# Miller Integrator with Feedback (i.e. DC Stablization)

Discussion in 'Homework Help' started by jegues, Apr 22, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
I don't understand how to obtain the last portion of the question,

Sketch output for a 0.1ms 1v pulse with the feedback resistor connected.

I was able to do it without the resistor,

$\frac{-1}{RC} \int_{0} ^{0.1ms} 1d\tau + 0$

but I don't know how to do once the feedback resistor is implemented.

Can someone explain?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Is the issue that you don't understand the solution provided?

The unity frequency gain without the feedback resistance will be given by the condition ....

1/(ωRC)=1 or when C=1/(ωR)=0.7958 nF

When the feedback resistance Rf is added, the response will be a first order exponential function with a time constant determined by the feedback resistor and the capacitance.

Given the input resistance will be 20kΩ the feedback resistor will be 2MΩ to give a DC gain of 40dB (100x).

So the exponential time constant τ will be 0.7958nF*2MΩ=1.592ms

The output for a constant DC input then has the general form

Vo(t)=-k*100(1-exp(-at))

where k=input DC voltage and a=1/τ=1/1.592ms

This must be true because the DC gain has been set to 100x and at t=∞ the value vo(t)=-k*100. For +1V DC input there would be -100V DC output at t=∞.

However, the 1V input lasts only for 0.1ms. So at 0.1 ms ...

Vo(0.1ms)=-100(1-exp(-0.1/1.592))=-100(1-0.939)=-6.088V

When the input drops back to 0V at 0.1ms, the output as a time function will then be

Vo(t>0.1ms)=-6.088*exp(-a[t-0.1ms])

A more rigorous proof requires you to solve a differential equation. The posted solution refers you to page 112 of the text, which presumably sets out the method of setting up & solving the DE.

It's not hard to set up the equation ....

If the input is Vi(t) then the input current is i(t)=Vi(t)/20k

If the output is Vo(t) the parallel {Rf||C} feedback network must satisfy the condition

$i(t)=-\frac{V_o(t)}{Rf}-C\frac{dV_o(t)}{dt}$

or

$\frac{Vi(t)}{20k}=-\frac{V_o(t)}{Rf}-C\frac{dV_o(t)}{dt}$

In accord with this problem, one would solve this DE with Vi(t) a constant DC value.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
The first order DE is what I remember I just couldn't remember how to get it into that form.

Since we have a constant input voltage between t1 and t2 the voltage will look like,

$V_{c}(t) = V_{c}(\infty) + [V_{c}(0) -V_{c}(\infty)] e^{\frac{-(t-t_{1})}{\tau}}$

correct?