miller capacitor - Page 1
Forum section: electronics / projects

author: rex1232
author_rank: Contributor
timestamp: December 03, 2019, 11:43:24 am
content: Hey guysI am wondering what C3 does in this circuit.As I can understand it is what you call a miller cap.I have read about the miller effect but I can't quite understand what the cap does.Thank you in advance

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 03, 2019, 02:52:47 pm
content: Yes.Normally, when you apply some base current to the transistor, it turns on and collector voltage falls. With the capacitor in parallel with the transistor, the falling collector voltage induces a current I = C dV(b-c)/dt between the collector and base. Which means: 1. some of the collector current is diverted to the source, and 2. the source must be able to deliver this current, otherwise it's not going anywhere. The overall effect is to turn it into a transconductance integrator, i.e., Vout = integral(Iin dt).Other ways to look at it: for a step input, the output is a ramp; for a ramp input, the output is a parabola; for a sine input, the output is cosine (a phase shift of 90 degrees); etc.The integral is imperfect, only holding true while the transistor is active; a negative step will turn the transistor off, and the only current flowing is the collector load current, which limits the collector-rising slew rate. (In this amplifier, the current comes from an active current mirror, so will be complementary to the input signal, which helps.) The transistor can be turned on quite hard, but no harder than it is ultimately capable of, limiting falling slew rate. (In this amplifier, the current is limited by the input stage bias current.) There is also a feed-forward term (a zero in the frequency response), due to the capacitor itself -- at frequencies higher than the transistor can respond to, or for voltage changes where the transistor is off, the transistor isn't relevant and there is simply a capacitor between input and output, coupling them together directly. That is, a step change of say -0.5V at the base will give an output step change of about -0.5V. This is most relevant for large (>50mV) and fast changes (fractional us to ns), while integrator behavior follows over a slower time scale (us to s).The relevance of an integrator to an audio power amplifier, is that it gives a precise output, i.e., low distortion and flat bandwidth, when used as the control amplifier. It has a major downside, that its phase shift is 90° to start with, so the rest of the amplifier can't take up much phase shift before the whole thing oscillates. We can free up some phase margin by connecting a resistor in series with the Miller capacitor. This adds a zero at a controlled frequency, easing the phase shift and allowing greater stability and bandwidth. Compare with this circuit:C1/C2 and R9/R12 serve as Miller compensation. The bandwidth from pretty ordinary components is in the 10s of MHz (at much lower power too, of course).Tim

author: rex1232
author_rank: Contributor
timestamp: December 03, 2019, 03:13:34 pm
content: Thank you for the explanationThat really made sense.So that means that if I include this transistor (as shown below) it would also work as a miller cap?Thank you in advance

author: T3sl4co1l
author_rank: Super Contributor
timestamp: December 03, 2019, 03:22:06 pm
content: Yes, and actually I would prefer doing that, to preserve symmetry.Tim

author: Kevin.D
author_rank: Frequent Contributor
timestamp: December 04, 2019, 05:41:53 pm
content: As Tim pointed out there a quite a few ways that one can look at this to understand it, having a couple of different perspectives in at your disposal can really be useful so to try to add to whats been said (trying not to repeat to much) .The position of that capacitor means it must be driven by the input but it's other side is connected to a node that has voltage gain so it requires a larger charge than say an equal sized capacitor that was connected to some relatively fixed potential like gnd would, this can be useful because it means a much smaller capacitor can be used (e.g to roll off the voltage gain of an amp stage for things like stability or to simply shape frequency response ) to replace what would otherwise require a much larger capacitor to achieve the same response (if placed say on the output).If you want to calculate that freq response then one way as you already read would be to use Miller's theorem to calculate anequivalent sized capacitor to replace an input/output straddling capacitor with an equivalent capacitor which can be connectedbetween the input and a fixed voltage like gnd (and it's capacitance will have a similar loading effect on the input) in the hope it will simplify circuit analysis (e.g input pole/zero position calculations), the single feedback cap is usually replaced with twoequivalent capacitors Cin(miller) and a smaller one on the output Cout(miller), keeping in mind that this miller 'approximation'assumes the voltage gain that was used to calculate them is not frequency dependent on something else ( i.e something else that causes shifts in the voltage gain like say a 'variable Rload or load pole that becomes the dominant pole' would also invalidate the calculated miller poles ).It's also useful sometimes to be able to view the function of such caps from a very different perspective using 'feedback looptheorems' instead, It's just another method to reach similar conclusions as the previous approaches. Viewing those caps from this perspective 'negative feedback path' then they sets the overall small signal voltage gain of the stages enclosed in the feedback loop ~ Zf/Rin (ignoring small error cause by finite gain of stages and other input caps) and from feedback theorems we can also conclude Zout is reduced by ~'loop-gain factor'and other such insights, personally I find one perspective will apply easier/better in certain scenarios than others and so change the way I look at it when required.Regards