error-amplifierswitch-mode-power-supplytransfer function
In case of switch mode power supplies, we uselly use "Error Amplifiers" (Type 1,2 & 3), my question is:
The output voltage of this amplifier (Verr), is it the DC error of the Vref and Vout (i.e: Verr=A(Vout-Vref), A is the DC gain of the error amplifier)? if is not, please explain to me what is it and what is the role of DC gain in that case??
simulate this circuit – Schematic created using CircuitLab
Let me label your circuit nodes for better understanding of your question
simulate this circuit – Schematic created using CircuitLab
To answer your question I must first clarify some definitions:
Now,
$$ y=k\cdot e $$ $$ e=r-y $$ $$ y=k\cdot (r-y) $$
So, the error gain function follows as
$$ k=\frac{y}{r-y} $$
So, you don't even need the transfer function to establish the value of k. You just need y and r.
Anyway the transfer function is
$$ TF=\frac{y}{r}=\frac{k}{k+1} $$
So, if you're given the transfer function (note that this is just a constant for this theoretical system) you can find k as
$$ k=\frac{TF}{1-TF} $$
Note also that, with negative feedback, in stable system conditions, and with finite k:
$$ 0\leq TF<1 $$
It's hard to explain a control system without control theory, but Vc will typically be the output of the compensator, which is the error signal modified by the compensation.
This goes into a modulator. So this voltage basically "sets" the duty cycle of the PWM. So working backwards, whatever the output voltage is requires a certain duty cycle. In the case of a buck converter (in CCM) it's Vout/Vin. Drawing current from the output would normally cause the output to drop due to series impedances, but the control loop will adjust the duty cycle by moving Vc up slightly to keep the output at a constant value.
The compensation adjusts the response of the controller so that Vc gets to just the right point in a reasonable amount of time without overshooting the mark or oscillating.
Best Answer
No. Now would be a good time to find some reference material on op-amp circuit design and do some studying.
The "error amplifier" is a PI controller with some band-limiting on the proportional term. The band limiting is almost certainly there to reduce the amount of ripple in the output and avoid sub-harmonic oscillation.
Moreover, the error amplifier is inverting: the higher that \$V_{out} - V_{ref}\$ gets, the faster \$V_{err}\$ will trend downward.
You can tell this by looking at the feedback network: there is not a DC path from \$V_{err}\$ to the \$V_-\$ input of the op-amp: both paths have blocking capacitors. That means that the output always integrates.
It is likely that the R4*C2 time constant is chosen to be longer than the PWM frequency, but shorter than the desired settling time of the power supply. C3 and R4 are chosen to stabilize the supply while having it respond as rapidly as possible to variations in the load.