Turns ratio = sqrt (impedance ratio)
Hammond Mfg. - REPLACEMENT & UPGRADES - Tube Guitar Amplifier - Output Transformers
Hammond Mfg. - REPLACEMENT & UPGRADES - Tube Guitar Amplifier - Output Transformers
The datasheet for a 6V6 power tube (as used in a Fender Deluxe Reverb amplifier) shows that the load resistance should be 5000 to 8500Ω; with an optimum power output and distortion at 6000Ω. The output transformer has a primary impedance of 6600Ω, with a center tap. If, in the push-pull configuration, each tube works into half of the primary impedance, that would be considerably LESS than the optimum. What gives?
The published value is the plate to plate impedance.
Commercial transformers also refer to same impedance, otherwise consumers would be very confused.
Now strictly speaking, each tube sees one fourth of the plate to plate impedance, not half.
Voltage ratio (end to end vs end to center tap) is 2:1 ; impedance is 4:1 , it´s a transformer, not 2 resistors.
Commercial transformers also refer to same impedance, otherwise consumers would be very confused.
Now strictly speaking, each tube sees one fourth of the plate to plate impedance, not half.
Voltage ratio (end to end vs end to center tap) is 2:1 ; impedance is 4:1 , it´s a transformer, not 2 resistors.
The effective plate-to-plate load resistance for Class AB1 operation (in the GE datasheet I have) shows 8000Ω at a plate voltage of 285 volts. I don't know how this translates to operation at the Fender Deluxe Reverb's B+ of 415 volts.
Yes, you are correct. One tube's load impedance would be at a 14.36/1 turns ratio (to the center tap); thus 1650Ω---one-fourth of the transformer's quoted impedance of 6600Ω. Still seems far away from the "optimum" shown.Now strictly speaking, each tube sees one fourth of the plate to plate impedance, not half. Voltage ratio (end to end vs end to center tap) is 2:1 ; impedance is 4:1 , it´s a transformer, not 2 resistors.
@dotneck335 - it’s against the Forum Rules to start multiple threads on the same/similar topic, and please keep all guitar-amp related threads/posts in the Instruments and Amps sub-forum.My RCA tube manual (RC-30) shows a section for 6V6 push-pull operation, and a section for single-ended operation. One operating point is fairly similar at Vp = Vg2 = 250V. The recommended plate to plate impedance for the push-pull is 10k, double the single tube 5K, mostly because the operation of the push-pull is nearly all class A (the Vg is only slightly more negative for the PP).
During push-pull operation, if neither output tube is cut off, each tube sees 1/2 of the plate to plate impedance. If either tube is cut off, the other tube sees 1/4 of the plate to plate impedance. This transition is central to AB operation and allows the operating line to 'wrap around' the knee created by the max plate dissipation. See this link for a nifty description and some nice pictures.
To answer the OP, the plate to plate impedance for any given push pull design is given by the tube characteristics, and may or may not be lower than you might expect. Generally speaking, however, the tube is often run somewhat near it's maximum plate voltage rating even when single-ended. When AB operation is designed in, the designer will tend to raise the plate voltage to the maximum and reduce the bias current. This often allows for a lower load impedance to be used and still miss the max plate dissipation. As a bonus the operation line will wrap around that knee (if designed properly), usually giving a substantial increase in output power.
During push-pull operation, if neither output tube is cut off, each tube sees 1/2 of the plate to plate impedance. If either tube is cut off, the other tube sees 1/4 of the plate to plate impedance. This transition is central to AB operation and allows the operating line to 'wrap around' the knee created by the max plate dissipation. See this link for a nifty description and some nice pictures.
To answer the OP, the plate to plate impedance for any given push pull design is given by the tube characteristics, and may or may not be lower than you might expect. Generally speaking, however, the tube is often run somewhat near it's maximum plate voltage rating even when single-ended. When AB operation is designed in, the designer will tend to raise the plate voltage to the maximum and reduce the bias current. This often allows for a lower load impedance to be used and still miss the max plate dissipation. As a bonus the operation line will wrap around that knee (if designed properly), usually giving a substantial increase in output power.
Absolutely true in the class B part of each cycle, when only one output device at a time is operating.
Then there is the class A part of each cycle, for smaller signal excursions around the operating point, when both output devices are working simultaneously. They assist each other to drive current through the load, so that the load appears to have twice as much impedance as it actually does.
So, instead of 1/4 of the end-to-end impedance, the load seen by each output device appears to have 1/2 of the end-to-end impedance in this class A region.
More details here: The Valve Wizard -Push-Pull
Leonidas probably didn't, either. If he was acting true to form, he was probably simply trying to use the cheapest transformers he could find, which happened to have those impedance and voltage values.
Tubelab George is getting 30 watts from a pair of 6V6, using a 3300 ohm Raa output transformer.
In his own words:
How is George getting 30 W from a pair of 6V6, on only 350 volts B+, with only 250 V on the screen grids? Unlike Leonidas, George knows exactly what he's doing. And he's using MOSFETs to drive the 6V6 grids, so they can go a little into positive grid current on peaks.Tubelab George said:
I don't think there is a single optimum load line. Even for a specified B+ voltage, there is one that is optimum for lowest full-power THD; there is a different one that favours even-harmonic distortion over odd-harmonic; there is yet another that is kinder to the output valves and likely to make them live longer.
If you also allow big variations in B+ (my Princeton Reverb reissue supposedly has 440V B+ according to the factory schematic
) the picture becomes even more blurry, and the optimum load lines shift again. Same for variations in screen grid voltage.We are all being rather naive when we use the factory 6V6 data sheets (with 285V on the anodes and screen grids in the Tung Sol data sheet), and expect the factory-specified load impedance (8k Raa) to work, while actual B+ in some Fenders is 440V and screen grids are at 425 V.
That much change in B+ and G2 voltage will move the tube curves quite a bit compared to the datasheet ones, and without a curve-tracer, I have no idea what the new max power load line impedance would be.
Fender PRRI schematic: http://www.thetubestore.com/lib/the...nder/Fender-65-Princeton-Reverb-Schematic.pdf
-Gnobuddy
In my attempt to understand the operation of a Fender Deluxe Reverb AB763 amplifier, I came across this explanation of the output stage by Rob Robinette:
"The output transformer's primary takes in a high voltage, low current signal (high impedance) and puts out a low voltage, high current signal (low impedance). Typically about 350 volts of swing (350 volts peak-to-peak and 124 volts RMS) from the power tube plates flow into the output transformer primary and about 13.3 volts AC RMS (37.5 volts AC peak-to-peak) flows out the secondary wires to the speaker jack and on to the speaker."
OK. But if the turns ratio of the output transformer is ~29:1, how can there be 124 volts AC rms on the primary and 13.3 volts AC rms on the secondary? It don't compute...
"The output transformer's primary takes in a high voltage, low current signal (high impedance) and puts out a low voltage, high current signal (low impedance). Typically about 350 volts of swing (350 volts peak-to-peak and 124 volts RMS) from the power tube plates flow into the output transformer primary and about 13.3 volts AC RMS (37.5 volts AC peak-to-peak) flows out the secondary wires to the speaker jack and on to the speaker."
OK. But if the turns ratio of the output transformer is ~29:1, how can there be 124 volts AC rms on the primary and 13.3 volts AC rms on the secondary? It don't compute...
You're quite right, it don't compute!
Let's set aside Mr. Robinette's numbers, and try it for ourselves:
13.3 volts RMS across the speaker - check (22 watts into 8 ohms)
Therefore peak voltage across speaker = 1.414 x 13.3 = 18.8 V
Transformer step down ratio (primary end-to-end to secondary) checks out at about 29:1.
Each output valve only drives one half of the output transformer. The stepdown ratio seen at each anode is therefore half of 29:1, or about 14.5:1.
A peak voltage of 18.8 volts at the speaker, going through a step-up ratio of 14.5:1 (OT working backwards), will turn into a peak voltage of approx 273 V at each anode.
Mr. Robinette's colourful schematic shows 415 volts DC at the output anodes. So it is certainly possible for the anode voltage to drop by 273 volts (leaving an extra 142 volts to account for the saturation voltage of the 6V6, and supply voltage drop at full power output.)
So the good news is that Leo's claimed output power is at least plausible, as long as the 6V6's can actually manage to sink enough current through that 1650 ohm anode load each one sees. (Anode current would have to increase by 165 mA to cause a 273 volt drop in a 1650 ohm anode load impedance. That 165 mA is on top of the quiescent current, which seems to typically be about 20 mA for pp 6V6 amps, so each 6V6 would be pushed to around 185 mA peak current on each half-cycle, at the full 22 W output.)
Now let's look at the voltages Mr. Robinette mentioned.
We've already calculated 18.8 volts peak at the speaker, so the peak-to-peak speaker voltage is about 37.6 V; close enough to Robinette's 37.5 volts.
We've also already calculated about 273 V peak swing at each anode. Peak to peak voltage at each anode is therefore twice that, or about 546 volts. This is very different from the 350 volts peak to peak Robinette states. Robinette got this wrong.
Worth noting: because of the push-pull transformer action, when one anode swings 273 V negative, the other swings 273 V positive. Peak anode-to-anode voltage is therefore twice 273, or 546 volts. And peak-to-peak, anode-to-anode, voltage is twice that, or 1092 volts.
Robinette says "...flow into the output transformer primary", which is confusing - does he mean from each anode to the OT centre-tap, or does he mean from anode to anode? Either way, the 350Vpp he cites is wrong.
Mystery solved!
-Gnobuddy
dotneck335: look at loadmatch4-pp-beamtetrodes, there is nice example with load lines and calculations and explanations.
Whoa, whoa, not so fast!! Just WHERE does this "1092 volts" come from?? The B+ is only 415 volts. And doesn't this circuit operate in Class AB mode? This means that one tube is "cut off" for a part of the cycle.
Imagine for a moment that we replaced each 6V6 output valve with a single-pole, single-throw, switch.
With both switches open, the transformer centre-tap is at 415 volts.
Close one switch; that end of the transformer (call it end A) falls from +415 all the way down to 0 volts (a drop of 415 V). The centre-tap is still at +415 volts.
And the other end of the transformer (end B)? It rises by +415 volts to match the 415 volt drop at end A. Since it was already sitting at +415 volts, and has now risen by a further 415 volts, end B will now be at (415+415) volts, or +830 volts.
So far, we have only turned on one switch. This corresponds to one half-cycle of the output waveform. So the voltage we are seeing across the transformer now is the peak (not peak-to-peak) voltage. PEAK voltage is 830 volts. End A is at zero volts, end B is at +830 volts.
Now open the first switch, and close the second. End B will now drop to zero volts. End A will rise to +830 volts. This is the second half-cycle of the output waveform.
End A has swung from 0V to +830 V. End B has swung from +830V to 0V (it's 180 degrees out of phase with end A). If you wired a meter between ends A and B, it would show a positive peak of +830 V at one instant, and a negative peak of (-830 volts) at the next instant.
So what is the peak-to-peak voltage between ends A and B? Twice the 830V peak voltage, yes? So that will be 1660 volts.
So: if our power supply never dropped below 415V, and if 6V6 anodes could drop all the way to zero volts (like the switches we used in our hypothetical example above), then we could get up to 1660 volts, peak to peak, across the two ends of the output transformer, using a power supply voltage of only 415 volts.
As we saw from the numbers I calculated previously, for the amplifier to put out its rated power, we need "only" 1092 volts peak to peak, measured from one anode to the other. This is comfortably less than the 1660 volts you'd get from ideal valves (with no on resistance) and ideal B+ voltage (perfectly regulated, no drop under load).
In practice, of course, the 415 volts will fall a good bit, and the 6V6 anodes will probably not be able to drop below, say, 50 volts. So we won't get 1660 volts peak-to-peak across the OT, but something considerably less. Since we only need 1092 V, and not 1660 V, there is plenty of room to accomodate these real-world losses.
Sure. That has no effect on the numbers we've just looked at. Earlier in this post, in our thought-experiment, we only closed one switch at a time - the other switch was open, representing the cut-off output valve. It is the voltage across the open switch that swings to +830 V in the thought-experiment.
In other words, the anode of the cut-off valve swings way up above B+ (to twice B+ if valves were perfect switches). This happens because of the action of the centre-tapped transformer, since the other end of its primary is being pulled low by the other valve, which is operating (not cut off).
Amplifiers without output transformers, such as virtually all solid-state audio amps, cannot do this. Their output voltage never swings above B+.
-Gnobuddy
I'm also having the same trouble as dotneck335 following the calculations - we know Robinette's figures were wrong, so far so good, but the "1092" does not seem right either... A quick back-of-the-envelope calculation shows that with 1092Vpp, the output power would be 45W, which is much higher than the rated power of the Deluxe, no?
It never hurts to check and re-check math, so let's do that.
End to end primary impedance of the OT is 6600 ohms.
If I'm right, end to end, peak-to-peak, voltage across the OT primary is 1092 volts.
The formula for calculating RMS power, given peak-to-peak voltage Vpp and load resistance R, is:
P(RMS) = Vpp x Vpp/(8 x R)
Note the factor of 8 in the denominator: this comes from the conversion of peak-to-peak voltage to RMS voltage [divide by 2 times the square root of (2)], and then squaring that to get RMS voltage squared. Squaring [2 x root(2)] gives you exactly 8.
So now that we have the right formula, and all the numbers, let's calculate the power:
Code:
P(RMS) = 1092 x 1092 / (8 x 6600)
= 1,192,464 / 52800
= 22.6 wattsI think what makes it hard to form a mental picture of the OT primary voltage is the fact that neither end of the OT is sitting at a constant zero volts. Instead, each end is flapping wildly up and down through 546 volts, but in opposite directions.
Perhaps this is an easier way to visualize it: We know each anode is flopping up and down through 546 volts. That means there is 546 volts peak to peak between one end of the OT, and its centre tap. That means there is 546 Vpp across one half of the OT primary.
So what is the voltage across the entire OT primary? It's gotta be twice the voltage that's across half of the primary, yes? And that is 1092 volts, peak to peak.
Is that making sense now?
-Gnobuddy
Thanks to Gnobuddy for the best explanation of a tube push-pull transformer-coupled output stage that I've seen. My math figuring took a different path, but with nearly the same result. 22 watts output into 8Ω gives 13.26 volts RMS. Transformer turns ratio is 28.72; giving 381.7 volts RMS on the primary--- 538.75 volts peak, 1,077.5 volts peak-to-peak.
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