Controlled Miller Effect

123

00:56:28

Trying to control the gain with negative feedback, but capacitance is about 2/3 what it should be. It's higher with lower bias voltage, but I thought I saw distortion. DC response is also very different.

published 7 years ago

jason9

7 years ago

Remove the 100ohm resistor, add a 2.26mA current source to the 1uF capacitor pointed so that it draws power from the capacitor, and reduce the 100nF capacitor to 10nF.

jason9

7 years ago

Also, the capacitor at ground isn’t needed and all it does is slow the startup time.

jason9

7 years ago

Also, for an instant startup with the modifications I’ve described set the phase of the voltage source to -9 degrees.

jason9

7 years ago

Here is your circuit with the modifications I’ve mentioned: http://everycircuit.com/circuit/6312745824944128

eekee

7 years ago

Thanks jason9. So, the transistor is partly acting in Miller effect, partly just sinking current. The reason it's sinking current is obvious but what to do..? Tomorow.

jason9

7 years ago

The sinking current is a benefit because it means that the capacitor can not only charge but also discharge, although at a limited rate.

jason9

7 years ago

But at a cost it also has to be supplied with some amount of current and once the current goes to zero it’s at maximum discharge rate, and it can’t supply current, only absorb it.

thebugger

7 years ago

It's not a miller effect at all. The 1k/1uF RC chain causes a phase shift at the given frequency. http://everycircuit.com/circuit/5053232337649664

thebugger

7 years ago

Also the output impedance of the amp is considerably high, so the RC chain stresses the amp and causes it to present a phase shift itself.

jason9

7 years ago

Dude, the amp is amplifying the capacitance causing it to act as a capacitor 100 times greater than the 10nF capacitor because of a gain of 100, and that makes it act like a 1uF capacitor as demonstrated on the right.

jason9

7 years ago

@eekee, you removed the ground capacitor, but you forgot to remove the 100ohm resistor interfering with the operation of the miller effect amplifier and you forgot to reduce the 100nF capacitor to 10nF and you forgot to add a 2.26mA current source on the right side to make it more identical to the left side in operation and you also forgot to set the phase of the AC source to -9 degrees, although that last one is least important.

jason9

7 years ago

@thebugger, I think you thought it was an amp rather than a miller effect thingie because you work in the amplification area and whenever you see anything that looks amplifier related you immediately think “Amplifier!” And, in truth, this is an amplifier, as is any BJT, and it’s current gain plus one is what is multiplied by the capacitance to give the effective capacitance. Therefore, it IS miller effect as given by the wikipedia definition. It works the same with resistors and inductors too, where it divides the resistors and inductors value by the gain rather than multiplying.

thebugger

7 years ago

Yeah I noticed the weird wiring up and that it doesn't work as an amp.

eekee

7 years ago

I forgot about it altogether. Anyway, lesson learned: can't control Miller effect with negative feedback.

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