Steve, The voltage gain of a common cathode tube stage is given by - G = u x RL/(RL+Ra) u is the tube amplification factor Ra is the tube internal plate resistance RL is the external equivalent load resistance. You can get u and Ra from the tube data sheet but tubes being the PIA that they are u and Ra vary with plate voltage and current. RL is the plate resistor Rp in parallel with the load resistance. The load resistance is the resistance to ground after the coupling cap. eg. A 12AX7 with a 220k Rp driving a 500k vol pot has an RL of 220k // 500k or about 150k. At 250V and 0.5mA plate current the 12AX7 data gives Ra=80k and u=100 so the gain is G = 100 x 150 / (150+80) = 65 Change Rp to 100k and RL is now 100k // 500k or 83k. The plate current will now be about 1mA and the 12AX7 data gives Ra=60k and u=100 so the gain is now G = 100 x 83 / (83+60) = 58 This is only true if the cathode resistor is fully bypassed. The gain will be less with an unbypassed Rk. Dave >> wouldn't the current be >> higher if the gain is higher? Not necessarily because we are talking about voltage gain. The output voltage is developed across the plate circuit resistance and that is greater in the 220k circuit. >> So how does the cathode >> resistor Rk fit into the >> equation? I was hoping you wouldn’t ask that. I can feel one of those headache coming on. Another way of calculating the gain is to divide the plate circuit resistance Pr by the total cathode resistance Cr so G = Pr/Cr Pr = RL // Ra Cr = Rk + 1/gm gm is the tube transconductance. (1/condutance = resistance) Rk is any external unbypassed cathode resistance. Rk=0 if bypassed by a large cap. You can get Ra and gm from the tube data sheet. For a 12AX7 operating at 250V and 1mA Ra=62.5k and gm=1.6mA/Volt. RL is the plate resistor in parallel with the load resistance as before. eg A 12AX7 has a 100k Rp driving a 500k vol pot. The cathode resistor is 1k5 bypassed by 25u. RL = 100k // 500k = 83k Pr = RL // Ra = 83 // 62.5 = 36k Cr = 1/gm = 1/1.6 = 0.625k G = Pr/Cr = 36/0.625 = 58 If the 25u bypass cap is removed from Rk then Cr becomes 1.5+0.625=2.125k so the gain is now G = 36/2.125 = 17 The circuit with a bypassed Rk has 3.4 times the gain (~10dB). If Rk is bypassed by a smaller cap like a 0.68u then the high frequencies will have 10dB more gain than the low frequencies. Dave moocow Wed Nov 4 20:32 I've never liked approximations, so I derived the gain equations for the small-signal triode model: (1) G = u x RL / [ (1+u)*Rk + Ra + RL] When Rk = 0, (2) G = u x RL / (Ra + RL) RCA Receiving Tube Manual RC-15 shows that Ra = 62,500 for a 12AX7 (u=100) biased for 1.2 mA plate current. With a plate resistor of 100k and a load of 500k, Equation (2) predicts a gain of 57 when Rk is bypassed. If Rk is not bypassed, Equation (1) predicts a gain of 28. The Fender Pro Jr. schematic shows that the first stage does not have its Rk bypassed. For an input voltage of 20 mV, the output voltage is measured at 545 mV, a gain of 27.3. This measurement agrees well with Equation (1). The Fender Bassman reissue schematic shows a first stage output of 213 mV for an input of 3.85 mV, a gain of 55. Since Rk is bypassed, this is as expected from Equation (2). I believe the G = Pr/Cr equation holds when Rk is 'large', or similar in magnitude to RL. The extreme case is when Rk = RL. Then we have the split-load phase inverter, and G=1. Hi Steve, Thanks for the information from Dan's book. I compared those numbers to the ones in the Resistance Coupled Amplifier charts in the RCA Receiving Tube manual, and it is appears that Dan borrowed them for his book. Nothing wrong with that, as long as proper credit is given, but I sincerely hope he included the column for Rg, the grid resistance of the following stage. The RCA chart shows the following circuit parameters for B+=180 and Rp = 100k: Rg-Rk-Gain 100k-1800-40 220k-2000-47 470k-2200-52 Notice both Rg and Rk increase as gain increases. At this point, it may seem unclear if the gain changes because of Rk or because of Rg. That's where the gain equation becomes useful. I am sure everyone is sick of this equation, but here it is once again, G = u*RL/(Ra+RL) Where RL = parallel combination of Rp and Rg u = amplification factor = 100 for a 12AX7 Ra = plate resistance = 80,000 ohms It is very important to realize that RL depends upon both the plate resistor Rp and the load resistance Rg. The equation predicts the following values of G: Rp-Rg-RL-G (calculated) 100k-100k-50.0k-39 100k-220k-68.8k-46 100k-470k-82.5k-51 As long as Rg is taken into account, the equation is consistent with the table values. The gain increase can be attributed primarily to the increase in Rg, not Rk. The RCA chart demonstrates this effect, and, believe it or not, every gain value in the RCA chart is consistent with the gain equation. By the way, neither Dave H. nor I derived this particular gain equation. It can be found in every tube manual and textbook printed during much of the 20th century. I know I'm not a very good technical writer and I'm sure R.G. could have written something that everyone could understand. But the bottom line is that these equations are valid, they just need to be used properly. Hello again, Dave, This time, it was my turn for a headache. Here are the long-tailed phase inverter equations. There are two of them, since a phase splitter has two outputs: Av1=RL1*[u/(Ra+RL1)]*[(1+u)*Rk/(Ra+RL1+2*u*Rk)-1] Av2=RL2*[u*Rk/(Ra+RL2)]*[(1+u)/(Ra+RL1+2*u*Rk)] Where RL1 = Rp1Rg (usually 82k220k) RL2 = Rp2Rg (usually 100k220k) Rk = total resistance from cathodes to ground u = amplification factor Ra = plate resistance Rk is the sum of the two cathode resistors, 470 and 6800 ohms in the case of the tweed Bassman. This was interesting to me because it shows Rk1 is chosen to set the grid-bias voltage. Only their sum affects the gain, and the larger it is, the better the balancing turns out to be. I did make a couple of assumptions during the deriviation. The 1M grid resistors are 'large' compared to Rk, so I left them out of the loop equations. At one point, I had to assume (Ra+RL1) / (Ra+RL2) = 1 and (u+1) / u = 1 to make the equations simpler to deal with. These are reasonable assumptions, and I did check the equations against a computer simulation. The results were not perfect, but they were pretty close. Since 'other' equations do not include Rk, I thought I should try and simplify the above equations to get rid of Rk. Assuming (2*u*Rk) >> (Ra + RL1), Av1 = -RL1 * u/[2*(Ra+RL1)] Av2 = RL2 * u/[2*(Ra+RL2)] Using the given parameters of: Ra=71k u=100 RL1=RL2=150k220k=89k the greatly simplified equations give Av1 = Av2 = 27.8 Which is almost exactly the same as the 27.5 given by Steve Bench's equation. However, the unsimplified equation accurately predicts the differences in gain (imbalance) between the two outputs. Once I had the equations, I tried several different phase splitters, like Vox, Fender and 20W Marshall. The equations did not match up perfectly with the simulations, but they do accurately predict the differences in gain between the two outputs. That's the whole point of the equations and simulations, to see how the different phase splitters perform. They had gains that varied between 24 and 28, and the matching between outputs varied as well. Wouldn't you know that the Vox circuit, with its large 1.2k/47k cathode resistors, is very well balanced ? I guess Dick Denny wanted the best balance possible, given that he did not want to rely on negative feedback to balance the two sides. A couple of things to note: Av1 will be negative, which denotes the signal inversion at this point with respect to the input. Also, the gain is calculated assuming no negative feedback network. If you try and verify these equations by measurement, the feedback resistor to ground (usually 4.7k or 100 ohms) must be shorted. Otherwise, negative feedback will change the gains and make my equations look wrong. THAT could never happen, right ??? Finally, do not attempt to derive these equations unless you have lots of free time and plenty of aspirin.