On 20 April 2011 13:04, V G wrote: > Hey all, > > I made a constant current source (for battery drain testing, charging, > pretty much any application). But there's so many variables that I don't > know how to calculate for it. So many of the components influence the > current through Q1 (power BJT) that I don't know how to select the values > and which one to vary to get the widest current selection range. > > Circuit attached. > > For the circuit above, I guessed the values and selected about a 1A curre= nt > source. > > -- > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist > > Just a couple of cents worth. The amplifier has a gain of 3 so to just turn the transistor on will mean that the voltage drop across the 0.39 resistor is about 0.2V, or a current of just over 1/2 amp. So you can either change the 0.39ohm resistor or play with the gain of the amplifier. I'd be tempted to do both - reduce the 0.39 resistor somewhat & increase the amplifier gain - and add a pot. for trimming the value. I don't know the details of the opamp you are using but presumably it will pull down to 0V OK (or very close to it)? Otherwise you could add a voltage divider at the output of the amplifier to reduce the voltage to the threashold point. Increase the amplifier gain to compensate. Effectively just add a resistor b-e on Q3. (~10k ??) Note that the current gain around the circuit is very high so you may get some instability. A capacitor in the right place (eg in parallel with the feedback resistor) might assist. The amplifier resistors at 1k and 2k are quite low values. If you go to higher values then if you do find you require parallel capacitance for stability you might be able to use more convienient (lower) values. RP --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .