According to RDH4, this transformer should be considered to have a 16:1 turns ratio. and a "plate-to-plate" impedance 4x that of one of the primary windings.
It is the sum of the two Va-k swings. The tubes are working into the windings in series, but you cannot simply pick two nodes to measure in between because you have two tubes with windings in between all in one series circuit.
I agree completely. I was merely pointing out that since it is well-known that the driver puts out somewhere in the neighborhood of 200Vrms max, you would only be able to get the output stage halfway to clipping if it had gain less than one. Attempting to draw a load line would reveal that.
Excellent. Glad to see some of us are seeing things correctly.
For the others, the answer will become clear in 50 years of further research, no matter when you ask.
Mac should be prosecuted for coming up with this dumb circuit, that doesn't fix anything useful, but inflicts endless torture.
For the others, the answer will become clear in 50 years of further research, no matter when you ask.
Mac should be prosecuted for coming up with this dumb circuit, that doesn't fix anything useful, but inflicts endless torture.
Last edited:
You want people to believe a number that is not measurable ????
You have no argument other then your adding up two numbers with your keyboard...
If you have 46.8V RMS AC developed in a circuit, then it is measurable with a diff probe... it's that simple...
If it is not measurable, then it does not exist...
You need to back it up...then tell me WHERE , what nodes to measure that develop this 46.8V between them...
I can measure and show that there is a CURRENT GAIN of 2 with current probes....
RDH4 has multiple mistakes and incorrect info throughout the book...
Last edited:
This is simply wrong. Do you understand how series circuits work? If I have a resistor, capacitor, resistor, and another capacitor all in series and I drive them with an AC voltage, there are not two points I can connect to in that circuit that will give me the total AC voltage across the two capacitors alone. It doesn't work that way.
Now consider what might happen if we replaced those two capacitors with two primary windings on a transformer and loaded the secondary. What two nodes would we connect to to measure the voltage across the windings alone? We can't. There is a resistor in the way (like there is a tube in the way in the UC circuit). We have to recognize that it is a series circuit, measure them separately, and then add them together.
Last edited:
Here is another way to measure it then...
If you remove one tube from the output stage... All the voltages will be the same in the windings as it would be with both tubes installed..
Now you have a GAIN of 1.5 ..... Because you have a single ended input of 15.5V and a full winding voltage of 23.4 .....
Now if you install both tubes... All the voltages in the windings will remain the same, however you will now have twice the AC current flowing , thus a CURRENT GAIN of 2 ...
Now the input voltage is with both tubes installed is at 31V..Voltage Gain not there..it is exchanged for Current Gain..
If you remove one tube from the output stage... All the voltages will be the same in the windings as it would be with both tubes installed..
Now you have a GAIN of 1.5 ..... Because you have a single ended input of 15.5V and a full winding voltage of 23.4 .....
Now if you install both tubes... All the voltages in the windings will remain the same, however you will now have twice the AC current flowing , thus a CURRENT GAIN of 2 ...
Now the input voltage is with both tubes installed is at 31V..Voltage Gain not there..it is exchanged for Current Gain..
Can you demonstrate they are in parallel? I can only see ONE current flowing hence they ARE in series. No doubt about it!
Then I repeat the math (I did in post n.45) once again. This time using the same ratio of the MAC.
Whatever is the step-down ratio to the secondary, the cathode and plate windings have the same turns. Hence the plate impedance once the cathode winding is connected will be 4x higher.
From this the feedback factor beta will be 1/2.
The closed loop gain (i.e. cathode feedback connected) will be:
Ac=Ao/(1+Ao*beta)= Ao/(1+Ao/2), where Ao is the open loop gain (i.e. with cathode feedback disconnected on the same transformer).
I don't know the gain of the KT88 in open loop in that particular application then let's make two extreme cases and a typical one:
1) Ao=3
Ac=3/(1+3/2)=1.2 >>> the closed loop gain despite being unrealistically low and never seen in pentode already results in voltage gain!
2) Ao=10 (more likely)
Ac=1.67
3) Ao=100
Ac=1.96
Last edited:
Yes, it is measurable...if the the circuit is truly in SERIES..
I take one set of diff Probes and measure the outside voltage above the top cap and below the bottom cap... and insert that probe into scope channel 1.... Now I take my second diff probe and measure between the caps, ie below the top cap and above the bottom cap...and insert this into Channel 2 on my scope..
I then use the math function on the scope to subtract CH2 - CH1 ... and this will be the SUM of voltage that appears across both caps in SERIES...
Just replace the caps for the windings using this same technique ...
As you say..drive them with an AC voltage...this AC voltage would then be quite large... what value is this AC voltage ??
Last edited:
Well, you didn't say you had two different probes and a math function available. 😀
Simply connect one diff probe across the full cathode winding and place the other diff probe across the full anode winding and use the SUM function. Then you have the full output voltage.
Technically that is invalid method, since I could put both Channels diff probes across the same Cathode winding and the Math function would SUM the same winding twice.. Which is technically what is happening when one diff probe is on CATHODE winding and one diff probe is on PLATE winding, because both the full CATHODE winding and full PLATE winding are wound B-Filar at the same time in parallel with each other..
Last edited:
Completely wrong! Supposing you can remove one tube (you can't in the real thing because there is no gap to allow DC current) in the best case you will get half heavily distorted power.....
Try it and you will see otherwise....make the measurements , then get back to me...I have done it many times....Otherwise your just arm waiving....
You would not do a full power output measurement.....no core saturation in real life, there is a small gap...
If your that concerned about the DC current then disconnect the driver input coupling cap ..this way you kill the AC signal but maintain the DC current..
Last edited:
Do it all the time.. I have designed extended DELTA water cooled 3-Phase transformers in the 8MW to 20MW range both 12 pulse and 24 pulse...windings are summed that are in SERIES... some are +15 degrees windings some are -15 degrees windings... they seem to add up and work fine for our military...
Last edited:
"If your that concerned about the DC current then disconnect the driver input coupling cap ..this way you kill the AC signal but maintain the DC current.."
That won't work either. Disconnecting the AC drive to one side will only leave the IDLE current nulled. But not the now half wave AC (near class B op) signal put into the xfmr. Half wave rectified. No wonder you are getting only half the meter reading expected.
That won't work either. Disconnecting the AC drive to one side will only leave the IDLE current nulled. But not the now half wave AC (near class B op) signal put into the xfmr. Half wave rectified. No wonder you are getting only half the meter reading expected.
Last edited:
- Status
- Not open for further replies.