Is the Claim that Ic = Ie and Veb = 0.7 Volts Accurate for this Circuit?

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The discussion centers on the accuracy of assuming Ic = Ie and Veb = 0.7 volts in a transistor circuit. Participants clarify that while the assumption simplifies calculations, the base current (Ib) is typically negligible compared to the emitter (Ie) and collector currents (Ic). The base current is often ignored in design, following the 1/10th rule, where the bias circuit current is chosen to be significantly larger than Ib. Calculations indicate that with a gain factor (β) of about 100, the base current remains much smaller than the other currents, validating the assumption. The final results of the calculations confirm the initial assumptions were correct, demonstrating the validity of the simplifications used.
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The current through the base of a transistor is always pretty small.

Often it is neglected to make the calculations easier.

I see you also have Ic=Ie.
These 2 currents are also not exactly the same.
It should be Ie=Ic+Ib.
So here the base current is also neglected.

How big do you think Ib is (approximately)?
 
I see you also have Ic=Ie.
These 2 currents are also not exactly the same.
It should be Ie=Ic+Ib.
So here the base current is also neglected.

The exercise tells me to assume Ic = Ie , that's why I wrote it :)

How big do you think Ib is (approximately)?

I'm told that the base current is negligible
 
Femme_physics said:
The exercise tells me to assume Ic = Ie , that's why I wrote it :)

I'm told that the base current is negligible

So if we treat Ib as being zero, then I1 is the same as I2...Btw, I believe you have a formula ##I_e = β I_b##, with β the amplification factor.
For a BJT transistor, β usually has a value of about 100.

In other words the base current is about 100 times as small as the emitter current.
 
What he said...

It's common for a circuit designer to choose the current flowing in the base bias circuit R1 and R2 to be at least 10 times the expected base current so that the base current can be ignored (engineers 1/10th rule). Sometimes worth assuming that they have done this, solving the rest of the problem to work out the collector current and then go back and check your assumption was correct.

In this case if we assume the base current can be ignored then IR1 is about 0.1mA

So the base is at about 8V and the emitter 8.7V

So Ie is about 12-8.7/R3 = 1.1mA

If the transistor is half decent it will have a gain of 100 so the base current about 1.1mA/100 = 0.011mA which is indeed ten times less than 0.1mA.
 
Ahh...yes... since IB is irrelevant the same current flows... true...

Thanks :) I'll solve it tomorrow and see if I get the same results.
 
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Looks ok. For completeness you should probably comment on the discharge time through R4. Negligible in this example but might not allways be.
 
Hey!
There is a little squirrel in there!
That looks alright. :)
 
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Thanks you two :)

And yea, everybody loves the Ice Age squirrel!^^
 

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