Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: JPK on December 25, 2022, 11:03:00 am
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Question comes from below link:
https://robrobinette.com/Drawing_Tube_Load_Lines.htm (https://robrobinette.com/Drawing_Tube_Load_Lines.htm)
If you scroll 75% of the way down to the 5E3 Deluxe 6V6GT Power Tube part, he talks about how to plot the class A line and Class B line. I'm struggling to understand why you use 1/2 of the RLoad for class A, and 1/4 RLoad for class B. I understand the impedance is the square of the turns ratio, what I don't understand is why class B doesn't also use 1/2 RLoad like class A. Because to me it seems like in class A each tube's plate to cathode is also only across 1/2 of the turns. I do get that in class B one tube is conducting, class A both are. Below in quotes is the part I'm referring to:
"During Class A operation both power tubes are conducting so each tube sees 1/2 RLoad: 8k / 2 = 4k.
We divide the plate-to-cathode voltage by the RLoad = 340V / 4000 ohms = 85ma. We draw the initial or temporary Class A load line (pink) from 340 cathode-to-plate volts at the bottom of the chart to 85ma on the left. This gives the correct line slope but we have to slide the line up, keeping the slope the same, to intersect the load line with our idle point (green). This is our actual Class A load line (red).
During Class B operation only one power tube is conducting so half of the output transformer windings "disappear". Since the impedance ratio is the square of the turns ratio the operating tube sees only Ľ of the transformer's plate-to-plate impedance so we'll use an RLoad of 8k / 4 = 2k for Class B operation. 340 / 2000 = 170ma. We draw the Class B load line (blue) from 340 volts to 170ma. Since 170ma is off the chart we have to estimate where the line intersects 170 on the left side of the chart."
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I don't understand why class B doesn't also use 1/2 RLoad like class A.
Because impedance is the square of the turns ratio, therefore 1/2 x 1/2 = 1/4
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In the class A area of operation, current flows through the whole primary winding over the full cycle.
When signal levels increases to push things into class B operation, current stops flowing momentarily in one half winding, then the other.
With half the windings out of circuit, the primary impedance falls to 1/4 of class A operation.
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Class A, both tubes work.
Class B, only one at a time.
Due to flu, Santa is short on reindeers. With Donner and Blitzen pulling, each feels a load of 0.5 Santas. When only Blitzen pulls the load is a whole Santa.
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I don't understand why class B doesn't also use 1/2 RLoad like class A.
Because impedance is the square of the turns ratio, therefore 1/2 x 1/2 = 1/4
I understood the turns to impedance relationship, I just didn't understand why you consider all the turns in class A, but only half the turns in class B.
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Because in class B, a valve is in cut off, ie switched off, no current flows, hence half the turns are out of circuit.
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What ^pdf64 said^
Also, transformers work by induction, so they only ‘transform’ between separate windings when there is changing current (to cause induction). Hence when each respective output tube is in cutoff, there is no changing current on that half of the primary winding that would otherwise cause induction - so, the OT only ‘sees’ the induction from the opposite half of the primary winding (and each half only has 1/2 the number of turns), so the Pri:Sec winding ratio is effectively halved (which means the Pri:Sec impedance ratio is a quarter of the usual plate-to-plate impedance ratio)
Let’s say we have an OT with 20 turns on a center-tapped primary (10 turns on each half) and one turn on the secondary; a 20:1 turns ratio (i.e. 20:1 VAC ratio), which is 400:1 impedance ratio. Now one half of the primary is idle, so the secondary only sees current induced from 10 primary winding turns, i.e. a 10:1 VAC ratio, which is 100:1 impedance ratio. 100/400 = 1/4.
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Now that you guys have explained it so well I get the class B part. One tube is off, 1/2 the turns. 1/4 RLoad because of the turns to impedance relationship. So now it has me asking why it's only 1/2 RLoad if both tubes are conducting in class A? As PDF64 mentioned in class A it uses through all the windings of the primary. Thanks for all the help. Amazing that I'm hung up on this, seems so simple. I get the rest of it.
Another question: During class A is the current for each tube in the same direction through the windings or opposite? I believe they are opposite and summed in the OT during the crossover transitions?
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So now it has me asking why it's only 1/2 RLoad if both tubes are conducting in class A?
But we’re talking about the ‘B’ part (of the cycle) of Class AB1 (when the grid of the ‘off’ tube is driven into cutoff by a sufficiently large driving signal from the phase inverter).
In Class A, (when the input signal isn’t large enough to cause cutoff on the negative side of the signal swing) both tubes are conducting signal at the same time, and so both halfs of the OT primary are causing induction, and the output impedance is ‘normal’.
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So now it has me asking why it's only 1/2 RLoad if both tubes are conducting in class A?
But we’re talking about the ‘B’ part (of the cycle) of Class AB1 (when the grid of the ‘off’ tube is driven into cutoff by a sufficiently large driving signal from the phase inverter).
I get the class B part.
In Class A, (when the input signal isn’t large enough to cause cutoff on the negative side of the signal swing) both tubes are conducting signal at the same time, and so both halfs of the OT primary are causing induction, and the output impedance is ‘normal’.
Still fuzzy why it's 1/2 RLoad if all windings are conducting in Class A. The only thing that I can think to explain it is the two plate currents are summed in the OT and each plate current is 1/2 of the total OT's "I L". Hence why they use 1/2 RLoad to calculate each tube's individual plate current. Ip = Vp x 1/2 x RLoad.
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Still fuzzy why it's 1/2 RLoad if all windings are conducting in Class A.
It’s not 1/2 impedance load in Class A. That’s only in Class B. In Class A there is full impedance. There are two load lines in Class AB. The A line and the B line. The A line occurs more than 50% of each cycle; the B line, less than 50%.
During the B line, the ‘on’ tube is conducting really hard (over Pmax), which increases the output signal amplitude (during B), but as it occurs only about 30% of each cycle (when the other tube is ‘off), the average dissipation (over time) is fine. That’s why you get more output power in AB. It’s more efficient.
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Still fuzzy why it's 1/2 RLoad if all windings are conducting in Class A.
It’s not 1/2 impedance load in Class A. That’s only in Class B. In Class A there is full impedance. There are two load lines in Class AB. The A line and the B line. The A line occurs more than 50% of each cycle; the B line, less than 50%.
I'll re-quote part of my opening post. This is a quote from Rob's web site:
"During Class A operation both power tubes are conducting so each tube sees 1/2 RLoad: 8k / 2 = 4k.
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Each tube sees full impedance in Class A, and there are two halves of a primary winding. For each tube, it sees ‘positive’ current going ‘clockwise’ in one half of the winding during one half of the cycle, and then it sees ‘negative’ current going ‘anticlockwise’ in the other half of the winding for the other half of the cycle. This is still all ‘load’ (positive current going clockwise is the same thing as negative current going anticlockwise, with the current source/sink being the primary winding center tap). And in PP Class A, the other tube is seeing the ‘opposite’ current cycle at the same time. So there is 2x as much current in the primary as there would be in a SE amp). So we talk about ‘plate-to-plate’ load across the whole winding.
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Each tube sees full impedance in Class A, and there are two halves of a primary winding. For each tube, it sees ‘positive’ current going ‘clockwise’ in one half of the winding during one half of the cycle, and then it sees ‘negative’ current going ‘anticlockwise’ in the other half of the winding for the other half of the cycle. This is still all ‘load’ (positive current going clockwise is the same thing as negative current going anticlockwise, with the current source/sink being the primary winding center tap). And in PP Class A, the other tube is seeing the ‘opposite’ current cycle at the same time. So there is 2x as much current in the primary as there would be in a SE amp). So we talk about ‘plate-to-plate’ load across the whole winding.
Thanks that was a very good explanation (everybody's was good, I just have a hard head).
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Class A, both tubes work.
Class B, only one at a time.
Due to flu, Santa is short on reindeers. With Donner and Blitzen pulling, each feels a load of 0.5 Santas. When only Blitzen pulls the load is a whole Santa.
Awesome analogy, thanks.
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Still fuzzy why it's 1/2 RLoad if all windings are conducting in Class A. ...
You figured it out already. But restating to drive it home:
In Class A, tubes on each side of the primary are conducting. However, our calculations are based on what the tube(s) on one half-primary see as a load.
In push-pull, there is a center-tap and it is connected to the B+ supply voltage. So the tube(s) only see the "half-winding" between themself and the B+ supply. Therefore "Class A Load = 1/2 the plate-to-plate impedance."
And as you've learned, when entering the "Class B part of the cycle" the tube(s) on one side are cut off. Since there is no current through that half-winding, the transformer sees it as though that half-winding doesn't exist. The resulting primary impedance is 1/4 the plate-to-plate primary impedance.