# Matched biasing transistor vs. emitter resistor in current mirror

If adding emitter resistors can reduce the temperature variation in $$\V_{BE}\$$, why do we still need to find a matched pair of transistors to construct a current mirror?

• So that temperature variations don't produce significant errors. Apr 29, 2020 at 11:19
• The main idea behind current mirrors is that the output transistor (Q2) behaves just like a VCCS (Vbe controlled current source). So by changing the Vbe voltage, we change (set) the output (Ic2) current. And the first transistor (Q1) job is to convert the "input current" into Vin (Vbe2) voltage and nothing more. And the output transistor job is to convert this Vin (Vbe) voltage into Ic current As for your question, in short because in a current mirror circuit, the Vbe mismatch is the main source of an error in the output current even if both BJT's are kept at the constant temperature.
– G36
Apr 29, 2020 at 13:43
• KMC, perhaps head towards an fuller answer by asking different questions proposing a specific current mirror circuit and help analyzing the expected range of differences between the set-current and the mirrored-current over BJT part variations, ambient temperature variations, and differences in temperature between the BJTs. These (and other effects) can be individually and collectively analyzed (in all cases including emitter resistors as variables which may or may not be set to 0 Ohms.) Several questions may be needed to take it piecewise (partial differential); then you can sum the effects.
– jonk
Apr 29, 2020 at 19:09

You are right about the emitter resistor reducing the temperature variation of the current mirror. However, not the collector current is mirrored but the base current / voltage is mirrored, meaning that you still have some dependency on the gain of the second pair of your current mirror.

In the following circuit, the current through $$\Q_1\$$ will be approximately given by:

$$I_{C,Q_1} \propto I_B\cdot \beta_{Q_1}$$

$$I_{C,Q_1} \propto I_{C, Q_2}\dfrac{\beta_{Q_2}+1}{\beta_{Q_1}+1} \dfrac{\beta_{Q_1}}{\beta_{Q_2}}$$

simulate this circuit – Schematic created using CircuitLab

EDIT #1

Consider the following circuit:

The base voltage which is common for both transistor is given approximately by (for the sake of simplicity the internal emitter resistance is neglected ):

$$V_B = V_{BE,1}+ R_{E,1} (I_{C,1}(1+\dfrac{1}{\beta_1})) = V_{BE,2}+ R_{E,2} (I_{C,2}(1+\dfrac{1}{\beta_2}))$$

For a matched transistor and resistor, the above formula yields:

$$I_{C,1}=I_{C,2}$$

Assumming now that, $$\\beta\$$ is not matched, the collector current of $$\Q_2\$$ is given by:

$$I_{C,2}=I_{C,1}\dfrac{\beta_1 +1}{\beta_2 +1}\dfrac{\beta_2}{\beta_1}$$

As already explained in this question, $$\\beta\$$, which is defined by collector to base current ratio, has some temperature dependence given by:

$$\beta = \dfrac{I_C}{I_B}=\dfrac{I_C}{I_E - I_C}$$ where $$\I_E\$$ can be written in terms of thermal voltage $$\V_T\$$, which is dependent on the temperature according to:

$$V_T=\dfrac{k_BT}{q}$$

The following simulation, can show that although not very significant, the beta of the transistor still play a role on the mirrored current despite the introduction of matched emitter degeneration resistors.

This simulation, modifies the beta of the second transistor, by adding a $$\\pm 50%\$$ tolerance to its nominal beta value. All other parameters are left untouched. Furthermore, the simulation is run for 3 different temperatures in order to accout their variations on the final collector curent.

As you can see in the above plot, the output current ($$\I_{Q,2}\$$) has a dependence on the temperature, and consequently on beta.

• The above answer is not correct. No influence of R1? No - the base currents are not mirrored...The working principle of a current mirror is based on the matching of Vbe1=Vbe2. Note that the BJT is voltage controlled. The result is: Ic2/Ic1 ~ exp(-Ic2*R1/Vt). This expression assumes identical transistors
– LvW
Apr 29, 2020 at 14:33
• I just wanted to emphasize that the base voltage is mirrored and not the collector current itself. Meaning, that the gain of $Q_1$ still plays a role in its collector current. That's what the question referred to. Apr 29, 2020 at 18:18
• What is the "gain of Q1"? Do you refer to the current gain? This would be wrong! The formula of the current ratio (as given by me) does not contain any beta-values. But in your formula the role of R1 ist not included. The correct formula shows that R1 can be used to match the current ratio to the desired value. Even the revised formulas as given by you are not correct. Where did you get these expressions from? The working principle of such a current mirror is very simple: Identical base-emitter voltages cause identical collector currents (without R1 and assumong identical transistors)
– LvW
Apr 30, 2020 at 10:39
• Thanks for your comment. I will update my answer, in orde to make it clearer. Apr 30, 2020 at 11:43

The reason for adding the emitter resistors, a technique known as emitter degeneration, is not for temperature reasons but to reduce the effect of any mismatch in the transistors' Vbes.

In a designed current mirror IC the Vbes would be matched but in say a power amplifier's input stage, which would make use of discrete transistors, the emitter resistors should be included, reducing the effect of Vbe tolerances and thereby matching the collector currents much more closely which reduces distortion in the amplifier.

• Yes - and more than that: You can (a) compensate the result of mismatched Vbe values (as mentioned by you, for identical currents) or (b) introduce a fixed factor between both currents (if we do not want two identical currents).
– LvW
Apr 30, 2020 at 13:54

The tilt of the collector-current versus collector-voltage (Early Effect) also causes errors.

Some current mirror topologies address that effect.