Problem with darlington circuit

[Les Jones]

Hi MikeMl,
I agree that with a real transistor an iterative solution would be required. The value for the hfe of the transistors quoted by the OP (443.3) is higher than the maximum value of 300 shown on the data sheet that I have. Also the value of hfe varies with the collector current so the method I used would not work with a real transistor. I don't see the point of this exercise as the result is so dependent on the hfe value which has quite a wide spread for any particular type of transistor.

Les.

 

[MikeMl]

This one is specifically for aliraza54 :

Everything you wanted to know or were afraid to ask when β=443.3:

Note that Ib(Q2) is only 0.22uA, much less than you thought...

 

[spec]

If you assume,
(1) perfect transistors (no leakage or hFE variation with IC and VC)
(2) that both transistors have a constant VBE of 0.6V
from Kirchhoff, an equation for the circuit is:

18V - (18- [0.6 + 0.6V) = (iB * 3.3 * 10^6) + (iB * hFEp * 390)
Which reduces to,
16.8 = (iB * 3.3 * 10^6) + (iB * hFEp * 390)
Which can be transformed to,
16.8 = iB * (3.3 * 10^6) + iB * (hFEp * 390)
Thus,
16.8/iB = (3.3 * 10^6) + (hFEp * 390)
Thus,
iB = 16.8/ (3.3 * 10^6) + (hFEp * 390)

Where:
iB : base current of the first transistor in Amps
3.3 * 10^6 : the 3.3M resistor from the 18V supply line to the base of the first transistor
hFEp : product of the hFE of the first transitor and the hFE of the second transistor
390; 390 Ohm resistor in the emitter of the second transistor

Which, if my maths is correct, shows that iB is dependent on the only variable in the circuit, hFEc. This can also be determined by an initial analysis of the circuit.

From this you can calculate the output voltage (voltage across the 390 Ohm resistor) for any hFEc and thus any combination of gains for the first and second transistors.

spec
(crossed posts)

 

[MikeMl]

If you assume,
(1) perfect transistors (no leakage or hFE variation with IC and VC)
(2) that both transistors have a constant VBE of 0.6V
from Kirchhoff, an equation for the circuit is:

18V - (18- [0.6 + 0.6V) = (iB * 3.3 * 10^6) + (iB * hFEp * 390)
...

Could you show how you got that equation?

 

[spec]

Could you show how you got that equation?

From Kirchhoff, it is the sum of the voltage drops across the 18V supply:
iB * the 3.3M base resistor
+ VBE of transistor 1
+ VBE of transistor 2
+ iB * Hfe tran 1 * Hfe tran 2 * 390 Ohm resistor.

I tried for hours to make the voltage drop across the 390 Ohm resistor the dependent variable but my head started hurting.

I would be very interested if any of you maths gurus could do the job.

spec

 

[Colin]

"It is the combination of the 3.3 meg and the two transistors that behave like a 16.79 ohm resistor. This is because the voltage drop across this virtual resistor will be the same as the voltage drop across the 3.3 meg resistor but the current through it will be the product of the two hfe values times the current that would be passing through the 3.3 meg resistor."

I have shown that you cannot say the 390R is like 16.79 ohms.
If you assume the maximum gain of each transistor, the current through the 390R would be something like 120mA.
So the transistors do not deliver their full capability.
But the gain of each is unknown. The only thing you can assume is the top transistor will provide a higher gain because it has the lower collector current.
Because there is more capability with the two transistors, then the circuit requires, almost any pair with a gain of 50 to 300 will provide about 16v across the 390R.
You don't have to go into any mathematics because the gains are completely unknown.

 

[MikeMl]

Try this: (using the reference designators as in post#22)

Ib2 = (18 -Vb2)/R2 = (18-Vb2)/3.3e6 (1)
Ic2 = β2*Ib2 (2)
Ie2 = Ib2+Ic2 (3)
Ib1 = Ie2 = Ib2 + β2*Ib2 = (1+β2)*Ib2 (4)
Ic1 = β1*Ib1 (5)
Ie1 = Ib1 + Ic1 = Ib1 + β1*Ib1 = (1+β1)*Ib1 (6)
Ve = Ie1*R1 = 390*(1+β1)*Ib1 = 390*(1+β1)*(1+β2)*Ib2 = 390*(1+β1)*(1+β2)*(18-Vb2)/3.3e6 = 390*(1+β1+β2+β1*β2)*(18-Vb2)/3.3e6 (7)

Equ 7 is the first of two equations in two unknowns. The second is:

Vb2 = Ve + 0.65 +0.65 = Ve + 1.3 (8) (I prefer Vbe=0.65V for Si transistors)

Substituting 8 into 7 gets:
Ve= 390*(1+β1+β2+β1*β2)*(18-(Ve + 1.3))/3.3e6 (10)

Suppose β1 = β2 = 443

Ve = 390*(1+443+443+443^2)*(18-(Ve + 1.3))/3.3e6
Ve = 1.181e-4*197136*(16.7-Ve)
Ve = 388.8 - 23.28*Ve
Ve = 388.8/24.28 = 16.01

Which, if you look back at post #22, is very close to what LTSpice predicted... (See V(e) = 15.9103V)

ps. I believe LTSpice more because it knows what Vbe2 and Vbe1 really are instead of our rule-of-thumb guess of 0.65V

 

[spec]

"It is the combination of the 3.3 meg and the two transistors that behave like a 16.79 ohm resistor. This is because the voltage drop across this virtual resistor will be the same as the voltage drop across the 3.3 meg resistor but the current through it will be the product of the two hfe values times the current that would be passing through the 3.3 meg resistor."

I have shown that you cannot say the 390R is like 16.79 ohms.
If you assume the maximum gain of each transistor, the current through the 390R would be something like 120mA.
So the transistors do not deliver their full capability.
But the gain of each is unknown. The only thing you can assume is the top transistor will provide a higher gain because it has the lower collector current.
Because there is more capability with the two transistors, then the circuit requires, almost any pair with a gain of 50 to 300 will provide about 16v across the 390R.
You don't have to go into any mathematics because the gains are completely unknown.

?

You will notice that I make no personal attacks like you did with Les in post #18

 

[spec]

Try this: (using the reference designators as in post#22)

Ib2 = (18 -Vb2)/R2 = (18-Vb2)/3.3e6 (1)
Ic2 = β2*Ib2 (2)
Ie2 = Ib2+Ic2 (3)
Ib1 = Ie2 = Ib2 + β2*Ib2 = (1+β2)*Ib2 (4)
Ic1 = β1*Ib1 (5)
Ie1 = Ib1 + Ic1 = Ib1 + β1*Ib1 = (1+β1)*Ib1 (6)
Ve = Ie1*R1 = 390*(1+β1)*Ib1 = 390*(1+β1)*(1+β2)*Ib2 = 390*(1+β1)*(1+β2)*(18-Vb2)/3.3e6 = 390*(1+β1+β2+β1*β2)*(18-Vb2)/3.3e6 (7)

Equ 7 is the first of two equations in two unknowns. The second is:

Vb2 = Ve + 0.65 +0.65 = Ve + 1.3 (8) (I prefer Vbe=0.65V for Si transistors)

Substituting 8 into 7 gets:
Ve= 390*(1+β1+β2+β1*β2)*(18-(Ve + 1.3))/3.3e6 (10)

Suppose β1 = β2 = 443

Ve = 390*(1+443+443+443^2)*(18-(Ve + 1.3))/3.3e6
Ve = 1.181e-4*197136*(16.7-Ve)
Ve = 388.8 - 23.28*Ve
Ve = 388.8/24.28 = 16.01

Which, if you look back at post #22, is very close to what LTSpice predicted... (See V(e) = 15.9103V)

ps. I believe LTSpice more because it knows what Vbe2 and Vbe1 really are instead of our rule-of-thumb guess of 0.65V

?

 

[MikeMl]

?

What didn't you understand?

 

[spec]

What didn't you understand?

Your circular statements.

spec

 

[MikeMl]

Your circular statements.

spec

Since I solved two equations in two unknowns, the statements cannot be circular. There would not have been a solution if they were...

Again, specifically, which of the equations (1) through (10) do you not understand?

 

[Les Jones]

Hi spec,
I can see no fault in MikeMl's equations. I realise now there was a mistake in my reasoning for my virtual resitor method. It's value should have been R2/(1+β1)*(1+β2) Not R2/(β1*β2) as I had based my calculation on.
This is the equivalent circuit that my method is based on.

Using the values in the above calculation Rv = 3300000/444 *444 = 3300000/197136 = 16.74 ohms. Using the vbe values of 0.65 volts we have 18 - 1.3 volts = 16.7 volts so the current through the resistors is 16.7/(390 +16.74)
= 16.7/406.74 = 41 mA so the voltage across the 390 ohm resistor is 15.99



Les.

 

[MikeMl]

...I tried for hours to make the voltage drop across the 390 Ohm resistor the dependent variable but my head started hurting.

spec,
Note that is exactly what I showed you how to do as equ. 7 in post #27, to whit:

Ve = 390*(1+β1+β2+β1*β2)*(18-Vb2)/3.3e6

I also showed that

Vb2 = Ve + 0.65 +0.65

The rest is pure algebra...

There is another circular statement.

So, are you going to leave us in suspense as to what it is?

Mike you do not understand- if you have some specific comment about my theories I would be very pleased to hear about it in specific technical terms.

I didn't even comment on what is wrong with the equations in your post #23. You however had no reservations about discounting my post as "circular" (which it is not).

In the mean time, you do what you want to do and leave my stuff out of it.

A truely rational approach to a scientific discussion; not!

If you are afraid that someone disagrees with what you post here, then don't post!

 

[spec]

Yes Mike- I completely misunderstood you post- you are quite right.

spec