Highly Precise Voltage Regulator

Why not just say 20mV instead of 2/100th’s of a volt? Also if it oscillates try adding some capacitance to the negative feedback. It will reduce the feedback for higher frequencies, but that’s the entire point really since those are the frequencies where the feedback creates oscillation. And if you don’t mind using OP-amps then they should be able to attain much higher precision (probably <1mV fluctuation for almost any reasonable load). Just make sure you provide a stable reference voltage that is resistant to power supply fluctuations. You could even use a separate, stable power supply for that purpose (e.g. alkaline battery).

Honestly I just phrased it that way cause I thought it might sound a little fancier at the time, I’m very familiar with scientific and engineering notation and use it on a regular, so I do indeed know the difference in case that’s what you were wondering. but I can change it back, doesn’t make much of a difference to me, just thought it might be more descriptive that way. Also, I didn’t realize OP Amps were that accurate… (does that really hold true regardless of what chip and load you use?)

It definitely was oscillating, I’m trying to see if there’s a way I can keep it clean DC without using such giant caps that the response to changes in the load is too awfully slow

I know NOTHING about MOSFETs, but could it be adjusted to prevent oscillating

You just need to find the oscillation frequency and put in a cap somewhere to attenuate the gain to less than 1 at that frequency. So if it oscillates at like 1MHz then even with the cap it’ll still have some negative feedback at frequencies up to 1MHz. To smooth ripple at frequencies too high for the damped negative feedback you can just add an output capacitor. Assuming a 1MHz ripple and a 10Ω load that’ll mean a capacitor somewhere around 100nF plus or minus a couple orders of magnitude. In terms of OP-amp accuracy, aside from bias it’s limited only by noise and gain and OP-amps have like 100kV/V gain so it only takes an input voltage difference of 100μV to swing 10V at the output, so noise is probably the dominant factor. But remember that OP-amps aren’t immune to oscillation either so they might need a capacitor or two too.

I’ve tried using a capacitor at multiple points that and it didn’t really stop the oscillations unless Use caps that are at least in the 10s of uF. @wyoelk, Unfortunately changing the mosfet itself cant really help here, and even if it did however, it would likely be at the sacrifice of performance. But @Jason9 knows much more about mosfets than I do I’m sure

Ok, I got to thinking a bit more creativity about it, and came up with what I think is a pretty good solution, and has been verified on my breadboard…. I decided to add a capacitor between the Mosfets gate, and the NPN transistor’s collector, my thinking behind why this specific spot is, to decrease the amount of negative feedback by instead introducing a little bit of positive feedback through the capacitor…. Now, many people might think this is not a good idea, because generally speaking, positive feedback is integrated into a design specifically to cause oscillations, however, that only works for harmonic oscillators and requires a specific phase shift (0°, 360°, 720°, etc.) in order to sustain oscillations; whereas relaxation (or negative feedback oscillators), require hysteresis as well as usually an upper and lower threshold. In doing so, placing this capacitor here should mean it can be smaller, and it shouldn’t affect the response time of the output voltage as much.

Well, what you want to do to eliminate oscillations is to make sure that at no frequency the feedback becomes positive with a gain greater than unity. If adding a positive feedback which then becomes negative at that frequency gets the job done, then great, but it could very well create a net positive feedback at some lower frequency instead. So just put a capacitor somewhere to reduce the feedback circuit’s gain at very high frequencies. One trick to do this is to put it between the collector and base of a transistor to amplify the capacitance via the miller effect to achieve greater dampening with a smaller capacitor. Of course here we’re not so much concerned with cap size, but it’s a great trick for ICs where die space is very limited. So don’t worry about cap size as long as it can kill the oscillations. The worst case scenario is that you need to add a 1mF cap or something to the output to smooth the high frequencies that the feedback can’t attenuate. Actually that alone might have chance of eliminating the oscillations.

Ok, doing some experimentation in the sim with the bode plot and stuff seems to show that the spot you chose for the cap is actually very good, but it probably doesn’t need to be as large as 470nF. Even as low as 10nF or even 1nF gets the job done for a low load resistance. The higher the load resistance the bigger the cap needs to be, but it probably doesn’t need to be as big as 470nF even for an open circuit, although there’s no need to make it any smaller.

Oh wow that’s cool, I seriously need to learn how to read a bode plot, how do you do that? …what all can you analyze with it, if you don’t mind explaining it?

Basically, you put in a frequency source somewhere and it shows you a plot of how strong that signal shows up and at what phase shift at some other point in the circuit and for all frequencies. So, you can put a frequency source at the feedback point (with proper DC bias so that the output is in the regulator’s range so that it is sensitive to small fluctuations at the feedback point) and then find the spot where the phase shift crosses some multiple of 360° (including 0° and negative multiples) and see what the amplitude of the signal is at that frequency. Assuming you put the frequency source at 1V (the 0dB point for EC) then if the amplitude is greater than 0dB that means the feedback is strong enough to oscillate but if it’s less than 0dB then it can’t oscillate. For this to work you have to disconnect the feedback from the output and connect the frequency source to the feedback, so you have to make sure that the impedance of the feedback circuitry (in this case ~50kΩ) is much larger than the load (1kΩ if I remember correctly) so that disconnecting the feedback like that doesn’t affect the output behavior too much. This means that you can’t really analyze it very well for high-impedance loads, but at 1kΩ the analysis works great. Also, in this case the the oscillations happen at the 0° line which is also the 0dB line and both curves go down with higher frequency so you just have to see which curve drops below that line first. If the blue one does, then great. If the grey one does, then you have oscillations.

This might be asking a lot, but do you think you could one day post a little bode plot playground or something, just to make it easer to see the proper setup for a few different conditions? I also still don’t yet feel that comfortable with dB, but I’m occasionally working on it

I’m not sure it’s about knowing the usual configurations really. Not for me at least. Like basically everything else in EC I learned it through experimentation, so for me it’s about understanding how it works and from there knowing how to apply it to any given circuit. And depending on what you want to measure, you may have to apply it in different ways to the same circuit. But a general rule of thumb is that if you have an amplifier or filter or other circuit with a distinct input and output, then it’s generally a good idea to put the frequency source at the input and measure at the output. This circuit here is actually an amplifier, just one with the output tied directly to the input for negative feedback. So, by disconnecting the feedback I turn it into a normal amplifier. For things like amplifiers you may need to pay attention to the DC bias. If the bias is too high or too low then the amplifier is in clipping range and a small change in the input will fail to make a large change in the output, so you have to make sure you bias it correctly so that it amplifies as it’s supposed to. In the case of this circuit, the bias range is very small, maybe even less than 1mV. Also, as I believe I mentioned some time before, the bode plot is a small signal analysis. This means that the bode plot measures the effect of infinitesimally small changes in the input and then scales that up linearly to match the actual size of the input. This means that even if an amplifier has an input range of +/- 2mV you can apply a 1V frequency source to that just fine as long as the DC bias is in that 4mV window. Because the bode plot assumes a small signal, it can ignore nonlinearities such as clipping and therefore tell you how strong a signal it would give if it didn’t have clipping. This means that it can give readings arbitrarily high (e.g. 100dB or 10kV) even if the power supply is only 10V. Anyway, as for reading dB, it’s a logarithmic measure of how far a signal is from 1V. -20dB is 100mV, -40dB is 10mV, -60dB is 1mV, and so on. Same in the other direction, with 20dB being 10V, 40dB being 100V, 60dB being 1kV, and so on. Also good to remember is that 10dB is ~3.16V and 2V is ~6.02dB. If you halve the decimal number, you take the square root of the voltage, so 3dB is about sqrt(2) volts, or about 1.41V. If you negate the decibel number then it takes the reciprocal of the voltage, so -3dB is the reciprocal of 1.41V or about 708mV. This means that if someone talks about a 3dB cutoff (or more accurately a -3dB cutoff) they mean the point where the output is only 70.8% of the nominal output. Anyway, just play around and experiment and stuff and you’ll get the hang of it with time. Try playing around and switching between bode and normal simulation to see how the bode plot relates to normal simulation. If you come across a reading you don’t understand then just switch to the normal simulation to see why it says that. For nonlinear circuits (most circuits containing semiconductors) make sure to keep the signal small too when using the normal simulation. And remember that the EC decibel system is relative to 1V and not to the amplitude of the input. It’s fairly natural to assume that 0dB means “unchanged” (after all, the decibel system is a relative scale) but that only applies in EC if the input source is 1V.

Ok thanks for explaining that more in depth, I’ll keep that in mind. Just the part where you mentioned you broke the negative feedback and placed the reference signal at the input to measure the output was very helpful, thank you