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Differential Amplifier not working properly.

  1. May 18, 2012 #1
    I want to learn more about Op-Amps and amplification so I am making a simple differential amplifier. The circuit looks like this...

    I set it up and it isn't working correctly. First I tried putting 3 volts for Vin+ and ground for Vin-, and the voltage accross Vout was around half a volt. I don't understand this result. Shouldn't it be an amplification of the difference in voltage? (Which is 3 volts correct?) How can I get half a volt? I didn't know what value of resistors to use so I used 100 ohm for RC1 and RC2 and 150 ohm for Re (Completely arbitrarily chosen).

    Can anyone help me out?
  2. jcsd
  3. May 18, 2012 #2
    You need to tell as about V+ ? , -V and Re = ?.
  4. May 18, 2012 #3
    In my original post I said...
    V+ = 3V
    V- = Ground
    Re = 150 ohms
    And the other two resistors are 100 ohms.

    EDIT: Ohh and I am using a 6.5 volt power source, if that is what you meant. Sorry!
  5. May 18, 2012 #4
    You need to be more specific. Do you have v+ at 6.5V and v- at 0V, or do you have v+ at 0V and v- at -6.5V (or something else). Also, when you say the input is 3V where exactly are you measuring. In other words, what node are you assuming to be 0V (where are you putting the black lead of your meter when you are putting red lead on Vin and seeing 3V).
  6. May 18, 2012 #5
    Sorry about the ambiguity. As you can probably tell I am new at this!
    The way I have it set up is. I put the positive lead of the 6.5 volt battery (It WAS a 9V but I have been using it for awhile..) for V+ and I put the ground for V-. As for the inputs, I put the positive lead of the 3V battery on one of the inputs and the ground to the other input. When measuring V+ and V- my black lead is on the ground for that battery, and likewise for the 3V power source.
  7. May 18, 2012 #6
    This is how you should set your differential amplifier.


    You need to use two battery one for Vcc and second one for Vee.
    Also kept in mind that the linear region for BJT differential amplifier is very small.
    If +Vin - (-Vin) = is greater then 150mV the one of a bjt will be in cut-off and the second one will be in saturation.

    Attached Files:

    • 12.PNG
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  8. May 18, 2012 #7
    If I understand your post correctly you have build this circuit


    If so. This circuit has nothing to do with real differential amplifier.
    Q2 is in cut-off and Q1 is in active region.

    Attached Files:

    • 13.PNG
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  9. May 18, 2012 #8
    Thank you very much! I will study this later and try to re-create it. I just have a few questions..
    1.) What roles to the values of the resistors play? What values should I use?
    2.) Same question, but for the batteries.
    3.) Why is it that one BJT will cut off and one will be in saturation after a 150 mV difference? Would that mean that one BJT is doing all the work while the other is doing nothing?

    Thanks again!
  10. May 18, 2012 #9
    First, for a differential pair to work properly, you have to have a matched pair for Q1 and Q2. But this is not your problem. Your differential voltage is too large. You have +3 at base of Q1, and 0 at base of Q2, Q1 is totally on and Q2 is totally off. You don't have a differential pair anymore.

    Differential voltage range between Q1 and Q2 can only be very small by equation:

    [tex]V_{B1}-V_{B2}=V_T ln\left[ \frac{I_{C1}}{I_{C2}}\right]\;\Rightarrow\; \frac{I_{C1}}{I_{C2}}\;=\; e^{\left({\frac{V_{B1}-V_{B2}}{V_T}}\right)}\;\hbox { where }\; V_T≈\hbox { 25mV.}[/tex]

    As you can see, if the input differential voltage is much larger than 25mV ( 250mV) the ratio is going to be very large and consider cocked to one side. Q2 is turn off and disappeared.

    This will hold true even in the more clearly defined biasing condition in the later posts.
    Last edited: May 18, 2012
  11. May 20, 2012 #10
    Well the resistor sets various of the amplifier parameters. The most important one is of course the voltage gain. But also the proper DC bias point is also important.
    For example if you use two 9V batteries then. It is good to have:
    Rc = (0.5*Vcc)/Ic and Re = (Vee - Vbe)/ (2Ic)

    So for example

    Rc = 4.5V/4.5mA = 1KΩ


    Re = (9V - 0.65V)/9mA = 920Ω ≈ 1KΩ

    I use Vcc = 9V and Vee = 9V

    We use symmetric power supply because it is a natural way to power differential Amplifier.
    Only in this case we can connect the BJT bases to gnd but we still have bout bjt to work in there linear region. So we made a true DC amplifier.

    http://www.engga.uwo.ca/people/adounavis/courses/ece235b/notes/ECE235b_L07_DifferentialAmps.pdf [Broken]

    Last edited by a moderator: May 6, 2017
  12. May 21, 2012 #11
    Thank you very much!!! The link you provided is proving to be very helpful! I just have one question. Why do they use an exponential relationship for the current in section 7.3.2? How would I derive or deduce that it is an exponential relationship?
    Last edited by a moderator: May 6, 2017
  13. May 22, 2012 #12
    When we star learn about BJT the told as that BJT is a current control (base current) current source.
    Ic = β * Ib
    But this simply model don't tell as the full story about BJT.

    In reality the collector current is equal to

    [tex]I_{C}≈I_{S}*\hbox e^{(\frac{V_{BE}}{V_T})}[/tex]

    And this e in the equation tell as that this is a exponential relationship.

  14. May 22, 2012 #13
    And if you use this equation for Q1 and Q2 and divide equation of Q1 by Q2, you get the equation I posted.
  15. May 22, 2012 #14
    I was asking on how would I derive the exponential relationship. (I am very well versed in mathematics, I will have a bachelors in applied physics in a year) Is there a differential equation I can solve that gives it to me?
  16. May 22, 2012 #15
    OK I understand what you want. But I don't know the answer to your question.
    But if you interest in this you should buy some book about semiconductor physics.
  17. May 28, 2012 #16
    Here is a simulation (first attachment) of a 2N5210 differential amplifier using LT Spice (free software). Vcc = + 9 volts, and Vee = -9 volts, with the base quiescent voltages at ground. The common emitter resistor Re is 420 ohms, to set the collector currents at about 10 mA. The collector resistors are both 420 ohms. The quiescent collector voltages are about 4.9 volts and the emitters about -0.7 volts. The input pulse is a 10 mV, 200 ns long pulse with a 2 ns rise and fall time. The full differential output is about 1.1 volts, representing a full differential gain of about 110.

    The second attachment from Fairchild Semiconductor http://pdf1.alldatasheet.com/datasheet-pdf/view/101319/FAIRCHILD/2N5210.html shows the 2N5210 gain-bandwidth product contours vs. collector voltage and collector current. With a collector voltage of 5.4 volts, and a collector current of 10 milliamps, the gain-bandwidth product is 175 MHz. (I don't think the simulation is accurate).

    Fairchild used to publish the gain-bandwidth plot with all its fast transistors.

    Attached Files:

  18. May 28, 2012 #17


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    your diff amp will have better common-mode rejection if you can make the Re as large as possible (which will require Vee to be even more negative). what you really want is a constant current source instead of the Vee/Re combination and you can get that with a transistor (also NPN) with a very low hoe parameter.

    if you do that and your transistor pair is well matched (and two Rc are well matched), you should have a very good diff amp.
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