How to compensate 3 poles voltage regulator?

[taofeng]

Here I have one low frequency dominant pole, around 16 kHz and it is fixed. The second (output of the voltage regulator) and third dominant poles(a diode connected transistor) are load current dependent ?

I am wondering if there is any effective way to compensate this.

 

[yxo]

Output pole dependent of load current because transconductance of the output transistors changes with load current. You need local feedback to stable gm of output transistors. For compensation you need local feedback too, or/and you must increase gm of output transistors.

 

[mario1980]

If you are worried about closed loop stability, than I belive the resistive load at the output is not too important, but the capacitive load is important because the pole of the output transistor, moves towards the origin, disturbing the phase at which the loop has unity gain, resulting in an oscillator. So, the point is to know maximum load capacitance, and starting with that, you compensate at the output of the opamp. The diode connected transistor doesn't interfere too much, in stability analyses, even if the current through it depends on the load, because the capacitance in that node isn't changing, right?
One more observation, the larger the output transistor, the harder to compensate the loop.
Hope that my understanding of your circuit is good.

Good luck!

 

[yxo]

output pole (second in our case) = gmout / Cload => More output gm, more far poles from origin

 

[mario1980]

"More output gm, more far poles from origin" True, but the overall loop gain increases and compensation becomes more difficult. So I recommend that the gm at the output to be as low as possible for WC conditions.

 

[spring1860]

"More output gm, more far poles from origin" True, but the overall loop gain increases and compensation becomes more difficult. So I recommend that the gm at the output to be as low as possible for WC conditions.

would u pls explain why more output gm , the overall loop gain increase and compensation becomes more difficult .
thx

 

[mario1980]

You have a loop gain that is equal to the sum of all gains in it. So, at the critical phase you need an overall loop gain smaller than 1 (0 dB). If you increase the output gm, the gain introduced by that transistor, requires a smaller gain at the output of the opamp (usually). This means more compensation capacitor for the opamp, which may require a LOT of area, raise poly density issues, and you might be forced to use Miller compensation, and this is a bit more complicated.

 

[krishnareddy]

miller compensation can be used to split the poles and adding zero in addition to pole splitting helps