Direct coupling circuit.. my attempt... I get too many equations...

Alice24

Joined Apr 22, 2011
44

MrChips

Joined Oct 2, 2009
25,018
I would begin by assuming that both transistors are fully saturated.
Hence you can calculate Q2 collector and emitter current, IC2.
This will give you the voltage across R2.
From this you should be able to determine IB1.

Jony130

Joined Feb 17, 2009
5,288
Try this approach

5V = Ib1 * R5 + Vbe1 + Vbe2 + Ie2 * R2 (1)

Ie2 = Ib2 * (β2 + 1)

Ib2 = Ib1 * (β1 + 1)

So finally

Ie2 = (Ib1 * (β1 + 1) ) * (β2 + 1)
(2)

And this is all we need to solve this circuit.

Alice24

Joined Apr 22, 2011
44
I would begin by assuming that both transistors are fully saturated.
Why? In this case Vce1 and Vce2 = 0 , although we do get to make the problem so much easier to ourselves, who says we're allowed to do it?

Ie2 = (Ib1 * (β1 + 1) ) * (β2 + 1) (2)
Ahh...I was unfamiliar with this formula!

Are there more formulas specific to directing coupling connection I might have missed?

MrChips

Joined Oct 2, 2009
25,018
What I said in post #2 would not work.
I did some calculations and assuming both transistors are turned on would not provide enough Vbe turn on voltage.

So here is another approach that will take perhaps two iterations.

Assuming Ib1 is small, and Vbe = 0.6V
we can estimate VR2 = 5 - 1.2 = 3.8V
Hence IR2 = 19mA
If we assume β = 100 for both transistors
we can determine Ib1

Use this and recalculate VR1 and VR2.

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Alice24

Joined Apr 22, 2011
44
But MrChips, you make a lot of assumptions. Thing is, we don't have to assume, we know VBE and Betas It's given to us.

As a student (me), I don't think I am allowed to take such privileges with assumptions...

I'm more curious about how to use the formulas I was provided!

I am trying to construct all possible formulas to this transistor... As well as:

Ie1 = Ib1 + ic1
Ie2 = ib2 + ic2
beta1 = ic1/ib1
beta2 = ic2/ib2
ie1 = ib1 (1 + beta 1)

What am I missing?

MrChips

Joined Oct 2, 2009
25,018
Sorry, I did not see that Vbe and beta were given at the top of the post.
Then use those values.

I think you have enough information now to solve for Ib1.

Last edited:

Jony130

Joined Feb 17, 2009
5,288
Your "special" equation in post 7 are mostly incorrect.

Ib1 = ( Vb - 2Vbe)/ (R5 + R2 *( (β1 + 1) * (β2 + 1) ) )

Ie1 = IB1 * (β1 + 1)

Ib2 = Ie1

Ie2 =
Ib2 * (β1 + 1)

Ic1 =
Ib1 * β1

Ic2 = Ib2 * β1

MrChips

Joined Oct 2, 2009
25,018
Typos on the two lines:

Ie2 = Ib2 * (β2 + 1)

Ic2 = Ib2 * β2