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I have a few UA741 op-amps which I need to use in differential op-amp configurations with Vref set so as to bias the input (2.5V AC pk-pk @ 50Hz) to a baseline of 2.5V. The output is to be connected to an ADC.

I have a 12V DC supply available for the supply rails.

I need the output to be able to reach at least 5V, therefore I want the positive rail to be around 7-8V. That leaves me with 4-5V on the negative rail. As stated in the datasheet, the minimum recommended negative supply voltage is 5V.

I have two questions:

  1. How much voltage should be supplied to the positive rail to ensure that the op-amp can output 5V? How can I find this value from the datasheet?
  2. Will there be any significant damage to the op-amp if I input less than the recommended voltage for the negative rail? (I don't need negative output)
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  • \$\begingroup\$ It will probably work fine. But to get a better answer, please specify the input frequency range, the gain of the circuit you are implementing, whether it is inverting or non-inverting, and the load impedance. Or you could just draw a schematic. If you don't know some of these things, then maybe try a general description of where the input signal comes from and where the output signal is going to. \$\endgroup\$ – mkeith Mar 11 '16 at 3:47
  • \$\begingroup\$ Sorry about that. I have provided the details. \$\endgroup\$ – Hassaan Mar 11 '16 at 4:01
  • \$\begingroup\$ It would also help to know the signal bandwidth. In other words, the frequency content of the input signal. \$\endgroup\$ – mkeith Mar 11 '16 at 4:30
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This is not generally a good idea (death to your ADC) so let's walk through the problem.

Looking at the DS on page 5: -

  • For a positive rail of 15 volts the maximum guaranteed output is +12 volts
  • But typically it could be +14 volts

Let's say you decide that the +12 V figure is best (you are guaranteed no more than a 3 V drop from the positive rail). You would use a +8 V rail and be able to generate a peak signal of +5 V.

Now, you build the circuit, put an input on the 741 and destroy the ADC. Why does it destroy the ADC you might wonder - it's because the 741 is able to provide between +5 V (guaranteed) and +7 V (typical) to the ADC and +7 V is likely to exceed the maximum rating on input voltages that your ADC can safely handle. No matter how careful you are at putting the right sized signal into the 741, you will eventually destroy the ADC (unless it has good input overload specs). You might even destroy the ADC in the first milli second you apply power to the (really crappy) 741.

The proper way to do this is use a rail-to-rail op-amp driving the ADC. Yes you lose maybe 50 mV at the top (+5 V) and 50 mV at the bottom (0 V) leaving you with a span of 4.9 volts but this is the price you must pay.

Same problem with the negative rail. Yes you can find a negative rail to work for your op-amp but you are going to likely destroy the ADC with excessive negative voltages.

From a different angle, try looking at the ADC data sheet and see what it says the usable range is - there may be a zero offset figure of several millivolts and this tells you that the lowest several mV of the ADC input is likely unusable. There may also be an ADC gain tolerance that prevents you using to top several mV so, even if you had the perfect input signal to the ADC, you can never guarantee that you can use the whole range.

How big a deal is this? OK so you limit your input range to say 4.8 volts out of 5 volts - that's 96% coverage. Let's say your ADC is 12 bit. Your coverage is 96% of 4096 = 3932 LSbs. Take the log of 3932 and divide by log2 to get the actual usable number of bits i.e. 11.94. Not much of a compromise is it?

If you look at most ADC specs, you will find that they specify SNR (signal to noise ratio or other AC performances) with an input signal that is typically 0.5 dB down on theoretical full-scale. How much is 0.5 dB down in real numbers for a theoretical 5V span?

It's 94.4% so don't get hung up on this and don't try overdriving your ADC.

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  • \$\begingroup\$ Thank you for your excellent detailed answer - I wish I could give you more than 1 upvote! \$\endgroup\$ – Hassaan Mar 11 '16 at 11:39
  • \$\begingroup\$ However, I have a question: wouldn't negative feedback in the differential amplifier configuration prevent the output of the op-amp from reaching greater than 5V if the correct combination of resistors is used? \$\endgroup\$ – Hassaan Mar 11 '16 at 11:41
  • \$\begingroup\$ Neg feedback dictates gain and won't prevent an over voltage situation. \$\endgroup\$ – Andy aka Mar 11 '16 at 12:30
  • \$\begingroup\$ What would you suggest to give over-voltage protection that doesn't interfere with voltage output of this circuit in the desired range (0-5V)? \$\endgroup\$ – Hassaan Mar 11 '16 at 13:06
  • \$\begingroup\$ As suggested in my answer, use a rail to rail opamp powered from 0 volts and 5 volts. You might get away with a series resistor of several k ohms in line with the adc input but this might cause other problems if your sampling rate is high. \$\endgroup\$ – Andy aka Mar 11 '16 at 13:26
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If you put 12 volts across the positive and negative supply inputs of the op amp, it's going to operate the same regardless of whether you choose to call those values +6 and -6, +12 and 0, or +8 and -4. The 741 can only output about 3 volts from each rail. (You can find this in the data sheet by noting that the maximum output voltage with a +15V positive supply is +12V, and similarly on the negative side.) You said you don't need negative output, so splitting your supply into +8V and -4V, or +9V and -3V, would give you the 0-5V output range you need.

Or get a different op amp. The TI 2471/2/4 for example will run on a 5V single supply and can output within a few tens of millivolts of both rails.

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  • \$\begingroup\$ How close it gets to the rail depends critically on the load resistance. The ADC input will likely be over 100k. So I believe it will work. However it also depends on frequency. See figure 4 and figure 5 in the datasheet. \$\endgroup\$ – mkeith Mar 11 '16 at 4:27
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    \$\begingroup\$ The output range may decrease as the load resistance decreases. Don't see a problem with 100k. He says 50Hz so no problems there. \$\endgroup\$ – Willis Blackburn Mar 11 '16 at 4:42
  • \$\begingroup\$ Oh, I missed the 50 Hz. \$\endgroup\$ – mkeith Mar 11 '16 at 4:51
  • \$\begingroup\$ It seems like a kind of boring signal. :-) \$\endgroup\$ – Willis Blackburn Mar 11 '16 at 4:53
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    \$\begingroup\$ I was being conservative and not getting into all those details because you only need 0-5V and you can get it with a 12V supply even if you allow for 3V on either side. \$\endgroup\$ – Willis Blackburn Mar 11 '16 at 12:20

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