Need help understanding constant current circuit.

peter taylor

Joined Apr 1, 2013
106
Woops, I meant ' Vbe1 and Vbe2 are below about 0.6 volts', and 'I3 =[ (V0 - V3 - V4) / R1 + B.V4 / R3 ] / (1 + B + B.B) ]' should read R2, not R3.
Why do I keep doing that ?
 

recklessrog

Joined May 23, 2013
985
Another point of view: R1 provides the only drive current to Q1. If Q1 has enough gain, it allows current to flow through the LEDs, through Q1, and through R2 until the voltage across R2 becomes enough to turn on Q2. As Q2 begins to allow more current to ground, the supply of current from R1 becomes less available to Q1 and the circuit arrives at equilibrium.

ps, you can't get 16 volts worth of LEDs to run on 10 volts. Tear it down to one LED and the circuit should start up with a bit less than 0.018 amps of current through the LED.
That was my interpretation of the circuit too :)
 

recklessrog

Joined May 23, 2013
985
If the Ice current of Q1 increases, then the voltage drop across R2 increases, which increases the Vbe of Q2, which increases the Ice of Q2, which increases the current in R1, which increases the voltage drop across R1, which lowers the base voltage on Q1, which lowers the Ice current in Q1. This forms what is known as a negative feedback loop in the circuit such that any change away from equilibrium starts a chain of events that reestablishes that equilibrium (or something very close to it).



Assume that the circuit is in equilibrium, which means that you will have Vbe of about 0.6V across R2. Size R2 to give the desired Ice you want in Q1.

Sizing R1 is a bit more complicated, but is also not nearly as critical. In this case let's say that the target current in Q1 is 20 mA, so R2=33Ω is a reasonable choice which would give about 18 mA. Assume that the transistors have a beta that is at least 100. That means that you need a base current of about 200 uA. The base voltage on Q1 is going to be about 1.2 V to 1.4 V, placing about 8.7 V across R1. Neglecting the current in Q2 for a moment, that would require R1 to be 43.5 kΩ. If the beta is actually higher, then we could use an even larger resistor. But we need to allow for some collector current in Q2, as well. So we make R1 quite a bit smaller (they chose 10 kΩ in this case). The negative feedback aspect of the circuit will simply turn Q2 on a bit more in order to sink the additional current. This has the effect of making the circuit very insensitive to variation in transistor beta.

To show how non-critical the actual choice of R1 is, if you were to use, say, 100 Ω for R1, then you would need about 87 mA of current in it. Nearly all of this would flow through Q2, which would then require a base current of about 0.9 mA. This would be supplied by the circuit increasing the drive to Q1 so that the collector current would be about 18.9 mA with 18 mA of that going through R2 (to keep the base voltage of Q2 at about 0.6 V) and the rest going to the base of Q2. With that much current in Q2, the needed Vbe would actually increase, possibly by 100 mV give or take, and this will increase the current in R2 by 10% to 20% as well. So even a radical 100x reduction in R1 will likely only increase the LED current by 25% or so.


Thats the difference between theoretical and real world.
 

peter taylor

Joined Apr 1, 2013
106
Voltage source, V0 = 20 V
VD = 8 x voltage across LED = 8 x 2 V = 16 V
R1 = R2 = 10 kΩ
B = 400
Base-emitter Voltage = Vbe1 = Vbe2 = V3 = V4 = 0.7 V

I1 = (V0 - V3 - V4) / R1
= (20 - 0.7 - 0.7) / 10 k
= 1.9 mA

I3 =[ I1 + B.V4 / R3 ] / (1 + B + B.B)
= [1.9 m + 400 x 0.7 / 10 k] / ( 1 + 400 + 400 x 400)
= 29.9 m / 160,401
= 187 uA

I4 = (I1 - I3) / B
= (1.9 m - 187 u) / 400
= 1.7 m / 400
= 4.3 uA

I5 = B.I3
= 400 x 187 uA
= 75 mA

I6 = V4 / R2
= 0.7 / 10 k
= 70 uA

V1 = I1.R1
= 1.9 m x 10k
= 19 V

V2 = V0 - V1
= 20 - 19
= 1 V

V6 = I6.R2
= 70 u x 10 k
= 0.7 V

I'm glad V6 works out to 0.7 V, because it should be equal to V4, which is just Vbe2.
Which confirms my equations.
And these are your quiescent voltages and currents.
Now we can see what happens when they change due to Temperature changes.
What else would upset this equilibrium ?
 

Thread Starter

dsharp02

Joined Aug 8, 2015
17
Everyone thanks for the responses! I'm still wading through all of the math. I'm sure I'll be back with more dumb questions in the future.

Dave
 

dannyf

Joined Sep 13, 2015
2,197
wading through all of the math.
There is very little math involved, to understand how it works, if you allow some reasonable assumptions, like the base current is sufficiently small, etc.

It is much easier to see the forest if you allow some "imprecision".
 

dl324

Joined Mar 30, 2015
16,845
There is very little math involved
I agree.

As I mentioned in post #17; LED current will be 0.7V/R2, R1 needs to provide sufficient base current to Q1 to accomplish that. One add and a couple divides are all you need to be concerned with.

You can approximate the base emitter voltage drops as 0.6, 0.7, or 0.8 and it will make little difference in your current calculations. Our response to light is logarithmic and differences of 50% in luminous intensity won't be objectionable, or noticeable.

However, I do advise you to not use a simulator until you have a better grasp of fundamentals. LTSpice will give you all the information you want, but very little of it matters for the circuit in question.
 

Jony130

Joined Feb 17, 2009
5,487
Voltage source, V0 = 20 V
VD = 8 x voltage across LED = 8 x 2 V = 16 V
R1 = R2 = 10 kΩ
B = 400
Base-emitter Voltage = Vbe1 = Vbe2 = V3 = V4 = 0.7 V
I5 = B.I3
= 400 x 187 uA
= 75 mA

I6 = V4 / R2
= 0.7 / 10 k
= 70 uA
I'm glad V6 works out to 0.7 V, because it should be equal to V4, which is just Vbe2.
Which confirms my equations.
And these are your quiescent voltages and currents.
There's something wrong with your solution. How can I5 (Ic1) be equal to 75mA if I6 (Ie1 - Ib2) = 70μA ??
Also your analysis is very hard to follow because you decided to use different designations than in original circuit.

And one of the possible solution can look like this:
Simply write a KVL for two loops.

(Ie1 - Ib2)*R2 - Vbe2 = 0
(1)
Vcc - (Ic2 + Ib1)*R1 - Vbe1 - (Ie1 - Ib2)*R2 = 0 (2)

And additional
Ie1 = Ic1+Ib1 = Ic1 + Ic1/β1
Ib1 = Ic1/β1
Ib2 = Ic2/β2

And after we substitute this into (1) and (2) we have this :

((Ic1 + Ic1/β1) - Ic2/β2)*R2 - Vbe = 0

Vcc - (Ic2 + Ic1/β1)*R1 - Vbe - (( Ic1 + Ic1/β1) - Ic2/β2)*R2 = 0

So now we can easily solve this simultaneous equations for Ic1 and Ic2.

\(Ic1 =\frac{\beta1*(R2*Vcc + R1*Vbe*\beta2 - R2*2Vbe)}{R1*R2*(1 + \beta2 + \beta1 \beta2)}\)

\(Ic2 = \frac{(R2*(Vcc - 2Vbe) (1 + \beta1) - R1*Vbe) \beta2}{R1*R2*(1 + \beta2 + \beta1 \beta2)\)

So for R1 = R2 = 10kΩ ; Vcc = 20V ; Vbe = 0.7V; β1 = β2 = 400 I got :

Ic1 = 0.0744634 mA = 74.4uA
Ic2 = 1.85981 mA
IR2 = (Ic1 + Ic1/β1) - Ic2/β2 = 0.07mA = 70μA


But no one with common sense do it this way.
In real life we simple notices that the voltage across R2 resistor is equal to Vbe.
So from this we already see that Ie1 = Vbe/R2 and Ic1 = (Vcc - 2Vbe)/R1.
And of course this equation "work" only if transistors work in active region, which is not the case for the circuit in post 1.
 
Last edited:

dannyf

Joined Sep 13, 2015
2,197
But no one with common sense do it this way.
The equations are good if you want to know the precise working points.

For someone noting understanding how the circuit works, the equations only add to confusion.
 

dannyf

Joined Sep 13, 2015
2,197
IR2 = (Ic1 + Ic1/β1) - Ic2/β2 = 0.07mA = 70μA
R2 provides the collector current for Q1 and the base current for Q2.

So IR2 = Ic1 + Ib2 = Ic1 + Ic2 / β2 ~= Ic1, assuming β2 being sufficiently big.
 

Alec_t

Joined Sep 17, 2013
14,280
The equations are good if you want to know the precise working points.
Not really (except as an academic exercise). Vbe is an unknown and is temperature dependent. Transistor beta is also unknown. So in practice you have to use approximate values and design circuits so that the values aren't critical.
 

dannyf

Joined Sep 13, 2015
2,197
Vbe is an unknown and is temperature dependent. Transistor beta is also unknown.
Pick a reasonable number for Vbe and beta, change the temperature within a reasonable range.

how much does the current through the diode change? how much does its luminosity change?
 

dannyf

Joined Sep 13, 2015
2,197
The circuit is simple to analyze numerically.

The voltage drop over R2 is Q2's Vbe. So the current going through R2 / LED / Q1 is 0.65/33=20ma;

The base of Q1 (=Q2's collector) sits one Vbe above R2 = 0.65 + 0.65 = 1.3v;

That means the current going through R1 / Q2 = (10 - 1.3) / 10k = 1ma.

Dahdah!

You can introduce other factors in but 10% of the work (=above) gets you within 90% of the correct answers.

That's the difference between an engineer (10% input for 90% of the output) and an engineering professor (90% input for 10% of the output).
 

dannyf

Joined Sep 13, 2015
2,197
BTW, the reason this circuit sinks a constant amount of current is right up there in those two short lines of calculation.
 
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