Is the a problem with this way of calculating triode gain over the other way? - Page 1
Forum section: electronics / beginners

author: Deactivated-1
author_rank: Frequent Contributor
timestamp: August 07, 2023, 08:55:55 pm
content: I usually calculate the triode gain stage gain by doing this:(ra || Ra || Zout) / (1/gm + Rk) = gainOfc plugging whatever the external impedance of the cathode is in place of Rk, taking into account the bypass cap.But I was reading the book on valvewizard.co.uk and he writes it like this instead:tubeAmplificationFactor * (Ra / (Ra + ra + Rk * (tubeAmplificationFactor))) = gain...That just seems convoluted and indirect. Is there any problem with the previous equation?The capital letter "R" and "Z"s standing for the external values, while ra or gm being the tube internal values

author: TimFox
author_rank: Super Contributor
timestamp: August 07, 2023, 10:50:17 pm
content: In the textbook equations for triode and pentode gain, it is common to characterize the triode by its amplification factor mu, since it is roughly constant over a range of operating conditions, along with a plate resistance that varies with the operating condition. In that model, the plate resistance is in series with the voltage generator -(mu)xVgk.For a pentode, the mu is so high as to be not very useful, and it is usual to characterize the pentode with transconductancegmand plate resistance, both of which vary with the operating condition. In that model, the plate resistance is in parallel with the current generator -gmxVgk.For either model, the plate resistance is the same for voltage generator or current generator.To add a degenerating cathode resistor, it is easier to think of it as a resistor in series with 1/gmto give a lower net transconductance.In both cases, adding the cathode resistor does not change mu, but the plate resistance increases by (mu)x(Rk) for the entire circuit including that resistor.I'm reciting this from memory: check that I didn't leave out a +1 or something in the equations.Your algebra proceeds from these equations.Of course, always (mu) =gmxRpfor small-signal operation at a given operating point for the triode (and the pentode with screen bypassed to cathode).

author: TimFox
author_rank: Super Contributor
timestamp: August 07, 2023, 11:37:44 pm
content: In the sense that I explained the basis for algebra you can do yourself.You found two equations in different form for the same calculation: I replied why they were done with those parameter sets.Should I just have answered the question in your subject line? Answer: No.

author: CaptDon
author_rank: Super Contributor
timestamp: August 08, 2023, 01:15:03 am
content: Actually, Tim, I do appreciate the calculations and explainations in your answer. That math is handy when I was building vacuum tube deflection amplifiers for 5" C.R.T.'s such as 5CP7A and the more improved 5ABP7A and 5ADP7A which were upgrade tubes for the old 500 series Tek scopes. I needed to build some dedicated purpose electrostatic deflected display units and went through deciding what amplifier tube to use. I ended up using 6SL7's because I could operate in an extremely linear region and still get large plate voltage swings. Plus I got two triodes in one bottle. NO would have been a direct answer to the question but sometimes we like the supporting facts. I notice the O.P. is always re-inventing the wheel but I am not sure anyone has broken down his door wanting to buy his new and improved wheels. Just my observation.

author: TimFox
author_rank: Super Contributor
timestamp: August 08, 2023, 03:44:11 pm
content: Quote from: CaptDon on August 08, 2023, 01:15:03 amYour use of a 6SL7 in a relatively high plate voltage application is a good example of the use cases for high-mu (6SL7, mu = 70) vs. medium-mu (6SN7, mu = 20) triodes.In general, high-mu triodes should be used at high plate voltages, while medium and low-mu triodes should be used for relatively high current at lower plate voltages.Approximately, the cut-off grid-cathode voltage for a triode isVco= -Vpk/mu.If you use a 6SL7 at a measly 100 V plate supply, the cut-off is only -1.4 V, and the grid-cathode voltage for normal operation would therefore be close to -0.7 V (depending on plate load), where there would be substantial grid current.The grid voltage required to cut off the grid current is sometimes called the "contact potential", which is a misnomer; however, the grid-cathode voltage must be substantially negative to reduce this current, due to the kinetic energy of the electrons emitted from a hot cathode.

author: TimFox
author_rank: Super Contributor
timestamp: August 08, 2023, 03:46:35 pm
content: Quote from: ELS122 on August 08, 2023, 02:34:53 amThere was nothing wrong with your question "Is there any problem with the previous equation?".My answer explained why both equations exist for the same situation.

author: CaptDon
author_rank: Super Contributor
timestamp: August 08, 2023, 08:22:06 pm
content: Both the stated equation by the O.P. and the traditional equation are the 'how', but it is also nice to know the 'why'. When trying to solve either equation do you get a similar answer? If the answers compare then you answered your own question.

author: EPAIII
author_rank: Super Contributor
timestamp: August 10, 2023, 05:59:44 am
content: Back in the day I was taught how to plot a load line on the tube's curves and then derive the gain and distortion from that. I could be wrong, but that seems to be the better way to me. Those equations rely on average values that are only valid at one point on the load line. This is particularly true with a triode.

author: TimFox
author_rank: Super Contributor
timestamp: August 10, 2023, 01:49:42 pm
content: For small signals, you need to know the effect of a small variation around the "quiescent" or "bias" point.At any reasonable quiescent point (characterized by the DC valuesVgk,Vpk, andIp), there are three parameters used to calculate the gain for small input variations:(mu) =gmxrp, as discussed above.For a triode, (mu) is roughly constant over a range of operating conditions, but the other two vary, especially as a function of current.With that relationship, you have only two independent parameters, since the third depends on them.For large signals, the parameters vary over the range of plate current and the graphical method (load line on set of characteristic curves) is required.