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What is the difference between intrinsic gain and just "normal" voltage gain of a circuit i.e Vout/Vin?

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As I understand it, in e.g. a transistor amplifier, the intrinsic gain will refer to the transistor beta, and the voltage gain to the actual Vout/Vin.

Usually for e.g. a common emitter setup, you would not rely on the unpredictable current gain of the transistor, and add some feedback in the form of an emitter resistor. Then the voltage gain becomes roughly the ratio of the emitter and collector resistors (unless you bypass the emitter resistor for increased small signal gain)
So the intrinsic gain of the transistor might be 200, but (with an e.g. 10k/1k ratio) the actual voltage gain of the circuit ~10.

Similarly with an opamp open loop gain, you could say this is the intrinsic gain of the opamp.

Here are a couple of related links:
MOSFET intrinsic gain
Opamp discussion (see page 3 second paragraph from bottom)

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  • \$\begingroup\$ so what you are saying is that the intrinsic gain is the gain internally in the device i.e a op amp or a basic gain cell, but the actual gain is the gain when you consider loads etc connected to those device? \$\endgroup\$
    – rrazd
    Sep 18 '11 at 2:01
  • \$\begingroup\$ @rrazd - not the load as such, the intrinsic gain is the "internal" gain yes, but the actual gain depends on the setup (ideally the load would not affect the gain, although you could specify e.g. a gain of x into an impedance of x) so for example an opamp with intrinsic (or open loop) gain of 100,000 could be setup for a (closed loop) gain of 10 using negative feedback. You do see unloaded vs loaded gain graphs in amplifier datasheets, but I would say the unloaded gain would be usually be used to describe the circuit itself. (unless the load is actually part of it) \$\endgroup\$
    – Oli Glaser
    Sep 18 '11 at 2:25
  • \$\begingroup\$ @rrazd - added a couple of related links that might be helpful. \$\endgroup\$
    – Oli Glaser
    Sep 18 '11 at 2:57
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According to the book of Prof. Behzad Razavi (Fundamentals of Microelectronics):

$$A_v = -g_m \cdot r_0$$

The “intrinsic gain” of the transistor to emphasize that no external device loads the circuit, \$g_m \cdot r_0\$ represents the maximum voltage gain provided by a single transistor, playing a fundamental role in high-gain amplifiers.

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As I understand it, the intrisic gain is the gain of the amplifier itself (or sometimes called the open loop gain, meaning there is no feedback loop). And the overall gain is the gain of the whole circuit with the loop (called closed loop gain)

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Intrinsic Gain is just the maximum possible voltage gain of a typical transistor, regardless of the bias point. it means \$ g_m r_o \$, that is the maximum gain of a CE (Common Emitter) amplifier.
by eliminating \$ I_C \$ it's equal to \$ V_A/V_T \$ and it hasn't any dependence to the bias point of transistor.

(Recall that \$ g_m=I_C/V_T \$ and \$ r_o=|V_A|/I_D \$, thus \$ g_m r_o=\frac{I_C|V_A|}{V_TI_C} \$.)

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    \$\begingroup\$ Could you explain why intrinsic gain doesn't depend on bias point? Intrinsic gain equals gm*ro and both gm and ro depend on bias condition. \$\endgroup\$
    – emnha
    Jun 6 '16 at 7:27
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I wanted to add the derivation for the intrinsic gain, that might help to understand it.

$$A = \frac {dV_{out}} {d_{V_in}} = \frac {dI} {dV_g} \frac {dV_d} {dI} = g_m \frac {1} {g_d} = g_m r_0$$

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