# current mirror

#### rahul

Joined Jan 10, 2006
3
hello friends,

In the current mirror arrangement our idea is to make the collector current in the second transistor equal to reference current in the first transistor, ie

I ref = Ic+Ib
in the first transistor, now we try to make Ib as small as posssible so that I ref is approximately equal to Ic which is the current in the collector of the second transistor as well.

my question is why do we short the collector and base of the first transistor, if we don't short I ref=Ic and we don't have Ib coming into the picture.

Sorry if the question is too simple or silly, I have just started learning.

#### CoulombMagician

Joined Jan 10, 2006
37
The current mirror is based om matched Vbe's( or the equivalent in MOS devices) in the two transistors and so is suitable for circuits in an IC( this packaging also makes for equal junction temperatures). I have done similiar things with discrete devices but significant emitter degeneration is required to minimize the effect of the transistor mismatch.
One way to think of the device is as a unity gain(almost) impedance transformer with current in and current out signals. The diode wired collector-base has a low impedance and the collector of the active transistor has a high impedance. Just what you want for current mode signals.

#### hgmjr

Joined Jan 28, 2005
9,027
A couple of things can be done to minimize the variations in a current mirror design due to temperature and Ib.

First, there are IC's available that contain matched transistor pairs. They come in both dual NPN and a dual PNP versions. This will help minimize the effect of temperature.

The second thing to do is select such a device with the highest beta you can find. This will minimize the effect of Ib.

hgmjr

#### n9352527

Joined Oct 14, 2005
1,198
Originally posted by rahul@Jan 31 2006, 07:30 AM
hello friends,

In the current mirror arrangement our idea is to make the collector current in the second transistor equal to reference current in the first transistor, ie

I ref = Ic+Ib
in the first transistor, now we try to make Ib as small as posssible so that I ref is approximately equal to Ic which is the current in the collector of the second transistor as well.

my question is why do we short the collector and base of the first transistor, if we don't short I ref=Ic and we don't have Ib coming into the picture.

Sorry if the question is too simple or silly, I have just started learning.
[post=13677]Quoted post[/post]​
If you refer to this page about current mirror, you could follow the rationale behind shorting out the CB on the bias leg. The purpose of the bias leg is to provide a matching constant BE voltage to the sinking leg. This can be done with a diode, but it would be difficult to match the BE junction with the diode junction.

Easier voltage matching could be achieved by using matching transistors (thus matching BE junction, to a degree), where one transistor is configured only as a forward biased diode on the BE junction.

Referring to the circuit diagram on the above page, the current you referred to as Iref is Ibias.

Analysing the circuit, Ibias = Ic1 + Ib1 + Ib2
Assuming Ib1=Ib2 for matching transistors, then Ibias = Ic1 + 2Ib
Assuming matching Beta, Ic1 = Ic2
Ic2 is the mirrored current

You would actually have 2Ib factor instead of only Ib.

Without shorting the CB, then we would need to set both the Ib with a resistor and the mirror current value would be Beta*Ib (depends on Beta). As you know, manufacturing a transistor with a specific Beta is next to impossible and therefore the mirror current value would be hard to adjust.

Shorting the CB, we would only need to match both transistors Betas (possible to some degree) and set the mirror current value through Rbias (the above equations, Imirror = Ibias - 2Ib). The effect of 2Ib would be small and could be easily quantified and also lessened by making Beta as high as possible.

So, the main reason is to decrease as much as possible the effect of Beta value variations to the value of mirrorred current.

#### hgmjr

Joined Jan 28, 2005
9,027

#### LRankin

Joined Jan 27, 2010
1
Although the responses were correct, they seemed to be a little too complicated for someone who is just learning the subject.
The shorting of the base to collector essentially makes the transistor into a diode. That diode is then used as a reference at the base of another transistor. Assuming that the diode is an ideal diode, the input to the 2nd transistor should not change, therefore the collector current (IC) in the 2nd transistor should remain constant therefore making it into a current source.
Hopefully this will give you added insight into the other, more complete, answers.

#### Wendy

Joined Mar 24, 2008
22,931
Uhhh, did you look at the time stamp on this thread? The OP is long gone, likely happily employed as an engineer or technician.