Current Mirror Application

[baseball07]

Hi. I have a question on the use of a current mirror. I understand what a current mirror does (transistor 2 mirrors current from transistor 1, depending on base-emitter voltage), however I do not understand what its purpose is in the following schematic:

http://www.sensorsportal.com/HTML/DIGEST/october_07/P_195.pdf

page 1627 and 1628 are what I have questions on. Basically the potentiostat is used to keep a constant voltage of .7V between working and reference electrode, and the output current (sensor current) of the solution is sent out to the current mirror (Page 1628). Here is where I get confused. All the current mirror is doing here is copying the current into transistor 1 to transistor 2 then sending it to the integrating capacitor. What is the purpose of the current mirror? All it is doing is making a copy. I can see how if it had more than 2 transistors it could be used as an amplifier, but that is not the case here. I assume it has something to do with protecting the signal so it cannot be altered???? Does anyone know what its puspose is?? Thank you for any help you may have!!

 

[IRstuff]

Why does anyone use buffers? It's to extract something from a source without loading it down.

TTFN

FAQ731-376

 

[baseball07]

So essentially this current mirror acts as a current buffer to produce an output current which is not a function the voltage (like an ideal current source). My question is, the collector currents of the transistors are a function of the the base-emitter voltage of the mirror. So if this changes then the current will change. Which loading are you refering to? What voltage will change which would have an effect on the output current? I am little bit confused on all of this.

 

[Noway2]

Current mirrors are used to create a constant current source. The one transistor "programs" the desired load current. The second transistor will then provide a current output, within its limitiations, as dictated by the first transistor. As long as the load on the second transistor does not create a situation where the transistor can't keep up, the load current will be determined by the programming and remain independant of the size of the load. That is what is meant by, not 'loading' the circuit.

As an other example, if you have a 5V source tied to two 1k resistors in series (to ground), the voltage at the node at the two resistors will be 2.5V. You may be tempted to use that 2.5V as a source. If you then put a 100 Ohm resistor as a load on that 2.5V you will find that the voltage drops considerably. In other words, you have 'loaded' the supply. Instead you could put a follower amplifier at that node and put your load resistor on the output of the amplifier. As long as the amplifier can source at least 25mA of current into that 100 Ohm load, then you will maintain the 2.5V regulation.

 

[IRstuff]

Part of this type of problem is that you don't know, and the author doesn't state, all the reasons for picking the topology he picked.



TTFN

FAQ731-376

 

[bogeyman]

The mirror works by establishing a Vbe on the input transistor that will provide a collector current equal to the input current, that is to say that the Vbe developed is appropriate for that particular transistor at whatever input current is supplied, it follows that if another transistor (At the same temperature and with the same base area) is connected base emmiter it also will have the same Vbe and will draw the same collector current that is being supplied to the input transistor but from another source, in your case the integrator cap.

The output current will not be exactly equal to the input current since:-
1 Iin is split between 2 bases and one collector hence the output current will be the 2 base currents lower than the input current.
2 The base currents for each transistor will be dependent on Vce of each transistor, in the input transistor its about .6v and Hfe will tend to be lower than it would with a higher Vce. In the output transistor Hfe will be higher than the input transistors and will be afected by Early effect as its Vce changes.

The small (1-2%) inacuracies of the 2 transistor mirror can be addressed by using a more sophistocated mirror circuit if needed.

The topology seems to have been chosen to isolate the changing voltage at the integrators input from the sensor cell, a condition that would create severe inaccuracy in an electrolytic cell, in fact it would render the measurement useless in the application shown. The upper block forms a current controled oscillator with a "fixed" input voltage of about 0.6v.

 

[Warpspeed]

I am unfamiliar with this particular application, but I can speculate...

The current mirror will give voltage independence between what is being measured and the voltage on the charging capacitor.

Voltage across the test cell will remain fairly constant over a very wide current range. Basically the polarizing supply voltage minus the Vbe drop of the mirror.

Voltage across the capacitor is free to rise from zero up to almost the full supply voltage.

The mirror reflects the current, but isolates any changes in voltage.

Without this current mirror, the voltage change across the charging capacitor would alter the polarizing voltage at the measurement cell. So as IRstuff has already said, this is a current buffer stage, to isolate voltage changes between output and input.

 

[baseball07]

ahh I get it now!! Thanks everyone.