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In a lecture I learnt a BJT VCVS which has gain = 1, i.e a voltage buffer as shown in the figure.

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The speaker said that it was the ideal VCVS model of a BJT. My question is that is it possible for a BJT VCVS to have a gain > 1. Thanks.

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  • \$\begingroup\$ Gain is close to R / (R + Re) where Re ~= 26 / ((Vin-0.6)/R). | Re is the effective internal resistance of the base-emitter junction and for silicon it is ~= 26/Ie_mA. so if eg Vin = 10.6V, Ve ~= 10V so Ie = 10/R mA so Re = 26/(10/r) . If R = say 100k then Ie = 100 uA and Re =~ 26/0.1 = 260 Ohms so gain =~ 100k/(260 + 100k) = 0.9974. Note that gain RISES with Ie. It would be 0.99997 at 10 mA with R = 100k BUT would need Vin = 1000.6V. [[[E&OE]]] \$\endgroup\$
    – Russell McMahon
    Commented Mar 3, 2015 at 13:00
  • \$\begingroup\$ Worse actually - Vout is less than Vin by Vbe which is typically ~~= 0.6V. The discussions re gain relate to AC gain but the DC gain is reduced by the 0.6V offset. So for eg 10.6V in the Vbe drop makes DC gain lower than AC by a factor of 10/10.6 = 0.94. For DC this modifies (reduces) the AC gain by this factor for DC. \$\endgroup\$
    – Russell McMahon
    Commented Mar 3, 2015 at 13:04
  • \$\begingroup\$ I wonder where this - in my view - very unfortunate term "effective internal resistance Re" comes from. I think, it is rather misleading (a) to use a capaital letter for this dynamic/differential quantity and (b) to call it "internal base-emitter resistance", which is also not true. It is simply the inverse of the transconductance gm=1/re, which connects input and output of the BJT with each other. Thus, it cannot be the "base-emitter resistance" and can be mixed-up with the real base-emitter resistance rbe (hie, h11) which is in the kohm range. \$\endgroup\$
    – LvW
    Commented Mar 3, 2015 at 13:27

3 Answers 3

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"My question is that is it possible for a BJT VCVS to have a gain > 1. Thanks."

I suppose, with your question you refer not only to the shown circuit (common collector configuration). Thus, my interpretation of your question is as follows: "Is it possible to realize a VCVS using one single transistor?".

My answer: A device which we call VCVS must have a very large (ideal infinite) input resistance and a very small (ideal: zero) output resistance. Both requirements cannot be fulfilled with a transistor. However, the shown circuit can be a rather good approach because we can realize an input resistance in the order of some hundreds of kohms and an output resistance of 50 ohms or so. However, as was mentioned already, the gain of this circuit cannot be larger than unity.

Of course, there are other circuit alternatives (common emitter) with gain values that are much larger - however, the required characteristics for a VCVS (input, output resistances) cannot be fulfilled as we could for the common-collector circuit.

As a consequence, if you need a "good" VCVS with gain>1 you need 3 transistor stages in series: : A gain stage (common emitter) and two other stages (common collector) at the input as well as at the output of the gain stage.

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It's actually got a gain that is a fraction below 1, probably nearer to 0.99 for normal transistors with average beta/current gain and an emitter resistor that isn't so low it forms too high a potential diver (thus losing a bit more gain) with the internal emitter resistance.

In this configuration (common collector) gain can approach 1 but never be 1 or greater than 1.

If you were to implement a common-emitter and take an output at the collector then gains far in excess of 1 are achievable.

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  • \$\begingroup\$ Thanks Andy. But in ideal VCVS form where input resistance is infinity and output resistance is zero can we get gain > 1. \$\endgroup\$
    – salil87
    Commented Mar 3, 2015 at 12:11
  • \$\begingroup\$ No, gain of 1 or greater is not possible. \$\endgroup\$
    – Andy aka
    Commented Mar 3, 2015 at 12:22
  • \$\begingroup\$ @Andyaka Gain is = 1 in his ideal case if you substitute infinite load" for zero Zout. FWIW. \$\endgroup\$
    – Russell McMahon
    Commented Mar 3, 2015 at 12:51
  • \$\begingroup\$ @RussellMcMahon I'm not going to get into arguments about this ideal form or that ideal form. One mans ideality is not the same as another. With an infinite load it becomes irrelevant to my way of thinking and where something is 1 or approaches 1 is splitting hairs because the op is clearly interested in gains >1 also. \$\endgroup\$
    – Andy aka
    Commented Mar 3, 2015 at 13:00
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    \$\begingroup\$ Ahhh @spehro, pedantry is alive and kicking and I agree of course! \$\endgroup\$
    – Andy aka
    Commented Mar 3, 2015 at 13:33
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What you have shown is a topology of a basic voltage buffer referred to as "common collector" or "emitter follower". It does not represent the model by itself. The transistor models can vary significantly over the voltage and frequency range you are working with. Assuming that the speaker was referring to a low-frequency small-signal operation, a famous model is the simplified hybrid-pi model.

Also, you can find more information on the emitter follower in this Wikipedia page.

In the wiki page, the voltage gain derivation comes down to this equation: $$ \frac{V_{out}}{V_{in}} = \frac{1}{1+\frac{R}{R_E}} $$ Now, the details of symbols \$R\$ and \$R_E\$ are irrelevant, they are values derived from internal resistor values of the transistor model and the load resistor. The important thing is they both are positive values by definition.

By looking at the simple equation, you can see easily that to get a gain higher than unity the \$\frac{R}{R_E}\$ value must be negative, which is impossible. So the answer is no, you can not get a gain higher than 1 in this situation.

EDIT: Sorry about not understanding your question fully. I thought you were asking specifically for the circuit you provided. Since the question already has an accepted answer, I won't change my answer, but it should be seen as a simple proof for common emitter configurations not having gains greater than 1.

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