Mesuring current on negitive rail

[markdem]

Hello,

I am tiring to get a voltage and current logger for my power supply, which has a positive and negative output. I am guessing the easiest way to do this is using an ADC and a PIC to a PC.
I have the following questions that I can’t seem to find answers for;
1. The power supplies output is 1-30VDC. Measuring the voltage is easy. I just used a resister divider to scale the voltage and measured it with an ADC. When I am measuring current, I will use a shunt (.1 10w resister). I will then use an op-amp to make the output single ended to feed to the ADC. If the voltage is 30v from the power supply, will I need to put it thought a resister divider before I feed it to the op-amp? From what I have read, the max voltage on the op-amp input (TL081) must be lower than the op amp supply voltage (15v). Is this correct, or can I just put the current sense resister straight onto the op-amp?
2. On the negative rail, I am thinking of just using an op-amp to invert the voltage for the ADC for voltage measurement. Is this an OK way of going about it?
3. Measuring the current on the negative rail has got me completely stumped. Should I first invert the voltage across the current shunt, and then use another op-amp to provide a single ended output for the ADC? Is there any other\better ways of doing it?

I have included a diagram of the current measuring circuits. The LEDs are just there to provide load for testing.
I don’t need any crazy accuracy from the measurements (.01 volts and 1 mA will done fine). I also don’t want to spend big $$ on in-amps and IC if I can get away with it.
Thanks for the help.
Mark

 

[Blars]

This has similarities to a project I'm thinking about. The op-amp only cares about the voltage on its inputs, but not all can handle rail-to-rail inputs. You may need two stages: first a diferencial amplifyer with a gain chosen to avoid exceeding the opamp's ratings, (You may need a gain of less than 0.5) followed by an amplifier stage to get a good range for the a-d converter. Unless you can get the resistor in the supplies feedback loop, it will make the voltage regulation on the load less acurate. If you don't have a common ground to worry about, it may be easier to put the sensing resistor in the negitive lead.

 

[markdem]

Hi Blars, thanks for the reply.

I think there is something I don't' understand here. With a op-amps limits in the datasheet, are the input voltage limits defined for each pin in reference tot he supply, or is it the differential input. In other words, the voltage across my current sense resistor will be a max of 300mV, but the voltage running thought the resistor will be a max of 30V. The rated max voltage of a TL081 is 15V. I am just not sure if I can connect the op amp directly to the shunt, or do I need to scale the voltage first.
I hope that makes sense.

Thanks

Mark

 

[Blars]

The op-amp limits are relitive to it's supply. See Operational amplifier applications - Wikipedia, the free encyclopedia, R1/Rf and R2/Rg form voltage dividers. When the gain is low enough, V1 and V2 can be quite a ways out of the opamp's supply voltage. It might be a good idea to add protection diodes from the inputs of the opamp to the supplies to handle startup conditions. Note that the ground side of Rg may be connected to a reference voltage to add an offset to the difference voltage.

 

[MikeMl]

Have you seen the High side/low side current monitor chips made by TI and Zetex (e.g. ZXCT1009).

You will need to match the resistors in the four-arm diff amp to better than 0.1% (hard to do).

 

[markdem]

Blars, After writing my last post, I decided to go and try what I want to do. Whats 30 cents for a TL081 anyway. I now understand the point about the input voltage, once I relised that it is not referenced to the supply, it all made sense.

Mike, That is what I am looking for. I did find some ICs like that form Analog, but they where way to expensive, so I decided to try with op amps. I have now found INA139 will do what I need, and the cost is good too.

Thanks for the help.

Mark