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I was measuring the current through current mirror, when I experienced a problem. But first, I want to mention that I measured this with two ammeters of same manufacturer and same model (Mustool MT826 Digital Multimeter). Both were set to mA (milli ampere) setting of measurement and through both of them, the current measured was about 0.5 mA for both collector currents through Q1 and Q2. Then I thought that it would be better to set both of meters to uA (micro ampere) setting of measurement, and so I did. But only one at the beginning; AM1 was set to mA setting while AM2 was set to uA setting - there I spotted a problem. Suddenly, the current through AM1 was around 2.5 mA while current through AM2 was decreased to 300 uA.

So this got me thinking for a while and I came to conclusion that at different settings for current measurement, the ammeter is "seen" as different load to the emitter of Q1 and Q2. So, my question is:

  • Is it normal for ammeter to change its internal resistance, when we switch between different settings of measurement (uA, mA, A) or is this due to bad measuring equipment bought?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Short answer: Yes, it's very normal. \$\endgroup\$ – Mike Mar 2 '18 at 20:15
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    \$\begingroup\$ Yep, perfectly normal. \$\endgroup\$ – Hot Licks Mar 2 '18 at 23:45
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Yes, it's completely normal. Often a fixed voltage (for full scale input) is presented to the measuring circuitry (the "burden voltage") so the resistance of the ammeter decreases (more or less) linearly with the scale.

In other words a 20mA range might look like 10\$\Omega\$ (200mV burden), but a 20A (called a 10A) range would be more like 10m\$\Omega\$. Plus lead, some internal, and switch resistances.

It's important to remember that an ammeter (when properly working) will tell you the current that is passing through it, however that may not accurately reflect the current that would be passing through the circuit if the ammeter was replaced by a short. You can think of the ammeter as something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The resistor Rm is usually fairly stable and fixed (for each range) however Rx can be unstable because it includes copper wires, the flaky rotary switch, trace resistances, the banana jacks and plugs on your meter and so on.

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If you build a current mirror with resistors of different values in the emitters, the output current (in a first order approximation) will be the input current multiplied by the resistor ratio.

Your ampmeters are acting like emitter resistors. Each range will have a different value of burden resistor, so what you see is not surprising.

You should put the ampmeters in the collectors if you want the current mirror to ignore their burden resistor.

Also you can't build a current mirror without emitter resistors unless both transistors are matched and at the same temperature. This works inside an IC, or if you use monolithic matched transistors, but it won't work with discrete components.

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  • \$\begingroup\$ Might be worth pointing out that the resistor ratio rule only works well if both resistors are fairly large (hundreds of mV drop). If one (bias) is zero and the other (output) is non-zero it's a Widlar source and the equation is a bit more complex. \$\endgroup\$ – Spehro Pefhany Mar 2 '18 at 22:15

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