practice runs of designing on paper would really help clarify a few things. Or even better, take an existing amp and work backwards from the finished design to see why choices were made in the first place. I tried that when working on a dual EL84 power section, and I ended up with something pretty close to modern designs.

One factor that needs to be considered when figuring the power and response is screen voltage and its power supply design. The dance between the screen and plate currents has a noticeable effect - that's the basis of the half power mode switch, ultralinear output transformers, and other related things.

16. ### RLee77Friend of Leo's

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I think what you meant to say here is that class A dissipates the most power when at idle. As signal signal levels increase, the tube dissipates less and less, until it drops to about half at full power output.

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17. ### Old Tele manFriend of Leo's

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The Child-Langmuir 3/2-Law equation easily illustrates 'how & why' screen voltage has a far greater effect on bias voltage than does plate voltage:

1) Ik = (Ip + Is) = G*( Vg + Vs/µ1 + Vp/µ2 )^(3/2)

...where:
Ik = Cathode current
Ip = Plate current
Is = Screen current
G = Tube Perveance, amps-per-volt^3/2
Vg = Control grid voltage
Vs = Screen grid voltage
Vp = Plate voltage
µ1 = Screen/triode mu (µ) amplification factor
µ2 = Plate mu (µ) amplification factor

Using the GE 6L6GC Class-AB1 data sheet(*) examples where µ1 = 8 and µ2 = 135; and the applied voltages are Vg = -37V (bias), Vs = 400V, and Vp = 450V, the equation now becomes (Note: G ≈ 0.000921 A/V^3/2):

2) Ik = 60.8mA = (58mA+2.8mA) = G*(-37Vg + 400Vs/8 + 450Vp/135 )^(3/2)

Thus, we “see” that the control-grid “bias” voltage of -37Vdc is “holding-off” or “offsetting” a large proportion of the combined effective screen (50Vs = 400Vs/8) and plate (3.33Vp = 450Vp/135) voltages.

Now, substituting the effective Vs and Vp voltages into the equation, it becomes:

3) Ik = 60.8mA = (58mA+2.8mA) = G*(-37Vg + 50Vs + 3.33Vp )^(3/2)

4) Ik = 60.8mA = (58mA+2.8mA) = G*(-37Vg + 53.33V )^(3/2)

Hence, equation 4 “shows” us that the screen voltage accounts for 94% of the combined effective Vs and Vp voltage, 53.33V (ie: 50V/53.3V = 0.9375 ≈ 94%), showing how little control the plate voltage has in the grand scheme. Thus, the control grid bias voltage is predominately controlling/offsetting screen grid voltage, not plate voltage.

5) Ik = 60.8mA = (58mA+2.8mA) = G*( 16.33V )^(3/2)

6) Ik = 60.8mA = (0.0608A) = (0.000921A/V^3/2 * 60.01V^3/2)

(*) sources:
http://www.r-type.org/pdfs/6l6.pdf

Last edited: Jul 21, 2017
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18. ### petebFriend of Leo's

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Old tele man, you sound skeptical, I'm not sure why, you have been following this thread all the way thru and contributing.

I too want the tube to be fully working but I'm exploring the usable bias are below the 50-60-70 guideline.

What i did is I measured how much the tubes are opening up on a few different amps at different bias points. What it demonstrated is that the tubes in the low biased amp are opening up just as much as the tubes in the higher biased Amps, so the bias point is acceptable, and it's based on the actual tube operation.

19. ### petebFriend of Leo's

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Thank you for that

20. ### petebFriend of Leo's

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Thank you Bendyha, I understand screen voltage plays a role. But it's a level of complexity I'm not ready for.

Do you have a method for evaluating the relationship between bias voltage and signal voltage?

I was estimating that if the max signal voltage can reach half of the bias voltage, then the signal voltage is strong enough relative to the bias voltage. Do you have any expectations for your own amps that you would like to share? How high SHOULD the max signal voltage get relative to the bias voltage?

Old tele man, thank you for linking the latest paper. I didn't get much out of it but near the end they said the signal voltage is is less than or equal to the bias voltage.

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