It was your mistake because it didn't make sense.
Mistakes are always a problem.
It's all in the Walker-Albinson paper. Changing the circuit, wrongly, won't help you understand anything.
The voltage is amplified by OpAmp (B), current flows through the resistor R2\Rn (B-C-Ground). Transistors V1\V2 amplify the current that the load L1\Rn (D-C-Ground) receives. #18
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But the current output by my transistor is several times smaller than the current of the resistor above.
An error current I2 flows through R2 (as long as V1\V2 are not turned on). After switching on, the main current I1 flows through L1 #18
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You really have to give more info instead of just some words that don't mean a lot.
Tell what you measured in the circuit, which circuit, why do you think it is wrong, what did you expect to measure?
Otherwise this can go on for a few more years with no progress.
Jan
Ok,Below are the parts I put on according to the original circuit diagram.
My question is, I think that in the case of conduction, the current flowing out of the transistor should be larger than the above resistance, but as shown in the simulation diagram, the situation is not the same as I thought.
Attachments
You really don't get it, do you?
" I think that in the case of conduction, the current flowing out of the transistor should be larger than the above resistance, but as shown in the simulation diagram, the situation is not the same as I thought. "
What case of conduction? what current out of what transistor, what 'aboce resistor', what situation? I have no idea what you are talking about.
Good luck,
Jan
" I think that in the case of conduction, the current flowing out of the transistor should be larger than the above resistance, but as shown in the simulation diagram, the situation is not the same as I thought. "
What case of conduction? what current out of what transistor, what 'aboce resistor', what situation? I have no idea what you are talking about.
Good luck,
Jan
A current cannot be either larger than, equal to, or smaller than a resistance. You are talking nonsense. It is impossible to help you if you can't use standard terminology correctly.
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Hey , that's just my opinion,i didn't say that was right.
I'm just an amateur student.
I really don't understand the principle.
It's all explained in the Walker/Albinson paper. If you can't understand that, it's far too broad to be answered here.
The point is that your posts are so unclear that it's impossible to know what you mean or to answer!
Has nothing to do with what you do or don't understand.
We're here to help people understand and progress, but you go a long way to make it impossible to get a meaningful discussion.
Please try to understand the point!
" I think that in the case of conduction, the current flowing out of the transistor should be larger than the above resistance, but as shown in the simulation diagram, the situation is not the same as I thought. "
What case of conduction? what current out of what transistor, what 'above resistor', what situation? I have no idea what you are talking about.
If you would say something like: "The Q2 collector current for the case that the dumpers are off would flow through L6 and generate a voltage between point A and B. How does that decrease the distortion?" we would have something to talk about.
Jan
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https://en.wikipedia.org/wiki/Wheatstone_bridge
When the bridge circuit is well balanced, the output voltage should be equal to the voltage of the negative input of the amplifier. (refer to the original post)
For an ideal amplifier, the negative input pin should be equal to the positive input pin. If so, there will be no distortion.
When the bridge circuit is well balanced, the output voltage should be equal to the voltage of the negative input of the amplifier. (refer to the original post)
For an ideal amplifier, the negative input pin should be equal to the positive input pin. If so, there will be no distortion.
https://ru.wikipedia.org/wiki/Мост_Максвелла
https://www.dadaelectronics.eu/uplo...nt-Dumping-explained-by-Peter-Walker-1975.pdf
The original current dumping concept from QUAD includes a very smart feedback network that changes it's feedback factor depending on whether the dumpers are on or not. It's a bridge type circuit with two arms formed by an inductor and a cap that have precise values.
https://www.diyaudio.com/community/threads/current-dumping-amplifier.243345/page-6
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