Exponential amplifier using LM741

Discussion in 'General Electronics Chat' started by summersab, May 10, 2010.

  1. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    First, let me say that my background is mechanical engineering. I have SOME EE coursework, but you may need to talk stupid to me :D

    Okay, so I'm trying to get an exponential amplifier using a LM741 op amp to work. I'm using this basic model:

    http://en.wikipedia.org/wiki/Operational_amplifier_applications#Exponential_output

    Attached is what I have built. The supply voltage is 9VDC (attached to the rails at the bottom) and the input voltage difference is 2VDC. The potentiometer is a 1 MΩ pot. I adjusted the pot by 50 Ω increments and measured the output voltages (measured from the positive input at A17 to the negative output at J10 as per the attached picture) as shown below:

    0 Ω 1.165 VDC
    50 Ω 1.138 VDC
    100 Ω 1.128 VDC
    150 Ω 1.122 VDC
    200 Ω 1.113 VDC
    250 Ω 1.102 VDC
    300 Ω 1.094 VDC
    350 Ω 1.086 VDC
    400 Ω 1.078 VDC
    450 Ω 1.069 VDC
    500 Ω 1.060 VDC
    550 Ω 1.051 VDC
    600 Ω 1.043 VDC
    650 Ω 1.035 VDC
    700 Ω 1.027 VDC

    Correct me if I'm wrong, but this isn't an exponential output. My eventual goal is to take a small voltage (0.02VDC or so) and increase it by a factor of about 250 (yielding 5VDC). If I can't get it to work with a very standardized setup, I'm not sure what to do. What am I doing wrong?

    **EDIT: Okay, so I realized some stupidity on my part. The output isn't exponential in relationship to the resistance. However, no matter what my setup, I cannot amplify a voltage ABOVE my input voltage. It's always <2VDC. Why?
     
    Last edited: May 10, 2010
  2. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    Firstly, it's an inverting circuit - it will only work with both positive and negative supplies to the opamp.

    The output range will be from zero to negative some volts.
    The 'R' value and sets the gain or scale factor.

    The input range is tiny, somewhere roughly in positive 600 - 700mV.
    Anything beyond that and you wil get either zero or maximum negative out.
    The exponential effect is caused by the diode's non-linear current to voltage curve.

    The 741 is an old device and the input & output voltages should be at least 2V less than the +/- supply voltages, preferably more. They were commonly used on +/- 15V to guarantee proper operation over a +/- 10V range.

    I suspect that as shown, it's more a concept than a practical circuit.
     
  3. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    I'm a little confused. You said that the input range for the 741 is around 600-700mV, but then you said that the input and output voltages should be 2V less than the supply and that the devices were commonly used over a 10V range. So, what IS the range for the inputs?

    Also, I know that it only works with both the positive and negative supplies connected to the op amp, but I do believe I have it all wired properly - the +/- for the supply and the inputs are there.

    Thanks for your help - I hope I can get this to work!
     
  4. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    The input range *for that circuit* is something like +600 to +700mV, it's the conduction voltage of the diode. Any less, the diode does not conduct at all, any more and and the circuit saturates.
    It also needs a low impedance input.

    *The 741 itself* needs 'headroom' in the supply. eg. If you want a 0 to +5V signal out, the power supplies would need to be at least +7V (5V+2) and -2V (0V-2), preferably more.

    As I said, for typical equipment that needs to work over a +/- 10V signal range (common in lab equipment and industry), they are used on +/- 15V power, to ensure the signal range is nowhere near the power rails.

    The +/- supplies are relative to the circuit ground, ground (0V) being what you refer the input and output voltages to.

    eg. A +/- 9V supply using batteries needs two 9V batteries, one from ground to positive and one from ground to negative, giving 18V total across the 741 power terminals.
     
  5. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    This is making more and more sense, but I still don't know what to do to make this work. Basically, my eventual goal is to amplify the output of a strain gauge/Wheatstone bridge setup. Since the voltage output of that is just a few tens of millivolts, I need to amplify it a LOT (and I want to do so exponentially so it's VERY sensitive). My supply is simply a 9VDC "wall wart" AC/DC plug. Is there a more appropriate op amp for this application? Is my circuit setup at least right?

    Thanks again for your help!
     
  6. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    Surely if you are amplifying a load cell signal you want good linearity so you can make sense of the resulting value?

    You want an instrumentation amp. My favorite general purpose one is the INA126, but there are many available.
    Datasheet here: http://focus.ti.com/lit/ds/symlink/ina126.pdf

    It even gives an example of a load cell setup on page 9..
    (The output is between pins 5 & 6. Pin 5 is the common of the next stage, it can be connected to 0V if you use a dual +/- supply for the INA126).
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    The INA128/INA129 is another good couple of inst. amps to look at.
    http://focus.ti.com/lit/ds/symlink/ina128.pdf

    Have a read through this Application Note:
    http://www.intersil.com/data/an/an1298.pdf
    It explains why you don't want to try to make an instrumentation amplifier out of several opamps, amongst many other fine points.

    These instrumentation amps might seem to be a bit expensive - but when you look at how accurate they are, and the huge number of hours you would spend trying to get individual opamps perform even close to their specifications - they're a real bargain.
     
  8. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Just another tidbit...

    I don't recommend trying to breadboard your circuit using an instrumentation amplifier.

    Breadboards will add too many parasitics; R, L, and C. You'll have a heck of a time trying to get it to work right.

    If you feel you must, use a socket in your prototype board - but remember that the socket will add a small amount of resistance. When you are dealing with such low signal levels, parasitics add up quickly.
     
  9. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    The 741 op amp is more of a teaching tool nowdays. They are pretty rugged, you can burn one up but you have to work at it, and every limitation that happens with any op amp happens sooner with 741's, which means students don't have to go to extremes to see what an instructor is talking about.

    It is a good bad example.
     
  10. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    This is not true. See the diode equation. The current is an exponential function of the voltage (VD) across the diode. For any forward voltage greater than zero, current will flow.
    This is true.
     
  11. rjenkins

    AAC Fanatic!

    Nov 6, 2005
    1,015
    69
    There will be leakage current at low voltage, but that is not relevent to the exponential circuit operation - see the curve above the equation you linked to and it's related text; quote:

    The current–voltage curve is exponential. In a normal silicon diode at rated currents, the arbitrary “cut-in” voltage is defined as 0.6 to 0.7 volts
     
  12. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    Actually, no. I'm not trying to measure the load and gather data or make sense of it. Moreso, I'm trying to use a load cell/Wheatstone bridge as a "switch" of sorts. Basically, I want the load cell setup to be hooked up to an exponential amplifier so small changes in the load will saturate the circuit, attach a 5V Zener diode to the output of the amplifier, and use it to drive a transistor. So, I need to amplify the load cell a LOT and with a lot of sensitivity so that it saturates the amplifier quickly to 5V. Basic ASCII diagram (the forum deleted whitespace, so I used periods as filler, sorry):


    ._________
    /\.......... __|>-----|>|-------|<
    \/______|

    (Bridge) (amp) (Zener) (transistor)

    So, no, I don't want to use an instrumentation amplifier. I don't want anything linear. I want a very high gain exponential amplifier that can take a small voltage (50mV or so) and give an output of 5V. This can be a very rough circuit. The 741 is giving me troubles, so what can I do to make this work? A different IC? A different wiring? Thanks for your patience!
     
  13. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Why don't you just use a comparator?
     
    summersab likes this.
  15. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    The problem is that the load cell will already be under a load and calibrated to it. So, for instance, it will already have 25 lbs applied to it, and the output of the Wheatstone bridge will be calibrated to a 0VDC output accordingly. Then, any increase in the load above the current load will trigger a response from the amplifier mechanism.

    I considered a switch, but it simply won't work mechanically for my application. Basically, I'm putting load cells under the end bedposts of my bed, so a load is already applied. The circuit will be tarred to 0V for when I'm not in the bed. When I get in the bed (or even sit down), it will trigger the amplifier circuit. That is why this needs to be sensitive exponentially - any slight change in voltage from the load cells should trigger a 5V output. I COULD get microswitches that work specifically to my bed, but I want to be able to take this circuit with me when I travel and tare it to different beds. I have the tare part of the circuit figured out - the amplification is the part that I CANNOT get to work.

    So, anyone? An exponential amplifier circuit that takes 20-50mV and amplifies it to 5V? Please???
     
  16. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    As I said,
     
  17. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    I'm failing to see how a comparator would work. If my input from the load cell is 0V when no load is applied and 20-50mV when a load is applied, comparing it to something like 5mV would STILL only give me 20-50mV output at most. How is this helping me drive a transistor with a digital 5V input? There's no amplification going on. At all. It's just comparing and giving me the larger input as an output. That doesn't help.

    Anyone? High-gain exponential amplifier that works?

    **EDIT: Ron, it's possible I'm completely not understanding what you're suggesting. If so, my apologies for shooting you down like that.
     
  18. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    A comparator is a high gain (100,000 or more) amplifier, optimized to give a digital output - a one or a zero - (5V or 0V in your case) when the inputs differ by millivolts or less. 20 to 50 mV is a piece of cake for a comparator.
     
  19. summersab

    Thread Starter Active Member

    Apr 8, 2010
    132
    0
    I've been Googling like mad since your last post to make heads or tails of what you're suggesting. Since I have mechanical engineering background, I'm not really understanding how this works. I saw the Wikipedia article on comparators, but I didn't follow it. Is there a good circuit diagram out there that shows me what to build? Will the basic 741 work?

    Talk stupid to me. I won't be offended. Thanks for your help - you seem to understand what I need more than I do!
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    LM393 is a common, good, inexpensive comparator. Others are available with even better specs, typically at somewhat higher cost.
     
Loading...