Relation between Iout and Vcc in this peak current source

[Tangos]

hello

It is a peak current source.I want to find the relation between Iout and Vcc.Coule you tell me how to analyse the circuit in this picture?

Thanks in advance
regards

 

[A.Anand Srinivasan]

Re: peak current source

please tell whether the two transistors are identical because the equation is too complex and cannot be simplified without details about the two transistors...

 

[sunking]

Re: peak current source

Vbe=0.7

 

[Ghitani]

Re: peak current source

Does such circuit really works ?!! I think that Vce1 < Vbe1 so Q1 may be in saturation and it can't drive the other transistor

 

[xcoder]

peak current source

to: Ghitani
I think Q1 works in soft saturation region[1], the transistor is ok.

to: Tangos
The ckt you refered is a current source, it's an improved version of Delta-VBE current source[2]. but i don't know the ckt's name.

to: sunking
vbe=VT*ln(Ic/Is),
so,vbeQ1 is a function of Iin,VbeQ2 is a function of Iout, the equation is too complex.


can anyone give some mathematical description of the ckt?


[1] razavi, fundamental of microelectronics
[2]Hans Camenzind,Designing Analog Chips

 

[mbanak]

Re: peak current source

I just stumbled across this, and couldn't resist an answer, here 2 years after it was posted.

Here's a quickie approach.

The odd resistor in the collector of the left hand transistor, should be set to 1/gm1. Then a cool voltage division occurs for smal variations in collector current for the left hand side. A slight increase in Collector current will cause VBE1 to rise by deltaI/gm1. But the resistor pushes it back down by deltaI*gm1. Thus they cancel, and you gain some slight protection against small vaiations in Current on the left side. For example, the setup resisistor may have a tolerance or TC drift. To a first order, this cancels out much of it.

Of course, gm1 = Ie/Vt ~ Ic/Vt

The transistor on the right has 4x the area of the one on the left. Since VBE changes by about 18mV for every doubling in current desity, you can anticipate the current of the right hand side if you simply assume you want the input current to match the output current. Then the device on the right has 2*18mV less of Vbe (two doublings). That extra 36mV is soaked up by the emitter resistor on the right. if you want 100uA coming out of this thing, set the input to 100uA (using the bias resistor R1) and select the emitter reesistor at 36mV/100uA = 360 Ohm. If the TCR of the emitter resistor is positive, you can expect some happy temperature compensation, as the delta VBE is also postive.

If you do not want the left and right currents to match, the equations already posted will stand.

- Michael

 

[snafflekid]

peak current source

get rid of the 4x and R3, and the equation becomes

Ic2 = Ic1 exp [-(Ic1R2/Vt)]

I have seen this called a peaking current source. Do a google search on that. The approximate value of Ic1 will be (Vcc-0.7)/R1, of course. I have not seen the 4x and R3 as part of this circuit when used as a current mirror.

The advantages of this circuit is that it can reliably generate small currents without having to do a large mirror ratio. It also has a peaked operating region where it rejects current variation, which has been pointed out. I have seen a circuit similar to Tangos that is configured as a bandgap. I think it uses the negative resistance region of operation (I am on vacation so cannot get to my stuff to tell.)

I have a gut feeling your circuit will have a transcendental solution and therefore give you a solution that provides no insight.