I would like to measure th bias current of an amplifier when it is "active".
Probably a good starting point would be a current sense amp which measures the voltage over both emitter resistors.
Has anyone ever tried this, or knows some amplifiers/schematics where the bias current is converted into a DC voltage (during normal amplifier operation) ?
Probably a good starting point would be a current sense amp which measures the voltage over both emitter resistors.
Has anyone ever tried this, or knows some amplifiers/schematics where the bias current is converted into a DC voltage (during normal amplifier operation) ?
To measure bias current you need a special Converter current to voltage. They are installed in the circuit of the emitter or the collector of the output transistors are called "resistors". If you measure the voltage across the resistor, you can calculate the bias current according to Ohm's law.
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That I am aware off and is easy. But obtaing the bias current while playing music (so completely disregarding the current to the speaker, which flows through the same emitter resistor) is a different story 🙂
The output current is the DIFFERENCE between the currentsin the upper Re and the lower Re.
If the two Re pass the same current, then the output current is zero.
You can force the amplifier into this operation by disconnecting the speaker load. Now the upper and lower currents are equal.
If there is an output current, then the upper and lower currents are different and varying.
It would need some very clever circuitry to come up with a bias current answer, when the variable load is drawing variable current while a variable input signal is presented.
If the two Re pass the same current, then the output current is zero.
You can force the amplifier into this operation by disconnecting the speaker load. Now the upper and lower currents are equal.
If there is an output current, then the upper and lower currents are different and varying.
It would need some very clever circuitry to come up with a bias current answer, when the variable load is drawing variable current while a variable input signal is presented.
Now the other way around. The bias current flows through both Re and creates a certain voltage drop. Let's assume the ideal situation where we play a sinewave.
During the top half the voltage drop over the upper Re increases During the bottom half the voltage drop over the lower Re increase as well but in the other direction. Thus taking the voltage over both Re and feed that in an rms converter would cancel out the audio and leaves the static bias current. I expect the biggest issue here is the non ideal world, such as cross over artefacts. Or am I overlloking something else ?
During the top half the voltage drop over the upper Re increases During the bottom half the voltage drop over the lower Re increase as well but in the other direction. Thus taking the voltage over both Re and feed that in an rms converter would cancel out the audio and leaves the static bias current. I expect the biggest issue here is the non ideal world, such as cross over artefacts. Or am I overlloking something else ?
The concept of semiconductor biasing and DC operating point applies only to small-signal conditions. An output stage works in large-signal conditions, so there is a subtlety.
If our two output transistors have emitter resistors called Re1 and Re2, then the bias current is the current that goes through both transistors, but not into the load. If we measure our currents in the same direction in both resistors (ie, top to bottom, not from transistor to load) then :
IBias = min( I(Re1), I(Re2) )
When load current is zero, I(Re1) = I(Re2) and both are equal to IBias.
However, when load current increases, and ultimately one transistor turns off, then there is no more bias. There is still a "bias setting", which is what the bias current would be if ILoad was zero, but at the moment, there is no more bias current.
If your amp is pure class A then the sum of voltaeges over both emitter resistors is constant, though.
If our two output transistors have emitter resistors called Re1 and Re2, then the bias current is the current that goes through both transistors, but not into the load. If we measure our currents in the same direction in both resistors (ie, top to bottom, not from transistor to load) then :
IBias = min( I(Re1), I(Re2) )
When load current is zero, I(Re1) = I(Re2) and both are equal to IBias.
However, when load current increases, and ultimately one transistor turns off, then there is no more bias. There is still a "bias setting", which is what the bias current would be if ILoad was zero, but at the moment, there is no more bias current.
If your amp is pure class A then the sum of voltaeges over both emitter resistors is constant, though.
No. when the Ire increases in one resistor, then it decreases in the other. It's the DIFFERENCE in those currents that becomes the output current.
When the output transistions from ClassA to ClassB, one of the resistors passes a static current as outputs change, while the other passes the output current (remember it's the difference that becomes the output).
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I build a special test rig to measure that. The circuitry and the findings are described here:
https://linearaudio.net/article-detail/2227
Jan
I will order the volume of linear audio and take a look.
@peufeu I agree with you that Ibias is the lowest of the current flowing through the emitter resistors. Assuming only one half is supplying current into the load.
But I do not see your point of "no bias" assuming that the amplifier has a bias current then both transistors always conduct a little. Imho that's the point of having bias current.
@peufeu I agree with you that Ibias is the lowest of the current flowing through the emitter resistors. Assuming only one half is supplying current into the load.
But I do not see your point of "no bias" assuming that the amplifier has a bias current then both transistors always conduct a little. Imho that's the point of having bias current.
I think what peufeu is signalling is that the bias current flows at no signal. When you have a large output current that causes a drop on one emitter R as large or larger than the set bias voltage, the other output device is necessary cut off. You can only 'measure' Ibias accurately at the zero-crossing of the output current. It's not as simple as it looks!
Jan
Jan
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Exactly !
Whatever Ibias at zero output current was, say 50mA or 200mA, when the load current becomes high enough to turn off one of the two transistors, say Iout=2 amps, then Iout flows in the transistor that is turned on. At this point, there is no way to know what Ibias was back when Iout was zero.
Another way to say it : whatever the set value of Ibias, the output stage will behave exactly the same once it enters class B. Ibias only changes output stage behavior in the class-A region where Iout is small enough that both devices conduct.
Excellent article btw Jan, I read it a few days ago. Your hardware setup is impressive !
Whatever Ibias at zero output current was, say 50mA or 200mA, when the load current becomes high enough to turn off one of the two transistors, say Iout=2 amps, then Iout flows in the transistor that is turned on. At this point, there is no way to know what Ibias was back when Iout was zero.
Another way to say it : whatever the set value of Ibias, the output stage will behave exactly the same once it enters class B. Ibias only changes output stage behavior in the class-A region where Iout is small enough that both devices conduct.
Excellent article btw Jan, I read it a few days ago. Your hardware setup is impressive !
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