5 volt success on the breadboardPosted: 2005-August-4
I got around to the basics of the circuit below. I set up the MOSFET in the configuration below and drove it from the op-amp. I added the PNP in a similar configuration to ensure that maximum current would get to the gate — I need to get that gate to 12 volts, and the 20 mA output on the op-amp wouldn't cut it. I got nice square square-waves but, using a simple pulse-width modulation to the capacitive output, I was getting the same problem of current-in equals current-out. I set up the buck configuration again and finally got some success: 6.6 watts in and 4.2 watts out for an efficiency of 63%.
At this point I realized the output-as-driven would end up as 5-volts from the positive battery rail. I remembered having 7905 negative-rail 5-volt regulators around (and always thinking, "what the hell will I ever use these for?") I switched to the terminology where the positive battery terminal was ground (calling it 0-volts) and the negative rail was -12 volts. I spent some time diagramming the circuit (using CADintosh from Lemke Software, GMBH.) I went back and set things up like I had drawn, made a few changes, and did some final tests on the breadboard. By varying resistors, and using the 51-ohm load I got 97mA out (4.95 volts at 470 mW) with 52 mA at the input (624 mW) for an efficiency of 75%. I tried the 8-ohm load and got 620mA out (4.96 volts at 3.08 watts) with an input current of 419 mA (5.03 watts) which gave me an efficiency of 61%. I think this looks pretty good.
So, after three days of slaving for a total of 20 hours or so, here's the circuit …
You can look up explanations for a twin-T oscillator and a buck converter for the oscillator and the inductor on the Internet, but one thing that I don't think is too obvious is that the diode after the op-amp is there so the base of the PNP transistor can actually go to the rail — the op-amp may not reach it, but the diode will be shut off so the 470-ohm resistor will shut the transistor down. The other thing is that the inductor has no value. I don't know what it is: it's a yellow torroid with red magnet wire that I pulled out of a dead computer UPS, so I don't know its value, but of the ones I had lying around, it worked best. I guess I should put parallel lines below it because it's iron-core … oh well.
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