Transistor Current source

[gogoman]

Hi I'm a bit confused.

the current source has a emitter resistor Re, If  Rc limits the transistor current Rc~= Re,
why is Re required?  If Re =1ohm  and Rc = 10K,  Rc dictates the current Ic.

thanks

[exe]

This doesn't look like current source to me. Anyway, Re is "emmiter degeneration resisor", it is used to provide some negative feedback to stabilize bjt and make the circuit less sensitive to temperature and gain : https://en.wikipedia.org/wiki/Common_emitter#Emitter_degeneration

[soldar]

Re limits base current and, in fact, fixes collector current. Just analyze the model.

[rstofer]

Consider a fixed voltage on the base.  This will try to hold the emitter at 0.7V (give or take) below the base and the emitter resistor times the current flow creates that emitter voltage.  If too much current tries to flow, the base-emitter voltage drops (because the emitter voltage rises and the base voltage is fixed) and the transistor shuts down a little bit.  If too little current is flowing, the base-emitter voltage increases and the transistor turns on a little harder.

[mikerj]

Hi I'm a bit confused.

the current source has a emitter resistor Re, If  Rc limits the transistor current Rc~= Re,
why is Re required?  If Re =1ohm  and Rc = 10K,  Rc dictates the current Ic.

thanks

You are looking at this the wrong way, the transistor is not operating as a saturated switch so Rc does NOT limit the collector current.  Collector current is determined by setting the emitter current  Ic ~ Ie = (Vb-Vbe)/Re.  Since some base current has to flow through the emitter, the collector current is a little lower but for a high gain transistor it's close.

[rstofer]

Hi I'm a bit confused.

the current source has a emitter resistor Re, If  Rc limits the transistor current Rc~= Re,
why is Re required?  If Re =1ohm  and Rc = 10K,  Rc dictates the current Ic.

thanks

Yes, you can pick values that won't work.  Vcc = 7000V ought to work for your example but you will have a current of 0.7A.

Above I gave you the theory, all you need to do is pick resistors that work for your transistor.  BTW, the base divider resistors that provide the base voltage probably need to pass 10x the base current so that variations in base current don't affect base voltage.

Start with the current you want and calculate the emitter resistor and base voltage (you can 'guess' somewhere between 0.6V..0.7V).  Now, as long as the load resistance times the design current is less than Vcc you should be good to go.  Give or take some round-off error.  Like Ie=Ic+Ib and we're calculating Ie knowing that Ic will be somewhat lower depending on H_FE

This paper wants to assume Vbe=0.6V.

https://www.electronics-notes.com/articles/analogue_circuits/transistor/active-constant-current-source.php

[pwlps]

Hi I'm a bit confused.

the current source has a emitter resistor Re, If  Rc limits the transistor current Rc~= Re,
why is Re required?  If Re =1ohm  and Rc = 10K,  Rc dictates the current Ic.

thanks

If Rc dictates the current then it is not a "transistor current source" anymore.   In this circuit you should always have  beta*Ib<<Vcc/Rc so that the load current is controlled (i.e. limited) solely by the transistor collector current Ic=beta*Ib and does not depend on Rc.

The Re is not needed in principle here, but if Re=0 then your Ib current (and thus the Ic current Ic=beta*Ib) will vary exponentially with VB=VBE, which is not practical.
With Re, as calculated in the cited ref.
https://www.electronics-notes.com/articles/analogue_circuits/transistor/active-constant-current-source.php
we have approximately
Ic=(Vb-0.6)/Re  which is much more practical and, more important, does not depend on the beta of the transistor.

Compare this with another version where instead of Re you put a resistor Rb in series between the input and the base to control the base current. Then
Ib=(Vb-0.6)/Rb  and
Ic= beta*(Vb-0.6)/Rb

so that Ic depends on beta. The role of the emitter resistor Re is to introduce a negative feedback making the final Ic/Vb gain roughly independent of the transistor parameters (look at the calculations: beta appears only in the factor beta/(beta+1) roughly equal to 1).  This is the same principle as in  the opamp circuits : the gain is determined by the feedback resistor network and to some extent is independent of the open-loop gain of the amplifier.