Please help with 3-transistor current source question.

Thread Starter

HunterDX77M

Joined Sep 28, 2011
104
Please refer to the attached image for the circuit and problem in question.

Hey guys,

I'm stuck on this current source question and was hoping someone could help. At first I made the assumption that the transistors all had the same turn-on voltage of VBE,on = 0.7 V but I quickly found out that doesn't make any sense, because the problem would be a simple middle-school plug and chug other wise.

I decided to go back to the basics and make some nodal equations, but I had more variables than equations so I scrapped that.

Any help please?

(The complete question and circuit is attached.)
 

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Jony130

Joined Feb 17, 2009
5,485
Normally, it is impossible to find analytical solution for this type of a circuit.
We need to use numerical method or use iteration method.
See the example problems
http://forum.allaboutcircuits.com/showthread.php?p=226670#post226670 (post 8)
http://en.wikipedia.org/wiki/Diode_modelling#Iterative_solution
http://socrates.berkeley.edu/~phylabs/bsc/PDFFiles/bsc3.pdf (page 4)

But in point B they want you to find R3 resistor.
And this can help you solve this circuit much faster.

Simply use β and Ic = β*Ib , Ie = Ib + Ic = (β +1)*Ib to find Ic5 and Ic6.

Ic5 = Ic6

Ib6 = Ic6/β

Ie7 = ( 2 * Ic6/β ) + IR3

Ib7 = (( 2 * Ic6/β ) + IR3 )/ ( β + 1)

And Iref = Ic6 + Ib7

So assume IR3 and solve for Ic6

Or you can try some different approach.

Assume for example Ic5 = Ic6 = 9μA then use shockley equation to find Vbe and select R3 that meets this requirements.
 
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Thread Starter

HunterDX77M

Joined Sep 28, 2011
104
If anyone cares to know, I asked my professor and he said that the three currents "I" in the diagram are all the same. It made the whole thing very simple indeed. :)
 

panic mode

Joined Oct 10, 2011
2,670
well that is what question suggested - all three are marked I.

btw circuit is not correct, 1k resistors are not connected to -Vee. only when they are connected this becomes possible (need junction placed at Vee, connecting all three resistors to negative supply).

btw, you are not supposed to assume value for any current, you should calculate it.
 

panic mode

Joined Oct 10, 2011
2,670
Normally, it is impossible to find analytical solution for this type of a circuit.
read the question (and understand it) and you may come up with some helpful conclusions. for example:

since all transistors are same, have same gain beta=200, and all collector currents are same, and all transistors are in same mode, they all must have same Q points. this means:
Ic=I same collector current for all of them
Ib = same base current for all of them
Ie=Iref same emitter current for all of them

using KCL we can write:

Iref=Ib7 + I
Ie7=Ib5+Ir3+b6
Ie7=Ie5=Ie6=Ib*(beta + 1)

or simply

Ie=Ib*201
I=Ib*200

since we know that
Iref=Ib+I=Ib+200*Ib=201*Ib
we can find base current directly:
Ib=9.4uA/201 = 46.766nA

and using diode model we can find Vbe for all transistors:

Vbe=Vt*ln(Ic/Is + 1)

where Vt is known (thermal voltage is assumed constant at roomtemp: Vt=0.0258V)

this means
Vbe=0.5341V

Ve5=Ve6=Ie*1k=Iref*1k=9.4mV
Ve7=Vb5=Vb6=Ve5+Vbe=0.5435V

Ir3=Ie-2*Ib=Iref-2*Ib=9.306uA
R3=Vb5/Ir3=58403 Ohm

Edit: ok Empire may have had a chance :)
 
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t_n_k

Joined Mar 6, 2009
5,455
Hi panic mode,

I would probably have used either the emitter current or the collector current to find VBE via the Shockley equation - rather than using the base current.

What do you think?
 
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