| Electronics > Beginners |
| how does blackdog's PSU work? |
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| Cliff Matthews:
--- Quote from: blackdog on March 19, 2018, 10:09:06 am ---..There are many measurements and information about this power supply on this page, use google translate is you cant read Dutch. https://www.circuitsonline.net/forum/view/110029 Kind regards Bram --- End quote --- Thanks Bram! I hope you don't mind it all complied into an English PDF? I did a Google translate on the whole thread and pasted it to a 118meg PDF: https://drive.google.com/open?id=1cEGFsj6kk3hI8KUnBXqF0GFBy2YFkUCL There were 35 linked images that were too small to be usable, they are here: https://drive.google.com/open?id=1iKNivHfgNfMc6xBsww707_LqQOR2TIin |
| T3sl4co1l:
--- Quote from: blackdog on March 19, 2018, 12:13:18 pm ---Yes an no, a part of the ennergy is comming from the output capacitors, and this picture is from one of my design stages. And if i want to test at a 9,5-Ampere level or even 15-Amp to see how the output capacitors are behaving during dynamic testing, i will do that. :) Kind regards, Bram --- End quote --- A part, yes. As you should always ask: "How much?" We can use the capacitor equation to find out. C is about 200uF, and pulse duration is 200us. Therefore: I = C * dV/dt (9.5A) = (200uF) * dV / (200us) 9.5V = dV The waveform shows less than 10mV of dV (100mV if it was 10x probe, I have no idea if that was accounted for or not), so this cannot be. Indeed, since the measured dV is nearly zero, we can say with quite good certainty that nearly zero energy is exchanged with the capacitors! Such is the begrudging truth about capacitors -- we almost always use them for dynamical stability, not for energy storage. Energy storage is only relevant when the change in voltage is large. --- Quote from: blackdog on March 19, 2018, 10:09:06 am ---Dynamic performance Tim explane, why i may not show a picture of the performance, at a peak current of 9.5-Ampere? And do you realy think i'am that stupid, that i measure the performance of this power supply at the dummy load input? This is one of my tests of the power supply, how it would reacts in a practical situation. --- End quote --- Please correct me if I am wrong. It doesn't seem to be a translation error; you specified a cable length and distant load capacitance, introducing dynamics other than the supply itself. :-// Tim |
| blackdog:
Hi Tim, 1e Can you explain to us, why you have such a negetive attitude? 2e And again you give the impression that I am a liar, 10mV/Div you know for sure that you're not lying ... 3e And of course I can not do a practical test, because according to your standards, you do not test a power supply in this way. 4e So ... if you do not show better behavior, I will not give you any answer anymore. Kind regards, Bram |
| Kleinstein:
The load test at 9 A is rather short - it's only 200 µs. The current limiting needs some time to respond and in this example it looks like the 200 µs are shorter than the response time of the current regulation. It depends on the application how much delay is good. In some cases a faster reaction might be required, but in many cases a slightly delayed response of the current limit is actually a good thing as the main application is as a voltage source. The LT1021 might be a bit of an overkill for a power-supply, especially if a NE5532 OP is used. For most cases I would consider something like 2xTL431 or an LM329 sufficient. It is possible to set the voltage via an DAC. In this case P2 is replaced by a fixed resistor and the reference voltage is supplied by the DAC. This way of adjustment also has an advantage for the control loop. It makes the control loop independent of the set voltage and thus gives very similar dynamic response at all voltages. In the current version with P2 to set the voltage, the response can be a bit faster at low voltages and slower at high voltage. This mainly effects the low kHz frequency range and might effect the performance for the case of a very large load capacitance. At a very low voltage setting there might be a slight chance to have instability with a lot of low ESR capacitance (e.g. something like 10000 µF of tantalum caps). In the extreme cases with the very fast loop also parasitic inductance of the layout and shunt can have an effect (both positive (especially for current control) and negative). The case of large, low ESR capacitance is kind of the worst case for a voltage regulator and some ringing is normal in that case. One can not expect to reach no ringing in that case - already having it not oscillating with any passive (that is RCL type) load is good. A response with little ringing is only feasible and required for a much more limited load range. One should do tests for both cases: a difficult load (where quite some ringing is acceptable) and the more easy ones that should not show much ringing. |
| Cerebus:
--- Quote from: blackdog on March 19, 2018, 04:46:14 pm ---Hi Tim, 1e Can you explain to us, why you have such a negetive attitude? --- End quote --- He doesn't, you just seem to react badly to any kind of criticism. Much of what Tim has had to say about the design is positive, but you choose to ignore that and over-react to the questions and criticisms. Vis: --- Quote from: T3sl4co1l on March 18, 2018, 09:36:53 pm ---Anyway: this is the only amateur linear power supply design I've seen that is actually reliable, and working, as shown. ... Overall, small complaints, and simple tweaks. This is a good design. --- End quote --- My experience of him tells me that Tim is both knowledgable and experienced, and generally quick to help. You can do one of two things, you can take advantage of Tim's interest and possibly improve the design or you can take offence, the choice is yours. |
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