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Current Mirror Load on Diff Amp

  1. Dec 4, 2011 #1
    I'm unsure what the purpose of this is. I have looked it up but can only find some explanations stating "role is to convert the differential current input signal to a single ended voltage signal without the intrinsic 50% losses and to increase extremely the gain."

    I'm not sure what double ended to single ended voltage signal means, or what the initial intrinsic losses are.

    Could someone clear this up for me?

    Thanks
     
  2. jcsd
  3. Dec 4, 2011 #2
    Are you referring to the current mirror on top of a differential transistor pair. eg. If you have two NPN to form a differential pair, you have a pair of PNP on top, one PNP has base connected to the collector and the other PNP has the base connect to the base-collector junction that form the current mirror?

    If so, the reason is the gain of the differential NPN pair is determined by the collector load. The gain would be [itex] \frac{R_c}{2r'_e}\;\hbox { where }\;r'_e \;\hbox { is the emitter resistance at the given tail current of the transistors where }\; r'_e=\frac 1 {gm}= \frac {V_T}{I_c}≈\frac {25mV}{I_c}.[/itex]

    To increase gain, [itex]R_c\;[/itex] has to be as high as possible. The collector resistance of a current mirror has the highest resistance so it give the highest gain. Also a current mirror as the name, reflect the current from one side to the other, that is the reason you don't loss half the gain if you just take one side of the differential amp and drive the next stage.
     
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