Convert linear output of current mirror to exponential

[Nucrino]

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 !

 

[barry]

Hey! Try this!!
Search “Log converter circuits”. PRESTO!!! 24 million hits!

 

[dick_freebird]

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.

 

[Nucrino]

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 ?

 

[BradtheRad]

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.

 

[DefconNowhere]

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

 

[crutschow]

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

 

[BradtheRad]

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.

 

[dick_freebird]

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

Maybe it's a current Fun-House mirror.

 

[crutschow]

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

How about this?:
One transistor, current output but no current-mirror.

 

[DefconNowhere]

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).

 

[Nucrino]

View attachment 189927

How about this?:
One transistor, current output but no current-mirror.

View attachment 189928

Thank you for your help
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 ?

 

[crutschow]

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.