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amplifying dc signal for measuring

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Junior Member level 2
Mar 11, 2009
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I have a low current circuit where I want to log the current every 10s for a long period of time

The current jumps from about 100uA to 25mA.

I have a data logger that can log from 0 to 20 V, so I can add a 1 ohm resistor with the volt logger to measure current but it needs to be much higher.

If I amplify the current with a gain stage of 200 from 25mA to 5A, how can I revert back to normal operating current of the circuit? do I need a buffer stage?

My focus is on being able to read when the 25mA signal is on with the datta logger, the 100uA can be shown as 0 and does not matter

Input appreciated

You can use a differential amplifier assuming that the input voltage is in the opamp range

The gain is set by Rf/R1

or you can use an Instrumentation Amplifier like AD620

You can use a shunt resistor that creates a voltage drop with the current and then use any of the above to amplify the voltage

Thanks for the reply

Ok, so I can use the op amp to make the gain into something noticeable for the datta logger.

I am trying to measure how many times a battery powered device goes into sleep mode (100uA) and transmit mode (25mA) over a long period of time with the data logger. So once I amplify the signal to eg 5V how can I lower it back to the 25mV range in order for the circuit not to get damaged?

This is what I am thinking of
**broken link removed**

I only have access to the two battery terminals which receive a constant voltage under normal operation.

So once I amplify the signal to eg 5V how can I lower it back to the 25mV range in order for the circuit not to get damaged?
What will be damaged?

For example with the 1ohm resistor and a differential amplifier with a gain of 50 you can get 0.025*1R=0.025*50=1.25v
this can be easily detected from your logger.


I mean something like

assuming that the battery - is also the ground
I was thinking using a low input offset comparator - if you just need to know low (uA) or high (25 mA) levels.

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