# Odd Transistor Circuit Solution

#### SamR

Joined Mar 19, 2019
4,334
Was given this question:

So, I worked the problem and almost came to the same solution. But, VsubCE = VsubCC - IsubC*RsubC. And I calculated 4.05V. It's not in saturation. IsubC is 16.1mA by may calculations and IsubCsat from the PDF is ~0.2V. What I don't understand is the key's VsubCE solution. They are taking the (what I calculated as) VsubCC and subtracting what I assume is the normal forward bias Voltage drop across the PN junction. But it is Reverse Biased! The circuit is not saturated so shouldn't the Voltage C to E be almost the full 3.35V? I did come to the conclusion that no parameters were exceeded.

#### MrAl

Joined Jun 17, 2014
9,333
Was given this question:
View attachment 273310
So, I worked the problem and almost came to the same solution. But, VsubCE = VsubCC - IsubC*RsubC. And I calculated 4.05V. It's not in saturation. IsubC is 16.1mA by may calculations and IsubCsat from the PDF is ~0.2V. What I don't understand is the key's VsubCE solution. They are taking the (what I calculated as) VsubCC and subtracting what I assume is the normal forward bias Voltage drop across the PN junction. But it is Reverse Biased! The circuit is not saturated so shouldn't the Voltage C to E be almost the full 3.35V? I did come to the conclusion that no parameters were exceeded.
Hi SamR

Looks like they made a mistake. If you have 4.05v at the collector and the emitter is grounded, then the Vce must be 4.05v and there is no way around that.
Maybe they got confused about the difference between the base and collector. The base is 0.7 above ground and you subtract that from the input voltage to get the base current but you dont do that with the collector voltage.
The only way that yellowed part could be true is if there was a diode in series with the emitter to ground and of course it was forward biased. That would also reduce the base current.

#### SamR

Joined Mar 19, 2019
4,334
K, I assumed it was erroneous but wasn't absolutely certain I hadn't missed something. Not the first error I've run across with this book... Yep, I'd already calculated IsubB to get IsubC. In fact, everything I calculated matched but the entire VsubCE part of the answer didn't jive. K, thx for the confirmation!

#### Audioguru again

Joined Oct 21, 2019
4,921
You cannot buy a transistor with "the maximum specified hFE". Instead you buy hundreds and test them all so you can be lucky and find one that has "the maximum specified hFE".

If you buy one transistor then it probably has "typical" hFE but it could have minimum or maximum hFE.

#### MrSalts

Joined Apr 2, 2020
2,043
You cannot buy a transistor with "the maximum specified hFE". Instead you buy hundreds and test them all so you can be lucky and find one that has "the maximum specified hFE".

If you buy one transistor then it probably has "typical" hFE but it could have minimum or maximum hFE.
Sounds like a new binning service could be implemented.
Back in the day, Motorola used to put various color dots of paint on the power RF transistors to indicate gain. Pink, Yellow, red, blue and green were the dot colors and could come with one or two dots. I have no memory of what range each color represented. You could order matched sets. Some distributor started the binning but Motorola eventually started offering matched pairs in-house for a premium price. You could buy pairs or quads of matched parts.

#### Audioguru again

Joined Oct 21, 2019
4,921
Transistors are rarely used at a certain hFE since the hFE of a transistor part number is a range of numbers. An emitter resistor adds some negative feedback so that the range of hFE works fine.

#### crutschow

Joined Mar 14, 2008
30,420
The equation would be correct for calculating Vcb, not Vce.

#### SamR

Joined Mar 19, 2019
4,334
The equation would be correct for calculating Vcb, not Vce.
I guess I'm having a hard time understanding exactly how the reverse bias Vcb barrier potential is 0.7V. Forward yes, but not reverse? Also, still a bit cloudy about when Vce(sat) is 0.2V and you have the V across Rc =IcRc that it doesn't add up to Vcc? In fact for Vce you have 2 barrier potentials Vcb and Vbe. Still hasn't gelled for me as yet.

#### MrAl

Joined Jun 17, 2014
9,333
I guess I'm having a hard time understanding exactly how the reverse bias Vcb barrier potential is 0.7V. Forward yes, but not reverse? Also, still a bit cloudy about when Vce(sat) is 0.2V and you have the V across Rc =IcRc that it doesn't add up to Vcc? In fact for Vce you have 2 barrier potentials Vcb and Vbe. Still hasn't gelled for me as yet.
Hi,

Vce doesnt have any barrier potential unless the voltage drops below the base voltage. When Vc drops below Vb, then the collector itself starts to draw current away from the base. That's why we often state that for saturation you should think of the transistor as having a Beta of just 10.
So for Vb=0.7v and Vc=4v with the emitter grounded, the CB junction is reverse biased so only the leakage current flows. When Vc drops to 0.7v nothing else happens, but when it goes below 0.7v then it starts to draw current away from the base. It actually wont draw too much current until the collector Vc gets down to around 0.5v (with base still at 0.7v) and then it may start to make a difference, but the more significant is when it gets down to 0.4v below the base, so in this case that would be 0.3v, and then it will start drawing more current away from the base. That action makes it look like the transistor now has less gain (Beta) and that's again why the rule of thumb is a Beta of 10 for saturation.

Note however really good quality bipolar transistors will behave better than this. They could hold the Beta higher due to the way they are made, and have significantly less Vsat..

I think maybe you would do well to look at a simplified model of an NPN bipolar Transistor.
Do you know how a current controlled current source works?
You could set up a circuit with two regular diodes like 1N4148 and a current controlled current source and see how this all works first hand in a simulator. It's very interesting to look at and think about. Once you do this i am betting you will never have another problem figuring out how this works.

I can post a schematic if that helps. It's just 3 components, and some input source like a current source or resistor and voltage source.

#### crutschow

Joined Mar 14, 2008
30,420
how the reverse bias Vcb barrier potential is 0.7V.
You appear to be having a little problem with Ohm's law.
0.7V is the forward Vbe voltage.
You subtract that from the Vce to get Vcb.

#### Danko

Joined Nov 22, 2017
1,392
I guess I'm having a hard time understanding exactly how the reverse bias Vcb barrier potential is 0.7V. Forward yes, but not reverse? Also, still a bit cloudy about when Vce(sat) is 0.2V and you have the V across Rc =IcRc that it doesn't add up to Vcc? In fact for Vce you have 2 barrier potentials Vcb and Vbe. Still hasn't gelled for me as yet.
It looks like that:

___
EDIT:
Since the CB junction is reverse biased, the transistor is operating in active mode vs. saturation.

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#### dl324

Joined Mar 30, 2015
14,897
So, I worked the problem and almost came to the same solution. But, VsubCE = VsubCC - IsubC*RsubC. And I calculated 4.05V. It's not in saturation.
The problem is $$V_{CE}$$. Did you write that or was it provided?

By the calculation, it's obvious that it's calculating $$V_{CB}$$. Since the CB junction is reverse biased, the transistor is operating in active mode vs. saturation.

If you use LaTeX, your text would be easier to read. This:
But, VsubCE = VsubCC - IsubC*RsubC
becomes this:
$$V_{CE} = V_{CC} - I_C*R_C$$

#### SamR

Joined Mar 19, 2019
4,334
Do you know how a current controlled current source works?
Not at all other than using a resistor.

You could set up a circuit with two regular diodes like 1N4148 and a current controlled current source and see how this all works first hand in a simulator
Unfortunately, the diode models for LTS are not available...

I can post a schematic if that helps. It's just 3 components,

I'm going to spend some time simulating and breadboarding and taking measurements. I need to better understand the Vce as the transistor comes out of cutoff and reaches full saturation.

#### dl324

Joined Mar 30, 2015
14,897
I need to better understand the Vce as the transistor comes out of cutoff and reaches full saturation.
A more accurate way to determine if a transistor is in saturation is to measure the CB voltage. When it's reverse biased, the transistor is in active mode. When it's forward biased, the transistor is in saturation mode.

The transistor is out of cutoff mode as soon as the BE junction is forward biased.

#### SamR

Joined Mar 19, 2019
4,334
$$V_{CE}$$ K. looks like I've got a start on latex...

#### dl324

Joined Mar 30, 2015
14,897

#### crutschow

Joined Mar 14, 2008
30,420
I need to better understand the Vce as the transistor comes out of cutoff and reaches full saturation.
Here's a simulation showing the characteristics of a typical NPN BJT from cutoff to saturation versus the base current (horizontal axis):
The transistor goes from the active region to saturation when the Vce collector voltage (green trace) becomes less than the Vbe base voltage (yellow trace) at about 48µA base current with a collector load current near its maximum of 10mA (blue trace).

Notice the (Vce - Vbe) or Vcb voltage (red trace).

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#### SamR

Joined Mar 19, 2019
4,334
This is AAC's LaTeX tutorial. I don't think it covers the \( shortcut.
That's strange because I used the AAC LaTex window to enter V_{CE} and it added the rest of the notation.

#### dl324

Joined Mar 30, 2015
14,897
I don't think I ever used the Insert->Math option.

For complicated things or many lines, I prefer to use Word or Writer.

This is from Writer:

As I recall, Word is easier. But I don't have it on the computer I'm using.