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How to use MCU on-chip ADC to measure negtive voltage?

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bittware

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Hello experts,
Could you give some hints on how to use MCU on-chip ADC to measure negtive voltage? In fact, I want to use ADC to monitor the negtive power rail output.

Thanks.
 

could u specify the MCU type u are using :)

thanks
 

Hi, Use a current mirror that is powered from the negative powersupply.

Paul.
 

Simplest - put resistive voltage divider with one side connected to stable voltage source second to your negative voltage output and middle to adc input directly or thorugh buffer .
Choose division factor to based on min negative voltage and voltage source resistors connected to .
 

artem,

Yep, that's what I was going to suggest.

V+ ---- R1 ----- ADCin ----- R2 ------- V-

Then just adjust the divider ratio.

To the asker:

Vx = (Rx/Rt)Vt

so (...where's the "therefore" character thingy??)

VADCin - V- = (R2/(R1+R2))(V+ - V-)

An example, using a +5v and -5v rail:

+5v ----- R ----- tap ----- R ----- -5v

Vtap - -5v = (R/(R+R))(5v - (-5v)) = (R/2R)(5v + 5v) = (1/2)(10v) = 5v

Solvng for Vtap,

Vtap - -5v = 5v

Vtap +5v = 5v

Vtap = 5v - 5v = 0v

And this makes sense, since 5v are dropping across R (the right side of the equation) and the total supply was +5 --> -5v, or 10v.

So, you can see that you only need decrease the R to the +v supply to send Vtap positive or decrease the R to the negative supply to send Vtap negative. Voltage dividers are neato. ...at least I think.

Oh, and what value is R? Start with 10k and 10k or 100k and 100k. That'll give you 0v, and then you can lower (or raise of course) one of them to get whatever middle voltage you want.
 

Hi guys,
What if the positive voltage supply collapses to zero under fault condition? The negtive voltage will damage the ADC input.
 

Hi
you must use a positive offset voltage to convert your input to positive voltage.
regards
 

It is always a good idea to add a couple of protective diodes per I or O (I would also add a small, serial resistor working as a fuse) .. see picture below ..

Regards,
IanP
 

    bittware

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