Push-pull output transformer primary impedance for a specific bias - Page 1
Forum section: electronics / projects

author: szoftveres
author_rank: Regular Contributor
timestamp: June 22, 2021, 05:10:31 am
content: Hi All,I have a design question regarding class A push-pull output transformer primary impedance for a specific bias.For the purpose of better understanding, let's assume that the tubes are perfect current sources, i.e. they present an infinite impedance on the plate, and can swing from B+ to ground. Furthermore, let's assume that the output transformer impedance is purely ohmic, i.e. non-complex. Even further more, let's assume that the transformer core can handle infinite magnetic flux and never goes into saturation.Let's assume the following biasing conditions: B+ = 200V, plate current = 20mA.If we design for a single-ended amplifier with only one tube, the ideal primary impedance of the output transformer is 200V / 20mA = 10kohm.If we're using only a single tube that drives one half of a push-pull output transformer (and leave the other side of the primary unconnected), then we get a class A single-ended amplifier. Assuming we're using the above calculated 10kohm as the impedance of one-half of the winding, the total plate-to-plate impedance of the transformer is 40kohm (given that the impedance quadruples).This amplifier will work equally well if we now insert the second tube, bias it, but don't drive it (AC short the control grid to the ground), since (assuming it's a perfect current source device) it represents an infinite impedance on its plate.What I'm trying to understand is what happens when we start driving the second tube with opposite phase. How will it affect the first tube, and the above described impedance-matching logic. Will the ideal plate-to-plate impedance still be 40kohm? Can someone shed some light on this?PS:The above example (and question) is purely theoretical and ignores many factors that would otherwise affect the above impedance calculation of a real amplifier.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: June 22, 2021, 03:03:59 pm
content: Quote from: szoftveres on June 22, 2021, 05:10:31 amTo be pedantic, we also need the condition that the CCS can only sink to GND, and that the output must be undistorted (the CCS is linear until this threshold).And, a real device won't saturate all the way to GND, but we can effectively handle that by subtracting it from B+ so we're good to go.QuoteSo you've got it perfect to here.What happens when the other one is driven? Well, the voltage doubles of course. They are acting in parallel, in complementary phase of course. We can transform this through a phase inversion at the device, and moving its driving point from the opposite side of the transformer, to the same side, and now the two CCS are exactly in parallel. And since the drive current is doubled, the impedance needs to be halved: 5k per anode, 20k total.The last trick is whether we allow class AB or B operation. If we modify the transfer function so that transconductance varies with output current, we might have a region ofgmat high currents,gm/2for low currents, and 0 at zero current. A two sloped piecewise linear transfer curve.With the breakpoints set up correctly, we can have it that, at idle, both devices sit atgm/2so act in parallel (into an ideal 40ka-a load, if this were the only operating regime), then exactly as one crosses into thegmrange, the other goes into cutoff, and now it's a single device drive situation. And it's no longer push-pull, it's just pull, one device acting alone for the top of the half-cycle. The same repeats symmetrically at the bottom of the next half-cycle, of course.Now we can set the impedance anywhere we want, as long as the CCS keeps driving harder and harder. In practical devices, sooner or later we'll run into power dissipation limits, or peak current limits, and we're done with the design.Current is easy; if our CCS can only go up to 200mA, then we need the low-gain range at 0...40mA, and high gain 40mA+, for our 20mA bias point. A peak of 200mA, dropping 200V across one half of the transformer, needs a 1k load, or 4ka-a. (The available class A power -- +/-20mA around the 20mA quiescent point -- is consequently smaller than in a pure class A design.)Calculating power dissipation, isn't so easy. I don't remember the method offhand, but I would recommend any of the classic methods to do so -- RC-26 and friends, RDH4, etc. In general, you need to integrate the waveform over the load line, and using a sine wave as a typical waveform, you can make some geometric simplifications. Maximum power output is a square wave of course, and will actually experience less power dissipation -- as you're pushing close to class D operation, without explicitly overdriving / saturating the devices.Note that the width of the "low current" region is arbitrary; it can be the full useful range, in which case, that's full class A operation; or it can be zero, in which case that's class B operation. It also doesn't need to be piecewise linear; it can be broken into finer and finer segments, so long as the breakpoints are symmetrical around the quiescent point, andgmalways sums to a constant. In real devices,gmchanges continuously with operating current, so we can always find an intermediate point like this, controlled by setting bias. Real devices inevitably produce distortion, as the gains don't quite match in a complementary manner, but it's just a few %THD, easily cleaned up with feedback.QuoteEh, it's pretty close, and we might as well go the whole way, eh?For which, plate resistance is probably the last thing? Simply acts in parallel with the load, reducing available output power. Most cases Rp is more than a few times higher than RL so we aren't too concerned about it.It's more pronounced with screen (ultralinear) or shunt feedback, and with triodes we can even approach the problem from the other direction, as a Thevenin rather than Norton equivalent source. But triodes don't have as much freedom, and, as it happens, a simple assumption of RL = 3*Rp (for class A, and lower for varying degrees of class AB to B) does pretty well.Incidentally, one catch about real transformers: the primary has significant stray capacitance acting in parallel with it, and significant series inductance (leakage) to the secondary. Where these resonate, the primary impedance peaks up. With a well-defined plate resistance, this isn't important (the capacitance is swamped by Rp and leakage simply acts to reduce available power output), but with a high impedance, it can cause early saturation, and if feedback conditions are just right (wrong) it can oscillate as well, destroying output tubes (most likely causing toaster screens, due to the high Ig2 in saturation). This is perfectly solved with an R+C damper across the transformer (typically a-a, but keep in mind there is leakage between halves of the primary as well, and individual R+Cs for each half may be desirable). Evidently, R = sqrt(LL / Cp) and C >= Cp * 2.5 are typical values for this network.Tim

author: szoftveres
author_rank: Regular Contributor
timestamp: June 23, 2021, 02:21:40 am
content: This is all very educative, very nicely explained, thanks for sharing your knowledge!