ADC Buffer Suggestions

Thread Starter

hardsoft

Joined Sep 7, 2009
13
I need to buffer a single sided signal feeding an ADC and would like to use a single sided op-amp at the same voltage of the ADC to protect it from over/under-voltage.

I need to be able to read the signal right down to 0V (upper rail not so important), and I'm afraid a rail to rail op-amp won't cut it as they usually can only get down to 20-50mV off ground.

One thing I was thinking was putting a diode on the output of the op-amp with a weak pull-down resistor and connecting the feedback after the diode to keep the op-amp output around 0.6V off ground to get 0V on the input of the ADC.

Another solution would be to use a difference circuit and subtract the input from a reference voltage so that zero on the input would be somewhere easily within the output range of the op-amp. My worry with this is that the input signal (which is coming from the outside world) won't see ground. It will be a floating voltage and I'm not sure I should assume outside signals will be able to work with that.

Any other suggestions that don't involve using a two rail op-amp?

Any help greatly appreciated!
 

MikeML

Joined Oct 2, 2009
5,444
Try just putting a 4.7K pull-down resistor from the output of your buffer opamp to ground. It'll burn more power, but should pull the output all the way to ground.
 

Thread Starter

hardsoft

Joined Sep 7, 2009
13
Try just putting a 4.7K pull-down resistor from the output of your buffer opamp to ground. It'll burn more power, but should pull the output all the way to ground.
I'm not so sure about that. According to the data sheet increased output current decreases the output swing, or pushes it further from ground.
 

MikeML

Joined Oct 2, 2009
5,444
I'm not so sure about that. According to the data sheet increased output current decreases the output swing, or pushes it further from ground.
Yes, but the "how low the output pin can go vs current" spec assumes the current flow is IN TO the output pin. With the pull-down resistor connected as I suggested, the current flow is OUT OF the output pin, which pulls the pin lower than it would be if ZERO current flows IN/OUT of the pin.
 
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