reasonable resistors for inverting opamp

Discussion in 'The Projects Forum' started by stirling, Jul 27, 2010.

  1. stirling

    Thread Starter Member

    Mar 11, 2010
    52
    2
    I read that the values for the input and feedback resistors should be kept to "reasonable" values but nowhere have I found a definition of "reasonable". My required gain is around 0.7 and I'm simulating with 680K and 470K in order to give me my required input impedance. Are these values "reasonable" in the real world? (hope this makes sense - I'm no electronics guru!!!)
     
  2. tom66

    Senior Member

    May 9, 2009
    2,613
    214
    It depends.

    For a precision application these are way too high. If you consider a typical op-amp, say one of the op-amps in the LM324, the bias current is 45nA and I'm assuming your 680k resistor is connected to the input signal, this practically means that you get an additional 30.6 mV(!) on top of your 2mV input offset... it wouldn't matter if you chose a virtually zero input offset op-amp with these values.

    But, for a typical audio level application, it is not so clear cut. For example if you are dealing with a 2Vp-p signal then 30.6mV is too small to really be noticed. But if you have a 20mVp-p signal, the 30.6mV offset can overwhelm it.

    Reasonable for an op-amp usually means less than 1 megohm. If you chose 6.8M for example, you would have a massive 0.306 volts!
     
  3. stirling

    Thread Starter Member

    Mar 11, 2010
    52
    2
    tom66 - thanks. I don't think I'd regard it as a precision application. I basically need to take a 0 to (minus)7 Vdc output of an existing (unchangeable) potential divider and convert it into 0 to 5Vdc for input to a atmel mcu ADC. I've yet to get the piece of kit that outputs the 0 to -7Vdc so I don't know what the output impedance is so I'm just playing with ltspice and trying to learn a bit about opamps in their various modes and what I need to do re: impedance and such. I was originally going to buffer and then invert but now realize I'd need a -ve supply for the buffer. I'm trying to keep things as simple as possible and learning what I can on the way. Some folks have been kind enough to help on previous incarnations of this but a few requirements have changed and I'm trying to understand how to change things for myself as I go along.
     
  4. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    There's competing factors in something like this.
    For a simple inverting configuration higher value resistors give you:
    -more noise (bad)
    -more offset (bad)
    -higher input impedance (usually good)

    Low value resistors (aside from the opposite of the above list) give you:
    -the risk of hitting op-amp output current limits (bad, but very avoidable)
    -higher power consumption (bad)

    This usually puts you in the kilohm to hundreds of kilohms range...

    If you need high input impedance and want lower value resistors you can use this modified inverting configuration with T-feedback (first circuit):
    http://users.ece.gatech.edu/mleach/ece3050/notes/ExCir/EX12.pdf

    The gain is:
    -\frac{R_2 +R_4}{R_1}(1 + \frac{R_2||R_4}{R_3})

    But the input resistance remains as R1
     
  5. stirling

    Thread Starter Member

    Mar 11, 2010
    52
    2
    Thanks Ghar. I'd read about T-feedback but discounted it as I'm going for less than unity gain. Is this correct?
    I've attached my test circuit. As I've said the potential divider is in the piece of kit I'm intending to buy and is not changeable by me for warranty reasons therefore I need to work with it. All I know is that it's a 50:1 giving a range of 0 to -7Vdc. (It's a plasma cutter with inbuilt torch control interface). I don't know its output impedance - is there a way to find this out empirically? i.e. could I just measure the resistance of the resistor to earth and figure it out from there? My 490K/10K is just a guess at the moment.

    Ian
     
  6. Ghar

    Active Member

    Mar 8, 2010
    655
    72
    You're right, if you want less than unity the T-feedback doesn't really help you. I missed that part in your original post.

    Looking at your schematic I'm a bit puzzled how you determined that you need 680k input resistance.

    The output resistance of that torch circuit you posted is about 10k, meaning that with 680k input resistance you're losing only 1.5% of your gain... which you're reducing anyway because you set the op amp to less than unity gain.

    Op amps also tend to get unstable with less than unity gain. The LT1490 says it's "unity gain stable". Looking at the frequency response plot you should technically be ok at 0.7 but you might be pushing your luck. Even when not entirely unstable you get ugly behaviour.

    You can make the op-amp unity gain and get your 0.7 by reducing the input impedance assuming your source impedance is constant.

    You can measure output resistance by comparing the output voltage under a known load and no load using the same input.
    http://www.bcae1.com/outptimp.htm
    You should try it with a few loads just to get an idea of how it varies or how consistent your measurements are.

    I'm also a bit worried about sensing the voltage below the rails (i.e. grounded op amp but -7V signal) but I'm not too sure about how that works out. The datasheet does talk about the input range being between 0 and 44V. I'm pretty sure SPICE models usually ignore this limitations.
     
    Last edited: Jul 28, 2010
  7. stirling

    Thread Starter Member

    Mar 11, 2010
    52
    2
    This is just one of the things that confuses me re: impedance. My reasoning was that as I *thought* it looked like the output impedance of the divider was around 10K that the input resistor should be around that also. However by trial and error it seemed like I needed to take it up to around 680K before it stopped "dropping" the -7V.


    OK - I've tried this and ended up with a 24K unity gain setup which gives me my 5Vdc. However it does it by pulling the -7Vdc up to -5Vdc (and then of course inverting it) - Is this OK pulling the -7V like this? I kind of thought that was an impedance mismatch and to be avoided. Confused!!!

    Thanks - when I actually get the kit I'll do this.

    Hmmmmm - that could throw the cat amongst the pigeons. Wow - I didn't think it was going to be this difficult to turn 0..-7V to 0..+5V.
     
  8. Norfindel

    Active Member

    Mar 6, 2008
    235
    9
    It's unfortunate that you don't know the output current capability, but i think you could make a voltage divider that has at least 1k impedance, and get the signal in the 0 to -5v range. Then using an opamp in unity gain with 10k resistors to invert the signal should be acceptable, or at least something to start.

    You don't need a negative supply, because the inputs of the opamp will never see the negative voltage, but will be at 0v at all times. You need an opamp that is able to take an input and provide an output at the negative supply 0v (in your case). If your Vcc is 5v, you need a rail-to-rail opamp.

    Another easier option if you're absolutely sure that the signal is floating, and there isn't any common gnd between the signal and your circuit, is to invert the cables from the signal. Then you only need a resistive voltage divider. But if there's already some common gnd between both circuits, you would be making a short circuit.
     
  9. stirling

    Thread Starter Member

    Mar 11, 2010
    52
    2
    Thanks norfindel for your reply

    I'm currently awaiting info from the plasma cutter manfacturer so that will help.

    Cool - that's what I like - nice and simple.

    Not an option I'm afraid - there is most definitely common grounds throughout.

    Thanks again to all. I'm beginning to see the light - I think!

    Ian
     
Loading...