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Please consider the differential amplifier (Q1&Q2) with current mirror active load (Q3&Q4) as shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

I know many questions have been asked already regarding this circuit on EE SE, but I didn't find an answer to my question.

My question is: what will the output voltage be when subjected to differential input WITHOUT ANY LOAD AT THE OUTPUT? Framed differently, what is the differential gain with no load?

In my understanding, the currents in the collectors of the 2 right transistors will not agree and thus the output impedances at the collectors will come into play. However, I do not know how to calculate this output impedances and how they will affect the output voltage.

Side note: If you think that asking about the output voltage with no load is meaningless than consider a load whose input impedance is comparable or larger to the output impedance looking into the collectors.

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  • \$\begingroup\$ Well, why don't you write down your equations with a generic R as load, and see what happens when said R gets very high? \$\endgroup\$ – Vladimir Cravero Nov 24 '16 at 19:56
  • \$\begingroup\$ My problem is, I don't know how to describe the behavior of the collector under heavy load (comparable to the output impedance at the collector). I saw some places described it as a current source in parallel with some Rout but I do not understand why and how to calculate the said R. \$\endgroup\$ – Harel Nahari Nov 24 '16 at 19:59
  • \$\begingroup\$ I think there is a misunderstanding: load your amplifier with a generic resistor R, calculate the gain, take R to infinity and see what happens. \$\endgroup\$ – Vladimir Cravero Nov 24 '16 at 20:00
  • \$\begingroup\$ Oh, sorry I misunderstood you. Anyway I get that when Rc becomes very large the differential gain becomes very large. I'm looking for something more quantitive that will help me understand the limitations of the current mirror active load. \$\endgroup\$ – Harel Nahari Nov 24 '16 at 20:03
  • \$\begingroup\$ Actually, it's not clear how to derive it from the Wiki mentioned by Mario. The Gummel Poon model makes it quite clear, as it explains the Early (and Late) Effects from a much more unified and physically sensible approach than the modified Ebers-Moll, which just hacks it in almost arbitrarily. It boils then down to this: \$V_A=\frac{Q_{B0}}{\frac{1}{V_{BC}}\int_0^{V_{BC}} C_{jC}\left(V\right)~\textrm{d} V}\$. \$\endgroup\$ – jonk Nov 24 '16 at 21:31
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Assuming that you are talking about the gain at low frequencies ("DC gain") a derivation of the gain will result in an expression like $$ Av = g_m (r_{o2} || r_{o3}) $$ where \$g_m\$ is the transconductance of Q2 or Q3 and the \$r_{o2}\$ and \$r_{o3}\$ is the output resistance of these transistors.

The gm of the transistor is easily found to be $$ g_m = \frac{I_C}{V_T} \qquad V_T = \frac{kT}{q} $$ The output resistance is a property of the transistor and usually described by the Early voltage \$V_A\$. Using the Early voltage the output resistance is $$ r_o = \frac{V_A + V_{CE}}{I_C} \approx \frac{V_A}{I_C} $$ since usually \$V_A \gg V_{CE}\$. Assuming that both transistors have the same ro the gain is $$ Av = g_m (r_{o2} || r_{o3}) = \frac{I_C}{V_T} \frac{V_A}{2 I_C} = \frac{1}{2}\frac{V_A}{V_T} $$ With an Early voltage of 100V and a kT/q of 26mV we get a gain of about 2000.

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  • \$\begingroup\$ can you elaborate about the early voltage and the output impedance? or refer me to a source where I can find the relation between the two? \$\endgroup\$ – Harel Nahari Nov 24 '16 at 20:56
  • \$\begingroup\$ sorry I meant the derivation of the relation, not the relation itself which is shown here \$\endgroup\$ – Harel Nahari Nov 24 '16 at 20:57
  • \$\begingroup\$ You should find everything here: en.wikipedia.org/wiki/Early_effect \$\endgroup\$ – Mario Nov 24 '16 at 21:01

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