Hello,

I'm experimenting around with Miller Capacitance Multiplication and was wondering if anyone could give me a realistic range of capacitive multiplier factor one could design. Right now I can reach 3x quite easily (in an R-C circuit, with the Miller C connected between the input and output of an inverting gain amp.)

Is 5x achievable? How about 10x? I'm looking for the best bang for the buck - to keep the main capacitor small.

Thanks,
Joe W.

it's a matter of amplifier design at the frequency of interest with 5 gain. Whats the problem?

Thanks,
Raj.

Hey Joe... As you know, the miller multiplier (MM) will be Vout/Vin+1. So suppose you had a gain of 100, in theory your MM should be 101. Well, what if Vin = 1V? Obviously the other side of the cap isn't going to be swinging 100V. Well not more than once anyway. Nope, it will swing to Vdd so the MM will always be less than (1+Vdd/Vin). At least that is how I remembered the way it worked.

Tell your boss he owes me $35 now  ;)

rg

hello Robg,
what is this 1v and 100v, miller multiplication is a small signal approximation (which has a very less dynamic range, of course it depends on the architecture).

Thanks,
Raj.

@RobG, the simple circuit will be a cap connected to a diode connected transistor & having this current mirrored by a factor of A times, to realize a cap of (A+1)time, but we will be burning a lot of DC current required for biasing the MOS. But there are some architectures, which do it in a efficient way ( like implementing the mirrored stage as differential stage). But the underlying problem is we will get cap multiplication in addition to Series & Shunt resistance, which is a headache.

Thanks Rob and Raja,

I'm not building an amplifier, but rather kicking around the idea of implementing miller multiplication to save capacitor layout area (at the cost of add'l current consumption.) But, not sure what a practical range that is achievable.

Joe W.

raja.cedt wrote on Dec 11^th, 2011, 6:23am:

No it is not a small signal thing - it is from I = C*dv/dt, where dv/dt is the voltage across the cap. If you use gain the cap voltage will be larger, thus dv/dt will be larger, making the cap *appear* larger for any signal big or small.

The point was that you need to remain within the output range of the amplifier, if it wasn't for that you could have as much gain as you need.

A is simply the gain.That is, the ratio of the voltage on one side of the cap to the other side (with sign inverted). If the voltage on one side of the cap is -A*Vin, the total voltage across the cap is (1+A)*Vin.  You could get this gain with an opamp or any amplifier.

It doesn't matter if the output is millivolts or megavolts, the equation gives the same result. HOWEVER, in real designs we are limited by how much the output signal can swing before it saturates the opamp.

Try it with a cap across an inverter being charged by a current source and you will see that it is the non-linearity when it hits the rail that kills you: Assume the inverter has a gain of A. When the input is at 0V, it will charge like a 1x cap because the output of the inverter is pegged at the rail. Once the input reaches (Vdd/2)*(A-1)/A the inverter's output will start to move and the cap will look like C*(1+A). But here is the kicker... it will only do that over a range of (Vdd)/A.

If you calculate the amount of time it takes for the input to get to Vdd/2, it is equal to the amount of time it *would* have taken if you used 2*C for the cap. That is, the inverter will only double the effective cap value if you start at the bottom rail! This is true even if the inverter has infinite gain. Pretty weird. This is because the inverter is pegged to the top or bottom rail for almost all of the signal except for that tiny range in the middle between (Vdd/2)*(1-1/A) and (Vdd/2)*(1+1/A). If you can keep the signal inside that range you get the multiplication, but you lose it once the amplifier rails.

Personally, I'm not sure that Miller multiplication beyond 2x is practical outside of a feedback loop because of this - the input signal has to be so small to keep the output of the amplifier from pegging at the rail that the applications are limited. But if your desired gain is less than (1+Vdd/ΔVin) I guess it could work.