What if I use 47k for R2 & R4! Theoretically by the voltage dividers both pin2&3 will get less than 12v when V1 is about 37v. The common mode voltage will then be (v(pin3+v(pin2))/2<12v. Is this process ok?
Hi,
I like to give credit where credit is due, and that's good thinking on your part. Not only do you meet the CM range that way, you also keep the resistor values at 100k or below which reduces the inherent noise as well as the chance of external noise pickup. Less that 20k even better if that is possible because this is part of a switcher.
But i am afraid i have to open still yet another can of worms. The two op amps will be in the feedback path (judging by your previous posts) to limit the current. That effectively puts two more integrators in the feedback path. The integrator time constants will be relatively fast, but might not be fast enough. There is also the GBW to consider for the stage that has a gain of 100, which will probably change to a gain of 300 once you reduce the gain of the first op amp. But even at 100 it could cause a problem, and even at a gain of 1 it is still a possibility.
I am not sure how familiar you are with feedback systems, so i'll try to explain this in time-sequential manner, as if each step is viewed in sampled time. This helps to get the idea what could happen more intuitively.
First, assume the circuit is working normally and is not in current limit yet. Then a higher than normal current comes through the 1 ohm resistor. When that happens the first op amp starts to ramp up. When that happens the second op amp starts to ramp up too, but it's ramp might not be a linear ramp like the first because a ramping function of a ramping function is a time exponential function. The time exponential function means the output ramps up a little at first, then the ramp increases. This can easily cause oscillation. But even with out the ramping squared feedback function, the second op amp has to ramp up too, and it's gain bandwidth will be quite limited due to the high gain. What that means is the switching regulator IC feedback pin has to wait for the output of the second op amp to respond favorably, and in the mean time the output of the switcher is still above the current limit set point. Now if the damping factor of the entire network is such that the time response is totally over damped, then the output might simply shoot up and stay higher than normal for a while, then come down, then settle out. So eventually we reach current limit. But if the damping factor is not over damped or just too slow, then the output goes high, the op amp output eventually pulls it down a little, then it pulls it down too far, then it reacts a second time to start to limit the current, then starts to allow it to go high again, etc. This causes unceasing oscillation.
If you like we can go through a calculation using some assumed gains and see what we get, but one general rule usually applies: The feedback circuit should be faster than the error amplifier. In the case above the feedback is much slower than the error amp, so we could see problems at some load or all loads that cause current limiting.
So the first thing to check is basic open loop functionality, then when closing the loop check carefully for oscillation. It may depend partly on the load current or load resistance or whatever is connected as the load, especially capacitor type loads.
Another way to think about this is you drive your car down the highway and step harder on the gas, and you can not see the speedometer, but a friend in the back seat can see it. He has to tell you what speed you are going. It takes time for him to see it and then verbalize the value of the speed, then tell you, and that takes time. In the mean time you've sped up way over the speed limit. By the time he tells you, your going too fast, but then you take your foot off the gas, then slow down too much by the time he tells you you are going too slow, then the process keeps repeating. Now if you know enough to limit your foot movement a little each time, it will be over damped so you'll eventually reach the right speed even if it takes 100 tries, but if you step too hard each time it will never damp out to a constant speed.
On the other hand, if your friend was using a computer connected to your brain that could also read the speedometer, the computer would tell you faster than the engine could respond to your foot. That would get the car to the right speed very very quickly before the car hit a curve in the road going too fast and flew off the road.
I am not sure what you intend to connect to this thing as load, but there are other ways to limit the current too that are much faster. There is a simpler way too but what i dont know is how accurate do you need to set the current limit on this thing. For example, if you set the current limit to 100ma, can you live with 90ma to 110ma, or how about 95ma to 105ma? If so there are faster more stable ways to do this that are guaranteed to work.
I also assumed that your max current would be about 120ma or so, is this correct?
Also, can you stand a little higher resistance for the current sense resistor (drop a little less than 1 volt at full current).
Also, does the current limit have to be adjustable, and if so, how fine tuned does the adjustment have to be?