How does a differential amplifier work?

Discussion in 'General Electronics Chat' started by cooded, Aug 3, 2011.

  1. cooded

    Thread Starter Member

    Jul 20, 2007
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    Hi ,

    I was working on a design when i stumbled upon a very basic question. I have attached the schematic.Please pardon me for the poor drawing since I do not have a schematic drawing software.
    I have drawn a basic 1st stage differential amplifier of an OP-AMP.
    The OP-AMP is connected with the signal source as given in the smaller diagram whose equivalent circuit diagram i have drawn.
    My questions are:
    If the negative terminal of the signal source and GND2 are not connected to each other then does this circuit make any sense,since the voltage at q1 and q2 bases are not with respect to GND2.
    IF the equivalent circuit i have drawn, is wrong then please can someone correct it for me?
    Please can someone explain me how an OP-AMP works (not according to the widely followed equivalent model) when the a differential signal source whose ground is not common to opamp ground is connected to the input terminals of OP-AMP?


    Regards
    Rahul
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Read up here: http://en.wikipedia.org/wiki/Differential_amplifier
    Look at the "long tailed pair" schematic on the right, about 1/3 of the way down.
    Your circuit is missing the current limiting resistors; without them, there is no way to get an output from your circuit.

    An explanation of the circuit accompanies the schematic.

    In the differential circuit, there is a difference of potential across the two bases, which is multiplied by the transistors. However, since you have no resistors in your circuit, there will be no output.
     
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  3. Adjuster

    Well-Known Member

    Dec 26, 2010
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    There does need to be some relationship between the input voltages to a differential amplifier and its power supply voltages.
    In the specifications of an op-amp, limits will be given for the so-called common-mode input voltage.

    Some amplifiers only work with inputs within a quite limited voltage span, others cover the full span between the supply voltage rails - in some cases multiple input stages are connected in parallel to provide extended range.

    Clearly, as well as other problems, your simplified circuit with a floating voltage input has no means of supplying the input bias current for the two transistors. A real circuit would need to have some means of holding the absolute input voltage within acceptable bounds.
     
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  4. mik3

    Senior Member

    Feb 4, 2008
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    For the op-amp to work, one side of the signal source shall be connected to ground.
     
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  5. cooded

    Thread Starter Member

    Jul 20, 2007
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    @SgtWookie: Yes, please assume that the transistors have been biased correctly by adding current limiting resistors. My question is the transistor q1 and q2 will only turn on when the voltage at bases of q1 and q2 is greater than 0.7 with respect to GND2. Here the voltage is the signal source which is w.r.t each other bases.

    @Adjuster : So in your opinion this circuit wont work even if the common mode voltage of the signal source is within the defined range, because the input side is not forming a complete path.If thats the case then how do you explain the OP-AMP with the configuration i have shown.

    @mik3: Thank you. But then the same question arises that OP-AMP which is connected in identical configuration, how does that work. I have just drawn the simplified equivalent circuit.


    Regards
    Rahul
     
  6. Hi-Z

    Member

    Jul 31, 2011
    157
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    Hi cooded,

    A number of points:

    You've shown an op-amp whose inputs are floating with respect to Vcc and GND2. This isn't a meaningful mode of operation - it's a bit like having a common-emitter stage with an unconnected base.

    In your transistor circuit, there will have to be collector loads, of course, but, more importantly, there needs to be a resistor (or current sink) in the "tail" of the differential pair; both emitters are NOT connected directly to the negative supply (which you've called "GND2"). In addition, the inputs cannot be left floating - they need to be referred to the supply rails.

    Usually, the op-amp will operate with positive and negative supplies, and voltages at input and output will be referred to 0V (ground). In practice, the input voltages (referred to 0V) are allowed to approach the supply rails, but certainly not exceed them. Differential input voltage range will in general be more restricted than this.
     
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  7. Hi-Z

    Member

    Jul 31, 2011
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    Actually, the better analogy is a common-emitter stage whose base is connected solely to a signal source, whose other terminal is left unconnected. Note that the transistor can't turn on, so the circuit is meaningless. In the case of the differential pair with floating inputs, you have a similar situation - neither transistor can conduct, whatever the differential input voltage.

    Now, if you connected the signal source to a power supply whose voltage was comfortably in the range GND2 to Vcc (plus you had the collector loads and tail resistor/sink), you'd have yourself a working differential amplifier...
     
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  8. cooded

    Thread Starter Member

    Jul 20, 2007
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    @Hi-Z
    Thanks for your reply. Yes I had similar opinion about the ckt but i wanted to confirm my thought. You gave a very nice explanation.I should have drawn a constant current source at the emitter common terminal and connected to -Vee.Also i should have drawn collector resistors too as load. But we have seen an OP-AMP connected in the way i have drawn. Does that mean that the OPAMP will not work since the equivalent 1st stage circuit is similar to what i have drawn(please assume the constant current source and other biasing resistors, since I want to concentrate only on various ways of applying input).Most of the internal circuit diagram of OP-AMPs have a similar 1st stage as i have drawn.

    In the new ckt i have drawn two 1MΩ resistors between base and ground. Here the 1MΩ resistors complete the input circuit path as well as provide sufficient biasing to torn on both the transistors according to the input voltage.Hence the current generated by the signal source will never enter the ground of the differential amplifier ckt.

    I hope this explanation is correct. Do correct me if i am wrong.

    Regards
    Rahul
     
  9. Hi-Z

    Member

    Jul 31, 2011
    157
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    I'm pleased you're finding my posts useful, and I'm happy to be of assistance.

    I'm very surprised that you've seen an op-amp connected as you've drawn. Are you sure it was exactly as you've portrayed? Don't forget op-amps aren't designed to be used without feedback of some kind, and the circuitry associated with the feedback may well be providing an operating voltage for the input circuitry (and base current for the input transistors). So, operating the op-amp without feedback is in itself rather meaningless, but having an undefined common-mode input voltage makes it doubly so.


    I'm having some difficulty in understanding just what is connected to what in your diagram (it's very small and blurry on my screen), but one key thing to understand is that, in general, the op-amp will operate from split supply rails, one positive with respect to our baseline (which is 0V/ground), the other negative. So the current sink at the "tail" will be connected to the negative rail. If the 1M resistors are connected to 0v (ground), then everything is fine: the input common-mode voltage is defined (at 0V), and it's within the supply rails - so it'll work OK.
     
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  10. cooded

    Thread Starter Member

    Jul 20, 2007
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    Hi,
    Well I have drawn a simplified equivalent diagram of an OP-AMP with the inputs as shown. I only wanted to tell you that the input going to OP-AMP is differential and connected in the fashion i have drawn. So if the op-amp first stage is like i have drawn(please see the schematic i have attached now) then the OP-AMP should not be giving any o/p.


    I have attached a redrawn schematic. I hope you will be able to see this one.
    Please do comment.

    Regards
    Rahul
     
  11. Hi-Z

    Member

    Jul 31, 2011
    157
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    Well, the diagram is clearer, but it would be an awful lot clearer if the tail current sink were actually shown! (Even if it's just a transistor with unconnected base - we'd know what you mean.)

    Anyway, assuming the 1M resistors are connected to the negative rail (as opposed to the transistor emitters), then the circuit won't do anything until the ac signal reaches a sufficiently high voltage to turn one of the transistors on. The two resistors act as potential dividers, so you'd need 4*Vbe peak-to-peak before the transistors start to turn on, and only one will conduct at any given time (for a brief part of the cycle), so you could hardly call the circuit a differential amplifier.

    The only way the circuit could work as a proper differential amplifier would be if the transistors were n-channel jfets (or vacuum tubes!), again, assuming that the 1M resistors are connected to the negative rail.
     
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  12. cooded

    Thread Starter Member

    Jul 20, 2007
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    Hi,

    Thanks for your explanation. Yes i should have drawn a tail current sink(-Vee). Anyways, correct me if i am wrong here. The current generated by the input(lets call it i/p current) AC source passes through the 1M resistors and back to the negative of the input AC source thus giving the current a complete path(This i/p current will not flow into the tail current source because that does not complete a path ). The voltage generated across the 1M resistance due to this i/p current current biases the Vbe junction of the transistors. The i/p current should be large enough to generate a sufficient voltage across the resistors for the transistors to conduct.Here the current path does not interfere with the differential amp circuit(kind of isolated) but the voltage produced affects the amplifier o/p. Hence there is a direct relation of the voltage at o/p and voltage at i/p and indirect relation of the current generated due to source and the voltage at o/p. This does make sense to me. I hope i have understood it right.


    Regards
    Rahul
     
  13. cooded

    Thread Starter Member

    Jul 20, 2007
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  14. Hi-Z

    Member

    Jul 31, 2011
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    I don't think we should be talking about the circuit which has the 1M resistors connected to the negative rail and a differential amplifier in the same breath.

    If you have the resistors connected to the negative rails, then the circuit will behave as I've described (but it ISN'T a differential amplifier, because it's not biased as one). And any base current which arises as one of the transistors starts to conduct WILL flow in the tail, the path being via the resistor connected to the side of the signal source which is negative at the time of conduction.

    I must say I don't understand why we're discussing such a strange use of a "differential" amplifier, particularly in the context of an op-amp. Are you sure you've seen such a configuration (i.e. an op-amp with bipolar input, operating with its inputs operating at the negative rail voltage)? It's certainly not a legitimate configuration.
     
  15. Akash Shesh

    New Member

    Aug 10, 2011
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    Ive got a Query. Does my voltage divider network across the transistor (q1 or q2) need a pair of equal resistances so that the node near the base is at 'Zero Potential ' ?
     
  16. cooded

    Thread Starter Member

    Jul 20, 2007
    28
    0
    Hi,
    I think i have not been communicating right. Ok. I have attached a diagram where i have drawn a double ended opamp. Now the signal source o/p has been directly connected to the pins of the opamp non-inverting and inverting terminal. The two terminals coming out of the signal source are connected to where to the op amp except the inverting and non inverting pins of the opamp. the op amp should act like a comparator and compare the voltages at it input terminal and give an output as +/- 15v. Will this circuit work at all?


    Regards
    Rahul
     
  17. Hi-Z

    Member

    Jul 31, 2011
    157
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    For a bipolar op-amp it is quite important for the source of bias current to be matched, otherwise an offset voltage will result. Normally, this means that the resistances "seen" by the inverting and non-inverting inputs should be made as equal as possible.
     
  18. Hi-Z

    Member

    Jul 31, 2011
    157
    17
    Well that's an unusual beast you have there! It's got differential outputs, which means that it's not a normal op-amp.

    Assuming that it is an op-amp which happens to have differential outputs, then there are two problems: the input is shown to be floating, and there is no feedback.

    As I've mentioned, you can't operate an op-amp with undefined input voltage. Now, it may be that the input voltage isn't actually undefined, but that would mean that some circuit details are missing from the diagram.

    As for the lack of feedback, I would say that op-amps aren't designed to be operated open-loop, so in general, it's bad design practice to do so.

    HOWEVER, if the "op-amp" in question isn't actually an op-amp at all, but is a specialised differential amplifier, then the circuit may be just fine. But I would emphasise that this amplifier would have internal feedback and bias circuitry hidden from view. If you have in mind a particular example, what is the type number of the amplifier?
     
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