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I fail to see why this circuit compensates for temperature changes with both transistors at the same temperature. If I understand correctly, the current through both transistors is the same, but I don't get why it is also independent of temperature. Can someone please clarify this to me?

Matched biasing transistor

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  • \$\begingroup\$ I'm not sure, but I'm assuming because as q1 opens, q2 closes, proportionally? \$\endgroup\$
    – Passerby
    Sep 7, 2015 at 13:21
  • \$\begingroup\$ picture credit? \$\endgroup\$ Sep 7, 2015 at 14:30
  • \$\begingroup\$ Horowitz AoE3 Fig.2.52 (2.3.5.BMatched Bias Transistor) \$\endgroup\$
    – KMC
    Apr 24, 2020 at 6:44

3 Answers 3

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Both transistor stages are already internally stabilized against temperatute changes. However, no stabilization can totally (100%) compensate the corresponding influence on the collector current. In this circuit, the influence of the Q1 stage can additionally improve the stabilization of the gain stage (Q2). This works as follows:

A temperature increase will only "slightly" increase the collector current Ic1 and, thus, somewhat reduce the voltage Vc1 at the collector node of Q1. This voltage Vc1 is the driving voltage for the base current Ib2 which reduces correspondingly. This small Ib2 reduction counteracts the temperature caused increase of the Q2 collector current Ic2.

For proper dimensioning one can show that, in this case, a nearly ideal temperature compensation for the gain determining transistor Q2 is possible.

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  • \$\begingroup\$ Does it mean that the current through both transistors is not the same, even if I put the same collector resistors? \$\endgroup\$
    – Rol
    Nov 1, 2015 at 16:04
  • \$\begingroup\$ Does collector current increase because \$V_{BE}\$ drop by 2.1mV/C per degree increase in temperature, and if temperature does increase, \$V_{BE}\$@Q1 drops and \$V_{E}\$ to ground increases causing both \$I_{E}\$ and \$I_{C}\$@Q1 to increase? If so, \$V_{out}\$@Q1 decreases, lowering current into \$I_{B}\$@Q2, and hence lowering \$V_{drop}\$ over Q2's resistor or increasing (compensating back to) \$V_{out}\$@Q2 (or +10V)? I want to make sure my understanding is correct before confirming it quantitatively. Thanks. \$\endgroup\$
    – KMC
    Apr 24, 2020 at 6:58
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Have a look at this document, circuit #2 figure 1B

The current gain beta of the transistors is temperature dependent, this together with the 10 kohm base resistor of Q1 gives a (somewhat) temperature compensated behaviour of the collector current Ic. Since Q1 and Q2 are connected as a current mirror, Q2 will have the same collector current as Q1.

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  • \$\begingroup\$ and the (probably less important) variation in Vbe (-2mV/k) also affects both transistors the same way. \$\endgroup\$ Sep 7, 2015 at 14:30
  • \$\begingroup\$ This cricuit is from chapter 2 of TAoE. Thanks for the explanation! It's much better then in the book. \$\endgroup\$
    – axk
    May 10, 2018 at 23:29
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As i see it this circuit keeps the bias to transistor Q2 temperature compensated. It does not take the signal variation in consideration. Basically Q1 is connected as a diode with the forward voltage in principal the same as the Vbe voltage of Q2. If the temperature increases the Q1 forward voltage drops with approximately the same voltage as Q2 Vbe voltage and therefore keeps the bias current the same over temperature.

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