OPAMP V-I Converter, help!

Discussion in 'Homework Help' started by radicalrad, Jan 18, 2012.

  1. radicalrad

    Thread Starter New Member

    Jan 17, 2012
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    I've not really spent much time on this forum, but I look forward to lending a hand and getting to know y'all. However, I have a problem for you:

    For my senior-level design course I have to construct a Voltage to current converter using a 741 OPAMP (datasheet). Here are the design requirements:

    1. Must utilize the general form of either figure shown below.
    2. Must provide a transconductance of 1 [mA/V]
    3. Must utilize +/- 12V supplies.
    4. Must be able to drive a maximum load of 2 kΩ, between the range of -3V to +5 V input.

    Here are my problems:
    1. I can achieve the desired functionality between -3V to +3.4V, but something is saturating and we can't go above +3.4V. I think it is voltage saturated, but I can't be sure.
    2. If it is voltage saturated, I don't know how to fix that problem.

    [​IMG]
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Which circuit did you use and what are the circuit resistor values?
     
  3. radicalrad

    Thread Starter New Member

    Jan 17, 2012
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    My group and I tried both topologies, but the one I worked on was figure B. I used R1=R3=R4= 1kΩ and R2,A = R2,B = 730Ω. Load resistance is 2kΩ.
     
  4. justtrying

    Active Member

    Mar 9, 2011
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    i don't really now much about it, but interesting configuration. Looking up some information on inmplementation, resistors have to satisfy a balance condition. Maybe that is causing the problem?
     
  5. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    A current source by definition has infinite output impedance. This means that, for a fixed input voltage, the load current will be independent of the load impedance. This will not be the case for the resistor values you used.
    This circuit is called a modified, or improved, Howland current pump. To get infinite output impedance, this relationship must be true:

    (R2A+R2B)/R1=R4/R3
     
  6. chobaugh

    New Member

    Apr 19, 2012
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    Ron,
    In the reply above, you say "A current source by definition has infinite output impedance." - I don't understand this. An infinite output impedance would be an infinitely high value, similar to, say, 900 Megohms - it's pretty hard to get much current through such an output. I believe you actually mean an infinitely LOW impedance. Any decent power supply, whether it's doing voltage regulated OR current regulated output, is always understood to be a low imedance source. Anything else would be less well-regulated...
    Please clarify this for us.
    Thanks large.
     
  7. jegues

    Well-Known Member

    Sep 13, 2010
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    I think what he means is that the internal resistance of an ideal current source is infinite.

    This makes sense because what does a ideal current source look like when it has a value of say 0 amps? An open circuit, right?

    With this idealization the voltage across the current source is purely a function of what you hook up at the terminals of the source.
     
  8. Ron H

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    Impedance of sources is generally defined as dV/dI: The amount the voltage changes when the current changes. In the case of a voltage source, you want zero output impedance, i.e., when the current changes, the voltage doesn't. In the case of the current source, it is easier to think of the conductance, dI/dV. You want dI/dV to be zero, i.e., when the voltage changes, the current doesn't. The reciprocal of zero is, of course, infinity, so dV/dI=∞.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    One of the first things I always try to do is look at the specs and see if they are remotely attainable.

    If you have a 5V input, that means that you want 5mA in the load. With a 2kΩ load, that means 10V across it. In both figures, the load is referenced to ground, which means the output of the opamp must be high enough to yield 10V to the load and drop some voltage across R2, while powered from a 12V supply. If I remember, the rule of thumb for a 741 was that it could get within 2V of the rails (or at least that you should count on it getting any closer than that). So that leaves you nothing to drop over R2.

    I would recommend going back to whichever circuit you chose to use and assume a 5V input, a 5mA load current in a 2kΩ load, and see what the necessary op amp output would need to be. Then compare that to what is reasonable.

    If I use your circuit values and the above assumptions, I get two different values of Vout depending on which way I go. I know that the load voltage has to be 10V and that there is then a simple voltage divider between the load voltage and Vin resulting in 7.9V at the V+ input of the opamp. But the V- input is simply 1/2 of the output voltage, so this would require the output voltage to be 15.8V. However, if instead I figure up the current flowing in R1 (which I get 2.9mA for) and add this to the 7mA load current, then I get a voltage drop across R2A of 2.1V resulting in a Vout of 12.1V. Either of these would exceed your limits, even if the 741 could operate rail-to-rail, but the fact that the two analyses don't agree means that the assumptions are wrong, namely that there are 5mA flowing in the load.,
     
  10. WBahn

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    Mar 31, 2012
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    How did you come up with these particular values.

    As I see it, what you need to do is analyze the circuit using an unknown load resistance and an unknown transconductance. Then solve for the transconductance and you should find that it is a function of the load resistance as well as the other resistors. However, hopefully you can separate the dependence on the load resistance into a single term and set it's coefficient to zero. This will impose necessary relationships among the other resistors. After applying them to the remaining terms, you will have the relationships that must apply in order to get a desired tranconductance. With five degrees of freedom to work with, you have a fair shot of pulling it off.

    Note that I suspect that the relationships given by Ron H will come out of the step where you set the coefficient of the load resistance to zero.
     
  11. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hi Wbahn,

    Your comments are undoubtedly well considered but you are probably addressing them to an OP long-since gone or satisfied with the outcome. The thread was started mid January and somewhat "hijacked" in post #6 which landed April 19th when the poster [chobaugh] was seeking clarification from Ron H.
     
  12. WBahn

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    Mar 31, 2012
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    Ah. Oh well.

    It's an interesting circuit and one that might be a good homework assignment or an extra credit problem on an exam. How I frequently do my extra credit problems is I give a review sheet with a number (typically 20) of fairly challenging problems that span the material covered and let the students know that some minimum fraction (typically 20% to 25%) of the points on the exam will come from the review sheet verbatim. I typically hand it out about three weeks before the exam. I also put on a short list of problems that are very challenging (in the time-limited context of an exam) and push noticeably beyond what we have covered explicity but that instead requires students to synthesize an approach based on several concepts that have been covered. That list never has more than three problems and frequently only has one. I tell them outright that one problem from that list WILL be the extra credit problem (typically worth 10% of the exam value) and that partial credit on that problem will be given sparingly.

    Now, what I would expect students to do (I sure as hell did it when I was a student!), especially when there is only one problem on the EC list, is to make sure they understand the problem (and its solution) so that they can walk into the exam, turn to the last page, whip out the solution, claim their 10pts, and then move on to the rest of the exam. Virtually never happens. In fact, hardly anyone ever attempts the problem at all. Two semesters, I even worked the problem in class AFTER handing out the review sheet and it made no difference. The handful of students that went through the motions of taking notes obviously never bothered to go back and look at them and since they probably hadn't looked at the review sheet yet, it didn't jingle any alarms. Then again, judging from the performance on the problems that came directly off the review sheet, I doubt many people even attempted to work any of them. A couple of times (this was in a C programming class), I made the second mid-term and the final all come off of the review sheet verbatim. Didn't notice any difference.
     
  13. t_n_k

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    Clearly you were a dedicated teacher. At least you set the bar high enough to challenge the students and no doubt the good ones benefited in their careers from your approach.

    The circuit is indeed an interesting one. The great and sadly missed Bob Pease wrote an app note on the Howland Current Pump when he was at NS. I've attached a copy FYI.
     
  14. WBahn

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    Mar 31, 2012
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    Very interesting app note. Thanks for bringing it to my attention.
     
  15. t_n_k

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    You are welcome.

    I notice it's about 1:40am in Colorado. Are you a light sleeper?:)
     
  16. WBahn

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    Nah, just a night owl.
     
  17. The Electrician

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  18. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Hello Electrician,

    Hope all goes well with you. Your wise counsel is missed.

    Thanks for the link to that thread on the forum. I do recollect there was a discussion at one stage. Interestingly I'd never heard of the Howland Current Pump until I joined the forum - most likely the time that thread was started was when it first came to my attention.
     
  19. WBahn

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    Similarly, I have never heard of it before, either, though the topology seems familiar for some reason. Probably a superficial similarity to a fundamentally different circuit.
     
  20. Ron H

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    For many years I have owned a copy of a 1971 book by John I. Smith called "Modern Operational Circuit Design", which has an analysis of the original Howland circuit.
     
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