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# Convert linear output of current mirror to exponential

#### Nucrino

##### Newbie level 4
Hello,
I am currently trying to change the output from my current mirror, which is linear, to a more exponential one.
The picture below shows the current mirror i'm using, which is intended to work from 8V to 28V.

What i would like is for the current to act like the graph below :

What changes could i do to the current mirror ?
Thank you for your help !

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Over a quite limited and temperature-varying range,
applying a voltage directly to the mirror base node
would give you that. If you convert current to stiff voltage
and apply that, ditto. Degenerate master side with R and
slave side with a diode, that will also look exponential.

Over a quite limited and temperature-varying range,
applying a voltage directly to the mirror base node
would give you that. If you convert current to stiff voltage
and apply that, ditto. Degenerate master side with R and
slave side with a diode, that will also look exponential.
Hey tank you for your help, I am not sure I understand the circuit you are explaining, could you show schematics ?

Demonstration (simulated) how silicon diode behaves. It starts conducting at ~0.6V. So as it turns out a diode traces an exponential curve.

Since a garden-variety BJT contains PN junctions, it displays similar behavior near the point of turn-On. So you can obtain a similar plot in an adapted similar fashion.

It is current mirror right? It mirrors what you feed to it so make sure that is exponential in the first place.

It's not a current-mirror if the output current doesn't match the input current.

The current mirror could come in handy because it avoids putting a resistor inline with the component you're measuring. It's a way to use Easy's recommendation:
"If you convert current to stiff voltage and apply that, ditto.'
You can take voltage-vs-current readings through a transistor diode junction.

Since you speak of a power supply 8-28 V, you can make a resistive-divider divide down what goes through the left transistor, so that you measure its current, whereas the right-hand transistor gets tested in its turn-On range of a few tenths of a V.

You can either a) apply voltage in tiny increments and measure resulting mA,
or
b) apply mA in increments, and measure what voltage is able to create each reading.

I once did similar tests on a real diode to discover what data is generated. I measured at different values. I had to carefully adjust my variable power supply, painstakingly. I had to read my meter, painstakingly. I plotted the results on a graph. My research took me a few evenings.
Good luck.

It's not a current-mirror if the output current doesn't match the input current.
Maybe it's a current Fun-House mirror.

What i would like is for the current to act like the graph below :

One transistor, current output but no current-mirror.

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A simple diode has exponential IV curve: apply a linear voltage across it and the current through it will be exponential (I=Is.(e^(V/kVt)-1).

View attachment 189927

One transistor, current output but no current-mirror.

View attachment 189928
Basically, the problem is to manage the current being the same as it was with the current mirror (0-150µA) and have the exponential curve on the whole 0-28V zone
Is there a way to have almost the two of them combined ?

Is there a way to have almost the two of them combined ?
Here's one way:
The offset voltage V3 can be generated from a reference voltage such as the programable TL431 shunt regulator.
Note that this circuit is sensitive to temperature.