How to calculate gain and draw loadline for tubes with cathode resistors?

[Alan0354]

As the title, how do I find gain of triodes and penthodes if you put a degenerate cathode resistor.......with NO bypass cap. I don't recall I ever seen any talk about this, if anyone can point me to some reading material, that would be very helpful.

In transistors, it's easy, we just use gm=1/r'e to calculate the gain of a common emitter stage. Can you use 1/gm on tubes to find the equivalent internal cathode resistor. Then if you add a cathode resistor, then the total cathode resistor = 1/gm + Rc. then you equate the new gm'= 1/(1/gm + Rc) and use the new gm' for calculate the u and gain.

 

[jharris]

http://www.aikenamps.com/index.php/designing-common-cathode-triode-amplifiers

According to that page, the way you work this into the same equation you'd use for a fully-bypassed cathode is to calculate a new value for the anode resistance (ra') using this formula:

ra'(unbypassed Rk) = ra + (mu + 1)*Rk

And for some reason this often works out to about 6dB decrease in gain, and that's often given as a rule of thumb for the gain difference between bypassed/unbypassed.

However I don't see any reason it should be exactly 6dB or should always be the same (for example with a large cathode resistor compared to the plate resistance, I'd expect a larger difference). So you can always do that math for a more precise result.

 

[kward]

A triode has that pesky dynamic plate resistance so it's not as easy as with a transistor.

I use the gain equation for a common cathode stage (cathode unbypassed):

Av = mu*RL/[Rp + RL + (mu+1)Rk]

Where
mu = gain factor of tube
RL = Load resistance
Rp = dynamic plate resistance
Rk = Cathode resistor

Rp changes as the tube operates, so approximate it at the quiescent point. Do this graphically using the plate curves for the given tube in question. Draw a tangent line to the grid curve that passes closest through the quiescent point. Rp is the inverse slope of that tangent line.

If your quiescent point is equidistant from two grid curves, you need to estimate an intermediate grid curve that does pass through your quiescent point, as well as the tangent line of that intermediate grid curve. This method has proven to be extremely accurate for me, when AC conditions don't load down the stage too much.

RL can usually be just the plate resistor in the above equation if you can guarantee the input impedance of the following stage will not load down the stage under investigation. As you know, AC loading can be both resistive and reactive. The resistive component is the equivalent resistance as seen from the plate of the stage in question looking to ground. The reactive component can affect AC gain too, such as Miller capacitance. The AC load line is always rotated clockwise from the DC load line. Origin of rotation is the quiescent point.

 

[Alan0354]

Hi

Thanks for the info, I got the article and read it already. How about the penthode? Using the same equations?

 

[kward]

For a small signal pentode with fully bypassed screen and cathode, gain can be approximated as

Av = gm * RL

For an unbypassed cathode gm is reduced from the above by approximately a factor of 1/(1+gm*Rk)

An unbypassed screen is yet another source of degenerative feedback, and it reduces gain by another factor of

1 - (Rg2/(Rg2+rg2))

Where
Rg2 = external screen resistance
rg2 = (dynamic) internal screen resistance.

(Original source: F.E. Terman, 1940, Calculation and Design of Resistance Coupled Amplifiers using Pentode Tubes)

The internal screen resistance at the bias point used can be estimated graphically using the mutual characteristics curves from the data sheet, if they are provided. Things are getting complicated now, and I think it might be better at this point to build a prototype stage and measure the gain directly to determine its suitability in your circuit.

Also, having one or two extra poles in the stage from bypassing the screen and/or cathode makes it more difficult to stabilize the amplifier when yet another source of feedback is applied--the global feedback loop. Careful placement of the time constants throughout the entire circuit now requires specific attention.

 

[Alan0354]

Hi Kward

Do you have articles on the derivation of the formulas you provided?

Thanks

Alan

 

[kward]

I'd love to see the original article also, if anyone has obtained a copy of it.

 

[Alan0354]

For a small signal pentode with fully bypassed screen and cathode, gain can be approximated as

Av = gm * RL

For an unbypassed cathode gm is reduced from the above by approximately a factor of 1/(1+gm*Rk)

An unbypassed screen is yet another source of degenerative feedback, and it reduces gain by another factor of

1 - (Rg2/(Rg2+rg2))

Where
Rg2 = external screen resistance
rg2 = (dynamic) internal screen resistance.

(Original source: F.E. Terman, 1940, Calculation and Design of Resistance Coupled Amplifiers using Pentode Tubes)

The internal screen resistance at the bias point used can be estimated graphically using the mutual characteristics curves from the data sheet, if they are provided. Things are getting complicated now, and I think it might be better at this point to build a prototype stage and measure the gain directly to determine its suitability in your circuit.

Also, having one or two extra poles in the stage from bypassing the screen and/or cathode makes it more difficult to stabilize the amplifier when yet another source of feedback is applied--the global feedback loop. Careful placement of the time constants throughout the entire circuit now requires specific attention.

I derived the penthode and verified your equation of penthode with degenerate resistor Rk is CORRECT. Here is my derivation:



My equation is exact first, then I make use of the high rp of penthode to simplify and get your exact equation.

What I did not see in any of the tube books is how people do in transistor circuit to equate gm=1/rk where rk is the dynamic resistance of the cathode. gain u = gm X rp = rp/rk.

I just find the new gm' = 1/(rk+Rk) and use it to calculate the gain.



I still need to derive the equation for triode, I'll post back if I get somewhere.


BTW, I cannot follow the article you linked. It used ra' ( which is the anode resistance, I called it rp in consistent with RDH4). I don't agree with the article. It just does not make sense as they calculate ra' is even higher than ra (rp). The anode resistance is already much higher than the load resistance. it's out of the picture like your equation. That's the reason I asked for where you got the equation to verify.

I'll work on the triode.

I think it's a good way to get linearity using degenerate resistor. gm ( 1/rk) is not constant, adding Rk will help a lot. You loss gain, but gain is easy to get.

 

[kward]

The triode model is nearly trivial. The only hard part (I think) is correctly modeling the degenerative nature of Rk.

To start with, the gain of a perfect triode would be u, or said another way, the output voltage Vout would be u*Vg, where Vg is the voltage applied to the grid. But there is an internal plate resistance rp, as well the cathode resistor Rk, which reduce the gain from the perfect trode. The tube is controlled by the voltage differential between grid and cathode, and there is a voltage potential created from the current flowing through Rk. Another way of saying this is Rk has the effect of reducing the effective voltage at the grid. The amount of voltage reduction is simply ip*Rk. Thus effective voltage at the grid is Vg - ip*Rk, and amplified voltage is u(Vg-ip*Rk). Then using Kirchoff's voltage law (voltage generated equals voltage consumed), the rest just falls out from there with some simple algebra.

 

[Alan0354]

A triode has that pesky dynamic plate resistance so it's not as easy as with a transistor.

I use the gain equation for a common cathode stage (cathode unbypassed):

Av = mu*RL/[Rp + RL + (mu+1)Rk]

Where
mu = gain factor of tube
RL = Load resistance
Rp = dynamic plate resistance
Rk = Cathode resistor

Rp changes as the tube operates, so approximate it at the quiescent point. Do this graphically using the plate curves for the given tube in question. Draw a tangent line to the grid curve that passes closest through the quiescent point. Rp is the inverse slope of that tangent line.

If your quiescent point is equidistant from two grid curves, you need to estimate an intermediate grid curve that does pass through your quiescent point, as well as the tangent line of that intermediate grid curve. This method has proven to be extremely accurate for me, when AC conditions don't load down the stage too much.

RL can usually be just the plate resistor in the above equation if you can guarantee the input impedance of the following stage will not load down the stage under investigation. As you know, AC loading can be both resistive and reactive. The resistive component is the equivalent resistance as seen from the plate of the stage in question looking to ground. The reactive component can affect AC gain too, such as Miller capacitance. The AC load line is always rotated clockwise from the DC load line. Origin of rotation is the quiescent point.

I can NOT verify this equation.

The gain equation for triode should be the SAME as penthode BUT without assuming rp>>RL. So one has to use the full equation as shown:



So you see I started with the exact equation from the penthode above. I expand the denominator out and I got 2 of the 3 terms the same, but there is no way I can make the last term gm X RL = 1



I am quite sure my exact equation is correct. I did not invent this, this is the standard way in analyzing transistors....For BJT gm = 1/r'e. This change the gm into the internal emitter resistor and the gain of the transistor is RL/r'e. This should be universally true and applies also to triode and penthode.

Also the fact that I verified the equation match the penthode equation already.


Unless someone can show the derivation of that equation, I am not buying it.

 

[kward]

We (apparently) posted simultaneously. See above.

 

[kward]

Also, you asked how to draw the load line. It's also quite trivial. If the tube is in cutoff, no current is flowing so the first point on the graph is along the X axis at (V, 0)., where V the unloaded plate voltage.

The second point on the graph can be determined by assuming a theoretical condition of zero volts drop across the tube, so all voltage is consumed across plate and cathode resistors. The current flowing is then: Imax = V(RL + Rk). This makes the second point along the Y axis of (0, Imax).

Then draw a straight line between those two points. The quiescent point must fall along that line somewhere (wherever you decide to put it).

 

[Alan0354]

The triode model is nearly trivial. The only hard part (I think) is correctly modeling the degenerative nature of Rk.

To start with, the gain of a perfect triode would be u, or said another way, the output voltage Vout would be u*Vg, where Vg is the voltage applied to the grid. But there is an internal plate resistance rp, as well the cathode resistor Rk, which reduce the gain from the perfect trode. The tube is controlled by the voltage differential between grid and cathode, and there is a voltage potential created from the current flowing through Rk. Another way of saying this is Rk has the effect of reducing the effective voltage at the grid. The amount of voltage reduction is simply ip*Rk. Thus effective voltage at the grid is Vg - ip*Rk, and amplified voltage is u(Vg-ip*Rk). Then using Kirchoff's voltage law (voltage generated equals voltage consumed), the rest just falls out from there with some simple algebra.

View attachment 1052854

You might be right, I literally take rp is parallel to RL, I use u = gm X (rp//RL) = gm X (rp X RL)/(rp + RL). I need to verify more.

 

[kward]

You might be right, I literally take rp is parallel to RL, I use u = gm X (rp//RL) = gm X (rp X RL)/(rp + RL). I need to verify more.

In the Norton model, rp is in parallel with RL. I used the Thevenin model, which places rp in series with RL, because I think the math falls out easier. Either model is correct if you accept the fallout of the math that results from it and if you model the degenerative nature of Rk correctly. I have not run through the math using the Norton model. If you get it to work out, I would be interested in seeing the algebra.

My bench doesn't lie. I know the formula is correct.

 

[Alan0354]

My problem with that is the equation will not work for penthode. rp of penthode is very high, if using Thevenin model that rp in series with RL, you don't have much voltage left. That's why I used rp//RL.

Unless of cause if the penthode model is not a voltage source = u X Vg. I am still trying to look it up. Can you confirm?

Don't quote me on this, I could swear with 12AX7, Rk=1.5K bypassed by 220uF, plate resistor of 100K( Rg=1M, so ignored) and +B=400V. The plate voltage can dip to about +100V. That's is about 3mA of plate current. rp = 62K for 12AX7, 3mA gives 180V drop across rp alone, so if rp is in series with RL for triode, the plate should never goes below +150V or so.

I know the 12AX7 in guitar amps can swing lower than 150V.

 

[kward]

That's where that article by F.E. Terman would be handy to have. I strongly suspect he uses the Norton method to model the pentode as a current source because of the term gm in the gain equation and ignoring rp because it is so high as compared to RL.

And in fact, the plate curves are flat (mostly) on pentodes once you get above the knee, so they do act like pretty good current sources under those conditions.

 

[kward]

Don't quote me on this, I could swear with 12AX7, Rk=1.5K bypassed by 220uF, plate resistor of 100K( Rg=1M, so ignored) and +B=400V. The plate voltage can dip to about +100V. That's is about 3mA of plate current. rp = 62K for 12AX7, 3mA gives 180V drop across rp alone, so if rp is in series with RL for triode, the plate should never goes below +150V or so.

That sounds about right to me, for a 12AX7. Of course for a hifi amp, I would typically run the quescient point at somewhere between 0.6 mA and 1 mA, so that voltage across the tube could drop to about 50V.

 

[Alan0354]

That's where that article by F.E. Terman would be handy to have. I strongly suspect he uses the Norton method to model the pentode as a current source because of the term gm in the gain equation and ignoring rp because it is so high as compared to RL.

And in fact, the plate curves are flat (mostly) on pentodes once you get above the knee, so they do act like pretty good current sources under those conditions.

I think that's where I got it wrong on the triode. Penthode is more a current source with very high rp, one really cannot use voltage source ( u X Vg) in series with rp and RL for penthode. I was trying to use the same equation for both triode and penthode.

Penthode is very much like transistor particular the FETs. You compare the plate curve and drain curves of FETs, they look very similar. The model for FETs and BJT are all current source where I = gm X Vin. BJT is just more perfect current, you look at the collector curves ( same idea as plate curves), it's very horizontal with very distinct knee. MOSFET is in between penthode and BJT.

I guess I have to use the voltage model for triode with voltage source = u X Vg and in series with rp.

 

[PakProtector]

Check the equations with the two extreme cases; no cathode resistor, and a split load, as in a phase splitter section where the plate is equal to the cathode load. If the numbers don't make sense the equations are not valid. Even if they do, it still might not be Right, but the chances are better...Yes?
cheers,
Douglas

 

[Alan0354]

I went back and check the equivalent circuits used for penthode and triode using COMMON CATHODE configuration( Rk fully bypassed or no Rk). The Thevenin and Norton equivalent gives EXACTLY the same result, so there is no difference, one can use either one and will work.

The key is this is an EQUIVALENT CIRCUIT, NOT THE PHYSICAL CIRCUIT. ALSO it is a SMALL SIGNAL model that cannot be used to calculate to the limit ( to clipping). This is not the physical equivalent circuit.

I use the numbers provided in the datasheet and calculated both ways on 12AX7 and 6L6GC, they came out exactly the same. Here is my work:



I need to go back and look at why I cannot get the same equation as Kward provided. Kward's equation is obviously correct. I have to double check my work again in post #10.