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Ordinary resistor as load in a transistor amplifier has a certain short coming…we cant have voltage and current gain simultaneously …..

In this circuit a current mirror is actings as an active load in place of RL …I wonder how this current mirror able to give voltage as well as current gain at same time??

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  • \$\begingroup\$ A resistor load AND transistor can have both voltage and current gain compared to the input. What are you trying to say? \$\endgroup\$
    – Andy aka
    May 18, 2015 at 9:33
  • \$\begingroup\$ i m trying to say that by using an ordinary resistor as a load resistance we can have one gain at a time..if Rload is high then its a voltage gain ..and if Rload is low then its a current gain..so text says that to compensate this problem we can use current mirror as an active load...and i want to know how the active load actually does so..... \$\endgroup\$
    – partykid
    May 18, 2015 at 9:37
  • \$\begingroup\$ Wrong, the current gain is determined and is largely fixed by the transistor across a wide range of load resistors. What are you trying to accomplish? \$\endgroup\$
    – Andy aka
    May 18, 2015 at 9:41
  • \$\begingroup\$ take a look at this link --->play-hookey.com/analog/current_mirrors/active_loads.html as they havent explained how the current mirror as an active load...solving out this dilemma....I want to know the working of current mirror as an active load..in reference to the circuit.. \$\endgroup\$
    – partykid
    May 18, 2015 at 9:49
  • \$\begingroup\$ Yes - poor explanation in the ref. link. Please note that VA is the Early voltage and the given ratio VA/IC is a rough approximation of the differential (dynamic) outpiut resistance of the BJT which is rather large. \$\endgroup\$
    – LvW
    May 18, 2015 at 12:04

2 Answers 2

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With a load resistor, the current through the resistor obeys ohm's law and varies with the voltage. However, an active current mirror allows a fixed current, even with large variations in voltage. This basically makes the amplifier work more effectively by presenting a much higher impedance. This high impedance increases the gain.

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  • \$\begingroup\$ I am getting it a little but could you please make it more precise in terms of working of above circuit.... \$\endgroup\$
    – partykid
    May 18, 2015 at 9:39
  • \$\begingroup\$ Perhaps the following helps: The current of a current mirror is not "fixed". Instead, it changes only a little as a result of the applied voltage variations. Hence the resulting conductance g=dI/dV is very small and the corresponding differential (dynamic) resistance is very large (if compared with a suitable ohmic resistor). \$\endgroup\$
    – LvW
    May 18, 2015 at 10:58
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In a standard common emitter design, the voltage gain is Rc/Re.

A current mirror (as a current source) is a device that maintains a constant current for variations in the voltage across it; so the transient resistance R = dV/dI which for a perfect source is infinity.

A real current source will not achieve this (perhaps some further reading on the limitations of current sources might be in order), but large resistances are normal.

With a large effective resistance in the collector, the voltage gain of (effective)Rc/Re can be very high, and much higher than a resistive solution.

This technique is very common (almost universal) in operational amplifiers, incidentally.

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