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Could somebody explain to me the principle of this current mirror? I don't understand why I throughout my R2 and R3 is not dependent on R1 (R bias)?

My goal is to have 0,8V voltage drop on R3.

Many thanks in advance.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ What's V2 doing? \$\endgroup\$ – Andy aka Mar 4 '20 at 15:23
  • \$\begingroup\$ The mirror is capsized, but not sure if it's correct. \$\endgroup\$ – Marko Buršič Mar 4 '20 at 15:46
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This isn't a proper current mirror as the current through R1 isn't copied to the other branch.

The reason for that is the fact that Q1 and Q2 have different things connected to their emitters. Q1 has a voltage source while Q2 has a resistor.

When the value of R1 is changed, the current through R1 changes but the voltage will not change much. The voltage across R1 is: V(R1) = 20 V - 3.3 V - 0.7 V = 16 V. This means that the voltage at the base of Q2 doesn't change either. That then means that the current through Q2 will not change.

To make a proper current mirror, replace V2 with a resistor of 3.31 kohm. Then we can change the value of R1 to make the correct current.

0.8 V across R3 (800 ohm) means 1 mA needs to flow. There should also be 1 mA through R1. There's 20 V across the 3.31 k ohm resistor + Q1 + R1. Q1 takes 0.7 V, that leaves 19.3 V for the 3.31 k ohm resistor + R1. 19.3 V / 1 mA means a total of 19.3 kohm, subtract 3.31 kohm

=> R1 = 16 kohm

Do note that the \$V_{CE}\$ of Q1 will be much smaller than the \$V_{CE}\$ of Q2. So Q2 will suffer much more from the Early effect than Q1 meaning the current through Q2 will be slightly higher than 1 mA. Just increase the value of R1 to compensate for that.

For easier reading, try to avoid using voltage sources that are "upside down" but then use a negative voltage. It's like double negative and not needed.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ "When the value of R1 is changed, the current through R1 changes but the voltage will not change much. The voltage across R1 is: V(R1) = 20 V - 3.3 V - 0.7 V = 16 V. This means that the voltage at the base of Q2 doesn't change either. That then means that the current through Q2 will not change." I don't understand why the voltage at the base of Q2 doesn't change \$\endgroup\$ – sancho Mar 6 '20 at 18:45

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