Tricky BJT Circuit

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Please post your best effort at solving the problem, so that we can guide you where necessary.

newbe_always said:

problem attached.It is very confusing to me

Click to expand...

The circuit is not "tricky".
Why is it confusing to you?
It should not be if you understand how BJTs work.

NPN or PNP?

The circuit is not "tricky" - it is crazy (with regard to the shown values and polarities)

I say that the circuit is "tricky", as in, there are several things which don't make sense.

The transistor is drawn as a PNP type.
A PNP transistor has the emitter connected to the +ve side of the supply and the collector to the -ve side.

The "supply" is drawn as a battery (a single cell actually), the long stroke of the battery symbol is usually the +ve side.
But here we have the +ve side connected to the collector and the -ve side connected to the emitter.
This is the wrong way around for the PNP transistor.

We are told that Icq is 3mA, I assume this means the quiescent collector current is 3mA.
I assume that the transistor has high gain so the base current is very small compared with the collector current, so we can say that the emitter current = collector current = 3mA.

We can calculate the voltage across the emitter resistor using V = I x R
so
V = 3ma X 500 Ohm = 1.5 volts.

We can calculate the base voltage, measured with respect to the common point of the circuit (the earth symbol).
So,
Base Voltage = Emitter voltage + Emitter- Base voltage

For a silicon transistor, base-emitter voltage will be about 0.5 to 0.6 volts when the transistor is biased into conduction, so we have:
Base voltage = 1.5v + 0.5v = 2.0 volts. (Negative with respect to the common point).

Looking at the circuit in the question, the base voltage is shown as -8volts (with respect to the common point.

So what is it 2v or 8v on the base?
Damned if I know!

As I have been typing, LvW has described the circuit as "Crazy", I think that he is correct.

JimB

Presumably it's one of the many completely wrong exam questions which have been doing the rounds for decades (long before the internet). In the final of the Sharp Electronics Engineer of the Year competition one year they had a similar completely wrong question, except the circuit was OK, but none of the four options were correct. I, and many others, added a fifth option with the correct answer - and when we came out of the exam room we all harassed the Sharp staff running it.

In a fine example of 'passing the buck' they all threw Jim under the bus - then ran away as we all surrounded Jim.

He admitted setting the exam paper, but had just culled questions from old exam papers, and never actually checked the answers were correct.

He then said he'd remove that question from the exam, where we then said what about those who added the correct answer?.

I can't remember what he said to that, but it doesn't matter, I won the Competition that year anyway

Generally speaking you start by mentally recording what you know
and do not know.



We know the magnitude but not the sign of the collector current.
We do not know if transistor is biased in linear mode.
We do not know sign of current in power source to right of transistor, yet.
We know a junction, if forward biased, is ~ .7V for silicon and moderate
currents.

So you know magnitude of current in collector, but you have a P-N-P,
and that transistors, generally speaking, want their E-B forward biased,
C-B reversed biased, to be operating in linear mode. Stated another
way to operate such that Ic ~= Ie, that Ibase is << Ic OR Ie

But if Vcb has C-B junction forward biased then Vce will not be biasing
the C-B junction so the current flow sign opposite

So is it ?

So that process step by step will either lead you to solution, or finding
circuit does not make sense.

JimB said:

We can calculate the base voltage

Click to expand...

It is actually shown as -8V (presumably with respect to ground).

Deleted

I don't see anything fundamentally wrong with the circuit, except that the Vcc voltage source is shown with a polarity indication. The student needs to decide whether Vcc will be positive or negative.

I'd interpret it as the forced -8V on the base causing the emitter to be at somewhere around -7.3 to -7.7V, depending on the transistor type. Call it -7.5V

The transistor will be fully saturated with something around 15mA base current and the only 3mA collector current, so the collector will be at virtually the same voltage as the emitter.

With 3mA through a 2K collector load resistor there will be 6V across that resistor.

-7.5V to the emitter then another -6V collector to power = -13.5V supply voltage.

newbe_always said:

Can you please solve the problem attached.It is very confusing to me

View attachment 149174

Click to expand...

This question is tricky. The reasons follow.

The design overdrives the base more than the collector, but this part is irrelevant. Even if it is a poor design, you must still answer the question.
The battery symbol, Vcc is shown backwards for a PNP, so it must be negative.

You may assume -0.1V drop for Vce or Vec=0.1 , although at this low current, Vec = 0 is closer to the truth. (20mV) as a very low current saturated switch.

The base current is irrelevant, but emitter voltage determines the KVL drop to Vcc.
You need to know that Icq is always determined by Vbe in linear mode and if saturated and we neglect Vce* then Icq= (Vcc-Ve)/Rc and Ie=Ic+Ib means Ic is not controlled by Ib.

Since Icq was given +ve , we will subtract it here only , Vbe = -0.7 V.

Thus all you need Ve = Vb + Vbe = -7.3 , Vc~ Ve - 0.1 ( -ve for PNP)
V(Rc) = 3 mA * 2k = -6V Thus Vcc = -7.4 - 6V = -13.4



However if Vb=-8 then assuming 0.65V, But for approx. use 0.7 or 0.6 for single precision.
Ve=-7.3 as the emitter is more +ve with P-N junction for emitter-base.

Then from Ohm's Law V/R=I= -7.3 / 500 Ohms 14.6 mA approx. for base current is much more than Icq.

Yes weird, unconventional, tricky, poor design, but still answerable.

Other info
Here as a switch although all data sheets by convention will use Ic/Ib = 10 for Vce(sat) specs, or = 20, 30, 50 and hfe must be ~10 times this ratio or more to see these specs..

*When saturated, the collector PN junction which was reverse biased as a current source is now forward biased as a voltage switch across C-E. Vbe (or Veb) is slightly more than Vce due to a higher bulk resistance. You can estimate Vce(sat) by knowing the approx. Rce = Vce(sat)/Ic which for small transistors ranges from 1 to 5 Ohms. Often we just assume,
|Vce|=0.1V before calculating but it is actually the difference in two diode voltages due to Rce * Ic

rjenkinsgb said:

I'd interpret it as the forced -8V on the base causing the emitter to be at somewhere around -7.3 to -7.7V, depending on the transistor type. Call it -7.5V

Click to expand...

Veb vs Ib depends on the device and temperature.
You suggested Veb= 0.3 to 0.7 @ 15 mA and chose an average of 0.5.
That would be true for << 1mA

I think Veb >= 0.6 V @ 15 mA and use 0.7 V for a general situation.

The typeface and typesetting (process used before word processing) as exemplified by the uneven subscript "cc" in Vcc, tells me this is from a very old textbook - possibly from when germanium PNP transistors with negative supplies (aka "positive ground") were the norm.

Tony Stewart said:

V(Rc) = 3 mA * 2k = -6V

Click to expand...

May I know how Vrc Comes -6V
Below is my solution .I made a mistake at this 6V

rjenkinsgb said:

I'd interpret it as the forced -8V on the base causing the emitter to be at somewhere around -7.3 to -7.7V, depending on the transistor type. Call it -7.5V

The transistor will be fully saturated with something around 15mA base current and the only 3mA collector current, so the collector will be at virtually the same voltage as the emitter.

With 3mA through a 2K collector load resistor there will be 6V across that resistor.

-7.5V to the emitter then another -6V collector to power = -13.5V supply voltage.

Click to expand...

May I know ,How the sign of 6V becomes -6V

newbe_always said:

Below is my solution

Click to expand...

Your solution is wrong. Note that Ve is more positive than Vb for a PNP transistor.

newbe_always said:

May I know how Vrc Comes -6V
Below is my solution .I made a mistake at this 6V
View attachment 149192

Click to expand...

If may help to redraw with conventional higher above lower or negative supplies.

But if using KCL or KVL it shouldn't matter as long as you recognize the forward bias condition causes the saturation condition for both Vec and both PN junctions are in "forward conduction mode."



My interactive simulation has more decimal places and assumes a primitive generic model with fixed hFE (beta) so I added a slider. I also added a slider for the negative collector supply, Vcc . https://tinyurl.com/28rl872j Don't worry about tolerances for now. https://tinyurl.com/28gtqdma

Ground is just a symbol for 0 V which can be defined anywhere when it is floating.