More details are needed, but in general I like to see 3dB or more headroom in the driver stage, and you definitely have not accounted for some extra margin. Also depends on whether you are planning to run class A as opposed to class AB, voltage drive requirements for class AB will be somewhat greater due to the chosen operating point.
kevinkr is right, more input is required here.
But you can do some good approximations. 211 @ 1200V in Class A will swing 0 to 2400V (a bit less in reality unless you want A2). 211 has µ=11=Amplification (actually it's less because Rout||RL ). So 2400/11 =~ 220Vpp
Thus..no. I would use a more powerful driver stage. (Assuming you want/need full output power)
But you can do some good approximations. 211 @ 1200V in Class A will swing 0 to 2400V (a bit less in reality unless you want A2). 211 has µ=11=Amplification (actually it's less because Rout||RL ). So 2400/11 =~ 220Vpp
Thus..no. I would use a more powerful driver stage. (Assuming you want/need full output power)
Let's rephrase what I've asked. Operating the 211 in 1200V for class A in PP and having Emission Labs 20A as driver tube in PP, how much step up the amplifier really needs?
This is the projected diagram.
Input Transformer (how much step up?) => EML20A PP => PP to PP Interstage transformer => 211 PP
EML-20A-20B-30A Data sheet. By Emission Labs.
Thanks!
This is the projected diagram.
Input Transformer (how much step up?) => EML20A PP => PP to PP Interstage transformer => 211 PP
EML-20A-20B-30A Data sheet. By Emission Labs.
Thanks!
The driver should be able to deliver at least twice the grid bias, which for a 211 operated in class A @ 1200V plate voltage will be around -80VDC. Thus you will need appx. 160Vpk-pk grid drive at full swing. It's good practice to leave some headroom above this to minimize driver generated ditorsion but remember that any higher drive will push the 211 grid in the positive area and the upper alternances of the waveform will be severely clipped due to grid current and bad distorsion will result, unless of course you want to design your amp for class A2 operation where you need a driver capable of delivering not only the right voltage swing but also some power.
In short, your driver may work for this application but with only 2dB headroom left it will be pushed hard and need to be very well designed for the lowest distorsion at max output swing.
In short, your driver may work for this application but with only 2dB headroom left it will be pushed hard and need to be very well designed for the lowest distorsion at max output swing.
Hi!
Staying with 3dB and 80V bias that would roughly calculate to:
1.4*2*2*80 which is about 450V peak to peak or 160V RMS
Not an easy task and this requires a beefy driver.
But I think the thread opener wants to know how much gain he needs.
This depends on the input sensitivity which he wants to reach.
Gain is normally calculated such that the amp delivers max power at the desired input sensitivity. Since the driver tube is given and he wants to use a step up, the necessary step up ratio is calculated as 160V RMS divided by the amplification factor of the driver tube divided by the desired input voltage for max power out.
However keep in mind that an input transformer with a step up ratio will need a low impedance to be driven properly.
Best regards
Thomas
That would be for single ended. He asked about PP which requires double the driver voltage compared to SE. Kevin stated 3dB headroom. I prefer 6dB.
Staying with 3dB and 80V bias that would roughly calculate to:
1.4*2*2*80 which is about 450V peak to peak or 160V RMS
Not an easy task and this requires a beefy driver.
But I think the thread opener wants to know how much gain he needs.
This depends on the input sensitivity which he wants to reach.
Gain is normally calculated such that the amp delivers max power at the desired input sensitivity. Since the driver tube is given and he wants to use a step up, the necessary step up ratio is calculated as 160V RMS divided by the amplification factor of the driver tube divided by the desired input voltage for max power out.
However keep in mind that an input transformer with a step up ratio will need a low impedance to be driven properly.
Best regards
Thomas
Not so. We're speaking about the peak a.f grid -to- grid voltage (which is the definition commonly used in most tube databooks) which is appx. twice the bias voltage in a PP class A amplifier.
In my exemple, when the upper tube is fully conducting his grid voltage will be at OV (referenced to ground) and at the same time the lower (less-conducting) tube grid will be at -160V. The difference (peak to peak) between these two values is obviously 160V. Don't confuse peak (with respect to ground) with peak to peak. I agree this can be misleading as most databooks use peak value (from respect to ground) in S.E mode operation and peak-to-peak (or grid to grid) in PP mode. Using RMS values at this point don't really help and only add to the confusion.
This can be found in many references, as per exemple RCA's RC15 tube manual (p.23):
Under maximum-signal conditions,the signal voltage swings from zero-signal bias voltage to zero bias for each tube on alternate half cycles.Hence, in the exemple,the peak af signal voltage per tube is 60 volts, or the grid-to-grid value is 120 volts.
(in the above exemple the gris bias was set at -60V for a PP 2A3)
Of course you don't need 450V peak to peak to drive a 211 based PP amp biased in class A mode.
Something else to point out here, the headroom discussion is moot if the driver cannot supply sufficient current to allow swing into the A2 region when loaded by the output stage grid current, but it is still not a bad idea to build in some headroom based on available open circuit swing.
Talking in terms of open circuit swing 6dB (320Vpp) is probably easily achievable with the EML - 20B, however I would trade off some of that swing for a lower driver plate voltage and more driver stage plate current - this may be a better compromise as you approach the grid current region in the output stage, hence my comment to design for perhaps 3dB instead.
If you have a good spice model for the driver tube and 211 you may be able to come to a good compromise, otherwise a bunch of experimentation is going to be required to figure this out.
I would not recommend anything other than a 1:1+1 interstage because of the already high rp of this tube, winding such a transformer is going to be difficult enough, and the source impedance increases as the square of the step up ratio which will get you into trouble with the miller capacitance of the 211.
I do NOT recommend an input transformer under any circumstances with this tube due to the rather high miller capacitance. There is another thread here about such an SE amplifier and it performed quite poorly.
You will need another stage to drive the 20B properly. This could be a dht of relatively modest mu, and a low(ish) rp like the 12A/112..
Talking in terms of open circuit swing 6dB (320Vpp) is probably easily achievable with the EML - 20B, however I would trade off some of that swing for a lower driver plate voltage and more driver stage plate current - this may be a better compromise as you approach the grid current region in the output stage, hence my comment to design for perhaps 3dB instead.
If you have a good spice model for the driver tube and 211 you may be able to come to a good compromise, otherwise a bunch of experimentation is going to be required to figure this out.
I would not recommend anything other than a 1:1+1 interstage because of the already high rp of this tube, winding such a transformer is going to be difficult enough, and the source impedance increases as the square of the step up ratio which will get you into trouble with the miller capacitance of the 211.
I do NOT recommend an input transformer under any circumstances with this tube due to the rather high miller capacitance. There is another thread here about such an SE amplifier and it performed quite poorly.
You will need another stage to drive the 20B properly. This could be a dht of relatively modest mu, and a low(ish) rp like the 12A/112..
Hi!
Maybe just a condusion of terms. I'm talking peak to peak which is different than peak AF voltage. At max excitation one tube is at 0V grid and the other at -160V grid which is 160V peak voltage. 180 degree later the first tube is at -160V and the other is at 0V which is a 320V difference, hence peak to peak. Add 3dB headroom to that and you are at 450V. To avoid confusion the required RMS voltage might be clearer.
Thomas
Maybe just a condusion of terms. I'm talking peak to peak which is different than peak AF voltage. At max excitation one tube is at 0V grid and the other at -160V grid which is 160V peak voltage. 180 degree later the first tube is at -160V and the other is at 0V which is a 320V difference, hence peak to peak. Add 3dB headroom to that and you are at 450V. To avoid confusion the required RMS voltage might be clearer.
Thomas
I think I see the confusion here. Peak to peak is not the same as grid to grid.
Grid to grid is the total AC voltage needed to drive both sides of a push-pull amp. The grid signals are out of phase, which in effect doubles the AC drive voltage vs. just one side.
A push pull amp for the same op point therefore requires 2X the drive voltage as a SE amp.
Peak to peak is the AC voltage measured from the most positive excursion to the most negative. In the SE example given above with -80V bias, the AC grid drive signal voltage goes 80V in the positive direction and 80V in the negative direction, or 160V peak to peak. Peak voltage is simply peak to peak divided by 2. 160V peak to peak = 80 volts peak.
If the SE amp needs 160V peak tp peak, the push pull amp needs 320V peak to peak because it's driving 2 grids out of phase. by definition that's 320V peak to peak, grid to grid. It's also 160V peak, grid to grid. It does make sense to always refer to grid to grid voltage for a push-pull amp.
Why should RMS be confusing? We use RMS all the time assuming a sine wave. It's a strict mathematical relationship RMS = peak to peak divided by ~2.82. So the SE amp needs about 57V RMS and the push pull amp needs about 113V RMS from the driver stage. Of course some additional headroom would not hurt but that's discretionary.
Cheers
Grid to grid is the total AC voltage needed to drive both sides of a push-pull amp. The grid signals are out of phase, which in effect doubles the AC drive voltage vs. just one side.
A push pull amp for the same op point therefore requires 2X the drive voltage as a SE amp.
Peak to peak is the AC voltage measured from the most positive excursion to the most negative. In the SE example given above with -80V bias, the AC grid drive signal voltage goes 80V in the positive direction and 80V in the negative direction, or 160V peak to peak. Peak voltage is simply peak to peak divided by 2. 160V peak to peak = 80 volts peak.
If the SE amp needs 160V peak tp peak, the push pull amp needs 320V peak to peak because it's driving 2 grids out of phase. by definition that's 320V peak to peak, grid to grid. It's also 160V peak, grid to grid. It does make sense to always refer to grid to grid voltage for a push-pull amp.
Why should RMS be confusing? We use RMS all the time assuming a sine wave. It's a strict mathematical relationship RMS = peak to peak divided by ~2.82. So the SE amp needs about 57V RMS and the push pull amp needs about 113V RMS from the driver stage. Of course some additional headroom would not hurt but that's discretionary.
Cheers
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