MOSFET Current Mirrors....

[Rohit Chatterjee]

In a simple current mirror consisting of 2 MOSFETS,why do we short the gate and drain terminals of one of the MOSFETS????And even if do it,shouldnt we do the same for the other MOSFET as well so that they r perfectly matched in terms of their terminal voltages???

 

[Hero999]

I didn't study this at college so my understanding is purely from Wikipedia.

For a MOSFET working in saturation mode, the drain current is proportional to the gate voltage.

The gate voltage of M1 is set by the drain current. Here's how it works: the voltage starts to turn on M1 which short circuits its gate to 0V lowering the gate voltage. An equilibrium is soon established, the higher the drain current, the higher the gate voltage.

If M2 has exactly the same electrical characteristics as M1, the drain current will be the same when it is supplied the same gate voltage.

**broken link removed**

Current mirror - Wikipedia, the free encyclopedia

 

[crutschow]

The idea in a current mirror is to have the current in the shorted transistor mirrored in the other transistor. The gate voltage of the shorted transistor will equal what is required to pass the current passing through it. This voltage then appears on the other transistor so it will carry the same current, if the two transistors have matched gate threshold voltages. If you shorted the gate and drain of both transistors then you wouldn't have a current mirror.

MOSFETs do not make good current mirrors because of the larger variance in the gate threshold voltage between devices. Bipolar transistors make much better current mirrors since their base-emitter voltage tends to be closely matched between transistors of the same type.

 

[Rohit Chatterjee]

Quoting frm wikipedia,"the output current is the same as the reference current when VDG=0 for the output transistor, and both transistors are matched."
VDG=0 implies shorting the D n G teminals of the o/p MOSFET.So why dont we do it???

 

[Hero999]

I didn't fully understand Wikipedia's explanation, in fact I didn't bother to fully read it, most of my understanding was gleamed from looking at the schematic.

If you shorted the drain and source terminals of the output transistor, it would cease to behave like a current mirror, assuming the voltage across the output is a perfect source, the MOSFETs will allow as much current at the threshold provided which might result in smoke if the power dissipation is exceeded.

 

[Rohit Chatterjee]

If you shorted the drain and source terminals of the output transistor, i

I was reffering to shorting the gate n drain actually......

 

[crutschow]

OK, now I understand your question.

For a current mirror to work correctly it's required that the drain-to-source resistance be very high. In practice this is achieved with typical transistors, since the drain current varies very little with a change in drain-source voltage at a given gate-source voltage. Thus even though the two transistors will have different drain-source voltages, their drain-source currents will be close (again for matched transistors).

 

[Hero999]

I was reffering to shorting the gate n drain actually......

That's what I meant. I made a mistake. If the gate and drain of M2 is shorted it will behave like M1 and if the voltage is high enough and the impedance is low enough, the MOSFET will be destroyed.

 

[Rohit Chatterjee]

Okkk...i think i get it now...