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[Question]0 -12V supply for op amp

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yes123

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Dear all,

I am new to this forum.
Recently i want to do a simple application which convert 0-3.6V into 0-12V analog output. So i use a simple Non-inverting amplifier to do that.
**broken link removed**
which R2 = 500R, R1=220R.

Theoretically, it should be working fine. But now there is a problem, my system only have 12V supply, there is no -12V for op- amp to function properly (No more space to fit in another negative voltage converter ic too).

So how to deal with this? any op-amp IC can support 0-12V power input, and still working fine? I tried basic 741, TL081... all doesn't work in 0-12V power range.

Thankyou all.
 
Use a single supply op-amp such as the LM358 or MC33171, the latter being better.
 
Use a single supply op-amp such as the LM358 or MC33171, the latter being better.

hi hero,
The OP is after a opa that gives Vout=12V from a 12V power rail.
 
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I will try that out, thanks for all your reply

hi,
If you look at the datasheet for these two OPA's you will see that with a 12V supply rail, the maximum Vout will be approx 10v.:)
 
Sorry missed that, he needs a rail to rail opamp.

You need a CMOS output op-amp, try the CA3160.
 
The gain is (500/220) + 1= 3.27.
He has an input up to 3.6V so the output should be as high as 11.78V.

But most opamps cannot drive feedback resistors as low as 500 + 220= 770 ohms because the output current into them would be 15.3mA.

The feedback resistors should be 500k ohms and 220k ohms.
 
You should use 560k and 240k or 910k and 390k which should give you a gain of exactly 3 1/3
 
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Sorry missed that, he needs a rail to rail opamp.

You need a CMOS output op-amp, try the CA3160.

So i should be looking for monolithic cmos ouput op-amp?
Monolithic means single supply?

CA3160 is obsolete product according to Farnel, any other better suggestion?

Thankyou.
 
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The gain is (500/220) + 1= 3.27.
He has an input up to 3.6V so the output should be as high as 11.78V.

But most opamps cannot drive feedback resistors as low as 500 + 220= 770 ohms because the output current into them would be 15.3mA.

The feedback resistors should be 500k ohms and 220k ohms.

Thanks for your remind,
But how can 500 + 220 become 770? shouldn't it be 720?
How you get the value of 15.3mA? 12V/770R should be 15.6mA
 
The buttons on my calculator get stuck so some numbers are wrong.
An opamp can barely drive such low value resistors because their current is too high. Use resistors with values that are 1000 times higher.
 
The buttons on my calculator get stuck so some numbers are wrong.
An opamp can barely drive such low value resistors because their current is too high. Use resistors with values that are 1000 times higher.

But as far as i know, opamp should theoritically having very high input impedance (nearly infinite). So due to its high input resistance, it should automatically limiting its input current in this case, am i right?

Why do we still need to bother its input current in this case?
 
But as far as i know, opamp should theoritically having very high input impedance (nearly infinite). So due to its high input resistance, it should automatically limiting its input current in this case, am i right?

Why do we still need to bother its input current in this case?
Your feedback resistors have such a low value that the current in them is too much for the output of the opamp to drive. You don't need a very high current in the feedback resistors because the input current of the opamp is extremely low.

Anyway, i just found another simplier solution in my textbook.
We can just easily add 2 more same value resistor to Vin+, this will do the job fine.
It is a different circuit. It has a voltage gain of only 1 and its output DC voltage is always 6.0V. Your circuit has a voltage gain of 3.27 and its output DC voltage should be variable.
 
All you figured out was how to bias the opamp when running it on a single supply. This connection will do nothing to solve the underlying problem that only a few opamps are capable of pulling their output pin nearly all the way to the same voltage as their positive supply pin is tied to. Several posters told you this, but so far you have missed it!

Go read the Voh spec on several opamp data sheets. All opamps are tested to determine the maximum voltage that they will pull their output pin to. It is usually specified as Vpos - Δ, where Δ is 2V for most opamps. In other words, regardless of how you bias an opamp, when operated off a 12V supply, it can never pull its output pin higher than about 10v!!!

There are some CMOS "rail-to-rail output" opamps which are cabable of pulling their output pin within a few mV of the postive supply PROVIDED THAT VERY LITTLE CURRENT IS BEING DRAWN FROM THEIR OUTPUT PIN. This is why you were told to raise the impedance of the feedback resistors to the fractions of megΩ.
 
So i should be looking for monolithic cmos ouput op-amp?
Monolithic means single supply?

CA3160 is obsolete product according to Farnel, any other better suggestion?

Thankyou.
Monolithic just means it's all etched onto one chip of silicon.

You need a rail-to-rail op-amp.

A rail-to-rail op-amp normally has a CMOS output stage which has a very low saturation voltage at low output currents.

Try OP295.
 
All you figured out was how to bias the opamp when running it on a single supply. This connection will do nothing to solve the underlying problem that only a few opamps are capable of pulling their output pin nearly all the way to the same voltage as their positive supply pin is tied to. Several posters told you this, but so far you have missed it!

Go read the Voh spec on several opamp data sheets. All opamps are tested to determine the maximum voltage that they will pull their output pin to. It is usually specified as Vpos - Δ, where Δ is 2V for most opamps. In other words, regardless of how you bias an opamp, when operated off a 12V supply, it can never pull its output pin higher than about 10v!!!

There are some CMOS "rail-to-rail output" opamps which are cabable of pulling their output pin within a few mV of the postive supply PROVIDED THAT VERY LITTLE CURRENT IS BEING DRAWN FROM THEIR OUTPUT PIN. This is why you were told to raise the impedance of the feedback resistors to the fractions of megΩ.


Thanks for your advise. I know i need rail-to-rail, i just don't understand why need of high value of feedback resistor since the input pin already having high impedenace by itself, so it should automatically lower the input feedback current, right?

Hero999 said:
Monolithic just means it's all etched onto one chip of silicon.
You need a rail-to-rail op-amp.
A rail-to-rail op-amp normally has a CMOS output stage which has a very low saturation voltage at low output currents.
Try OP295.
speakerguy79 said:
Order this:
Precision Amplifiers - Wide Bandwidth - OPA2725 - TI.com

Thanks for your advise , i will check them out.
 
I know i need rail-to-rail, i just don't understand why need of high value of feedback resistor since the input pin already having high impedenace by itself, so it should automatically lower the input feedback current, right?
No.
The current in the feedback resistors is not reduced by the input of the opamp. The output of the opamp must try to drive the high current in the low value feedback resistors but can't. If the output is +12V then the current in the 720 ohms of feedback resistors is 16.7mA which is too high to be driven from the output of an opamp and therefore the output voltage of the opamp will be reduced.

If you use 720k worth of feedback resistors then when the output is 12V the current in the feedback resistors is only 16.7uA which is easier for the output of the opamp to drive. Then the output of the opamp will be closer to +12V.

The input current of opamps is almost nothing. Some opamps have FET inputs with virtually no input current.
 
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