Drive

[mr2racer]

I know that gm and voltage swing are required to drive an output tube. What I don't know is how much current does/do the output tube(s) require?

 

[aardvarkash10]

almost none unless you go grid positive

 

[smoking-amp]

There is the bias feed resistor and then also the input capacitance of the tube. Triodes will have more Cin due to the Miller effect. (gain times C grid to plate + C grid to all else) Figure capacitive load current at the highest freq.

Going positive with the grid brings in grid current too.

 

[BinaryMike]

Even if you don't plan to drive the grids positive, as in class A2 or AB2, the driver still has to deal with significant power tube grid capacitance, which takes power to charge and discharge. Working grid capacitance is roughly the figure quoted in the datasheet for anode-grid capacitance times operating voltage gain (the Miller Effect). Slew rate is defined by dV / dT = I / C. Ferinstance, a driver circuit capable of pumping 1mA into 200pF can slew 5V/uS. The maximum sinewave frequency for a given slew rate is F = SR / 2 * Pi * Vp, so if you need 100Vpp of drive, you would hit slew limiting at 5e6 / 6.28 * 50 = 15.9KHz in this example.

 

[MagicBus]

Even if you don't plan to drive the grids positive, as in class A2 or AB2, the driver still has to deal with significant power tube grid capacitance, which takes power to charge and discharge. Working grid capacitance is roughly the figure quoted in the datasheet for anode-grid capacitance times operating voltage gain (the Miller Effect). Slew rate is defined by dV / dT = I / C. Ferinstance, a driver circuit capable of pumping 1mA into 200pF can slew 5V/uS. The maximum sinewave frequency for a given slew rate is F = SR / 2 * Pi * Vp, so if you need 100Vpp of drive, you would hit slew limiting at 5e6 / 6.28 * 50 = 15.9KHz in this example.

Hi,
Does this apply to direct coupled stages or it would be the same for AC coupling? I mean, a coupling capacitor should be taken in account or it's just the anode-grid capacitance that matters?
Thanks!

 

[DF96]

The coupling capacitor makes no difference. It is the output stage input capacitance which matters, much of which will be due to Miller effect.

 

[mr2racer]

I'm working with 7591 which lists grid to plate of .23 pf and 'input' as 10 pf. Bias voltage is about -17 volts. So current would be...

 

[SpreadSpectrum]

Formulas were given above. Here is another resource if you need another explanation:

slew rate explained | ken-gilbert.com

Or were you looking for someone else to do it for you?

 

[mr2racer]

The Tung Sol data sheet lists the grid to plate capacitance as .25pf for 7591. The mu for JJ 7591S is 16.8. 2*3.1416*20000*17 = 2136288, or 2.13V/uS. 17 x .25 = 4.25 and
.00000425 * 2.13 = .0000090525, and for headroom 5 x .0000090525 = @.000045 ma.

So now I can see why they say 7591 is a very easy tube to drive, assuming I don't have a decimal in the wrong place.

I'm using an input transformer as a phase splitter in a differential amp. So a 12au7 with more than a 17 volt PP swing should be able to drive 7591 to max output?

Thanks SpreadSpectrum for that website I bookmarked that puppy.

 

[tomchr]

assuming I don't have a decimal in the wrong place.

Use a scientific calculator which supports exponential notation and you won't have that problem.

I haven't checked your math. I'm just pointing out that there are more efficient ways to do scientific/engineering math than to count zeros after the decimal point.

Even the humble TI-30 calculator supports exponential notation.

Tom

 

[Merlinb]

That's a rather long-winded way of calculating it. Here's a short cut:
The peak current driven into the Miller capacitance is:
Ipk = 2 * pi * f * Cmiller * Vpk

The quiescent current of the driver valve needs to be greater than this figure. People here sometime make a mountain out of a molehill where driver slew rate is concerned, but the fact is that with as little as two or three milliamps of idle current in your driver valve you can drive most any power tube, including DHTs.

 

[aardvarkash10]

That's a rather long-winded way of calculating it. Here's a short cut:
The peak current driven into the Miller capacitance is:
Ipk = 2 * pi * f * Cmiller * Vpk

The quiescent current of the driver valve needs to be greater than this figure. People here sometime make a mountain out of a molehill where driver slew rate is concerned, but the fact is that with as little as two or three milliamps of idle current in your driver valve you can drive most any power tube, including DHTs.

You should write a book...

 

[SpreadSpectrum]

Thanks SpreadSpectrum for that website I bookmarked that puppy.

It's a pretty straightforward explanation but I didn't follow the last part where he states that 5x margin is a good idea because of the peak to average ratio of music. In his calculation, he is already calculating the peak slew rate needed in the amp, so no margin would be strictly necessary.

I'm not arguing against having margin (I would definitely include margin in any design I did), but his reason for including it was nonsense.

And you are right that the 7591 is a very easy to drive tube. It is about as easy as they get, low input capacitance and low input voltage swing. You would be hard-pressed to find a tube that *couldn't* drive a 7591.

Edit: I could check your calculation for you but before I do that, I need to know for sure if you are running the 7591 in "triode connection" or not.

 

[Hearinspace]

Slew rate is defined by dV / dT = I / C. Ferinstance, a driver circuit capable of pumping 1mA into 200pF can slew 5V/uS.

Math challenged me gets .001/.0000000002 = 5,000,000

How should I remember to input this (I/C) when doing this calculation?

That's a rather long-winded way of calculating it. Here's a short cut:
The peak current driven into the Miller capacitance is:
Ipk = 2 * pi * f * Cmiller * Vpk

If f is frequency then which frequency , lowest or highest ? and is C in F? uF? etc.

Thanks .

 

[SpreadSpectrum]

Math challenged me gets .001/.0000000002 = 5,000,000

How should I remember to input this (I/C) when doing this calculation?



If f is frequency then which frequency , lowest or highest ? and is C in F? uF? etc.

Thanks .

Hint: Convert your answer (which is in Volts per second) to volts per microsecond.

For frequency, do the calculation for 20kHz because that is the frequency that requires the most current.