Power supply display problem - Page 1
Forum section: electronics / projects

author: blackman
author_rank: Contributor
timestamp: January 14, 2012, 12:18:33 pm
content: Hello everyone.Watching dave build his PSU has pushed me to finish mine. Its a very basic crude LM317 Voltage regulated supply.It has current measurement and this is where im having a little dilemma. Im using a 1ohm current sense resistor before the V Reg so that my load regulation is good over the 2A output range. But the V Reg circuit is drawing 29mA so my current display is obviously reading 29mA at no load.I didn't want to put the resistor after the V reg circuit because I would have poor load regulation and I only have a 1ohm resistor handy. I was looking at some way of putting a 29mV offset on the display to zero the output but I dont have another supply in the box to feed it from. It is either that or I get/make a 0.1ohm current sense resistor which will be just enough to drive the display, if I go any lower ill need to amplify the sense voltage to drive the display. 0-200mV display. But at 0.1ohm I will still get 200mV drop at full current, which isn't too bad, but if I can go better with out too much problem I would like to.Thanks in advance people.

author: Rufus
author_rank: Super Contributor
timestamp: January 14, 2012, 01:23:39 pm
content: Quote from: blackman on January 14, 2012, 12:18:33 pmReducing the current sense resistor value will have no effect on the offset.A voltage divider from the +ve side of the sense resistor to ground could give you a reference voltage offset by 29mV. A constant current sink and resistor would give an offset independent of the supply voltage.Another LM317 with the same ADJ pin resistor and 1 ohm would do it.

author: blackman
author_rank: Contributor
timestamp: January 14, 2012, 06:15:42 pm
content: Thanks for the reply. I was thinking about reducing the sense resistor and moving it to the other side of the V Reg as well, there for eliminating the current draw from the V Reg.If I had a voltage divider from the +ve side of the sense resistor to ground wouldn't that only chance the span? I don't follow sorry. I can see how the constant current sink and a resistor raising the -ve terminal on the display would work.Thanks again.

author: Rufus
author_rank: Super Contributor
timestamp: January 14, 2012, 07:06:24 pm
content: Quote from: blackman on January 14, 2012, 06:15:42 pmI think (hope) you mean lower the +ve terminal.A resistor with a constant voltage across it is a constant current sink. If your supply voltage to the current sense resistor is fairly constant you can get away with voltage divider.If for example your supply is 20v a 16R and 11k divider to ground will give you 20v less 29mV, the same voltage as on the other end of the sense resistor with your 29mA regulator current draw.

author: IanB
author_rank: Super Contributor
timestamp: January 14, 2012, 08:36:17 pm
content: I was mulling over the same problem only yesterday.Here is a circuit diagram that illustrates what Rufus is saying:You need to subtract the 29 mV from the meter reading to make it read zero when there is no external current draw. In the circuit above this is done with a 10 mA constant current sink and a 2.9 ohm resistor. You can obviously adjust the values in any combination to get the same result.

author: blackman
author_rank: Contributor
timestamp: January 18, 2012, 10:42:36 am
content: Ahh I see now. Yes I was confused before in thinking that lowering the -ve rail would help, it would only double my problem. I really need to start drawing things and getting my head around them that way.I might have to use another transformer to get the isolation and using a voltage divider, as I think if I put the constant current sink in it will have problems with the ground reference.Thanks again for your help guys.