Problem using Opamp as buffer with single supply

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savvej

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I want to use OP-amp(UA-741 from texas instruments) as a buffer to read voltage from a potential divider network(using a potentiometer).
I am using a supply of 5V (single supply).Now this is what happens:
Initially when the output of potentiometer is zero the opamp output reads 1.85 volts instead of zero volts.Then I increase the input,the output starts following input after the input crosses 1.85 volts(at around 1.89 volts).Why is it happening so?I want to use it a as buffer in the range of 0-5 volts and using 5v supply only.Please help..
 
Read the datasheet for the 741 op-amp. It is not a rail-to-rail output amplifier. The output will only swing within about 1.5V from each supply rail. Either use a larger range of supply voltage or use a rail-to-rail op-amp.
 
savvej said:
kk.so using 9V instead(single supply),the output should then follow the input?
Click to expand...

hi,
You require two 9V batteries, one for +9V and one for the -9V supply to the 741, if you want the output to range from 0V to +5V
 
kk...can i use max 232 for providing supply to opamp?
I have done it that way before, you will around +/-9V
 
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Take a look at the amount of current the 741 draws, and if it's less then the current the max232 can give, the 741 will work properly.

kind regards
 
Cant be much current from 10uf caps!! BUT if it works let us know (handy little tip)
 
Use LM358 & supply it with a 9V single supply.No need dual supply.

Rail to rail op amps are harder to find.
 
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Check my tutorials for the way I solved the problem - but for a start don't use a 40 year old antique opamp.

I didn't even consider 'rail to rail' opamps, simply because they aren't what they say, they are only 'near' to the rails - my solution gives from exactly 0V to whatever you want.
 
Thanks @ Nigel and @be80be.
I used IC 555 along with diodes and capacitor to generate -ve supply though.
yet to try if max232 works.Hopefully it must.

@colin55 Your method is simplest .Will surely try it.And actually it serves my pupose directly as I need to clip of any negative voltage present in the input too.Though there will be a loss of 0.7 V across B-E.Thanks.BTW this is what I am exactly doing and needed a buffer to isolate the transformer output directly to uC and to supply greater current.I used a battery of 1.5V to level shift the sine wave which would'nt be needed in your case.Lets see how it works.The thread is here (though I removed buffer latter) :https://www.electro-tech-online.com...ormer-as-sine-wave-source.121343/#post1000410
 
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