# Make collector current in active region for BJT?

Discussion in 'Homework Help' started by ihaveaquestion, May 7, 2010.

1. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
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Can anyone explain the underlined part in this image?

http://img205.imageshack.us/img205/3047/img0003to.jpg

If you can't read it, it says:

"Since the maximum current that the collector can support while the transistor remains in the active mode is approximately 0.5 mA, it follows that the transistor is definitely saturated."

Thanks

2. ### Ghar Active Member

Mar 8, 2010
655
73
For the transistor to be in active mode you need EB forward biased and CB reverse biased.

The drop of EB, VEB is about 0.7 V.
The drop of CB is lower than this, and Sedra & Smith generally assume 0.5 when forward biased, 0.3 when 'on'.
You need VCB < 0.3 to be in active mode (to be reverse biased)
This is equivalent to VEC > 0.4 (because VEC = VEB - VCB)

The approximation they're doing is that VB = 0 and VE = 0.7, and then VC is about 0.7 - 0.4, which is near 0.
If VC = 0, then IC = 5 / 10k = 0.5 mA.

If IC is less than about 0.5 mA you will VC will become negative and VEC will increase, tending to be in active mode as IC decreases.

If IC is more then VC will become more positive, reducing VEC and the transistor will go closer to saturation.

Edit:
Made quite a lot of corrections... hopefully it's good now.

Last edited: May 7, 2010
3. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
You mention that Ve = 0.7 V (which I agree with), but if Vc becomes more negative, then Vce = Vc - Ve will decrease, not increase like you mentioned.

Why, though, does the transistor remain active only up to this special 0.5 mA number? There's something missing...

4. ### Ghar Active Member

Mar 8, 2010
655
73
It's Vec, not Vce, for a PNP, it's one of those things I changed... re-read my post. I changed almost everything...

The requirement comes from needing the CB junction reverse biased in active mode.

5. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Here's what you said that's currently still up..

6. ### Ghar Active Member

Mar 8, 2010
655
73
Yes, Vec = Ve - Vc
Vc becoming more negative increases Vec.

7. ### mik3 Senior Member

Feb 4, 2008
4,846
63
The book says that for this circuit the maximum collector current is about 0.5mA. From the calculations, it was found that Ic is about 4.3mA (assuming it is equal to Ie). The transistor will try to conduct this 4.3mA, however, because the maximum Ic is 0.5mA, the transistor saturates in its try to push the current to 4.3mA.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783

Ghar is correct.

If Ic=0.4mA then VEC≈4.6-(-1)=5.6V

If Ic=0.5mA then VEC≈4.5-0=4.5V

So as Ic decreases VEC increases - taking the transistor more into the active region.

9. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
0
Thanks for the clarification...

I suppose what I'm getting at is I'm not clear on 1) where they are getting the 0.5 mA number and 2) what dictates going from active to saturdation with what we are dealing with... how are 4.3 mA and 0.5 related... when do we know we're going between regions etc...

Thanks

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Hi Mik3 - did you consider the presence of the 10k resistor in the base?

The current in the emitter-base path is limited by both the 1K emitter and 10K base resistors.

11. ### Ghar Active Member

Mar 8, 2010
655
73
The definition of active vs saturated is the region of the junctions.

Active:
EB is forward
CB is reverse

Saturation
EB is forward
CB is forward

The junctions aren't the same, which gives you different voltages.
EB fully on = 0.7 V
CB fully on = 0.5 V

VEC = VEB - VCB = 0.7 - 0.5 = 0.2
That's VEC in saturation.

S&S generally assume that CB turns on around 0.3 V and you want CB off, so VCB < 0.3
This gives the requirement of VEC > 0.7 - 0.3 = 0.4 V to be in active.
That's the requirement - VEC > 0.4 for active.

In this case you assume VB = 0 for simplicity, it'll be slightly different obviously.
Then you get VE = 0.7.
So if you have VEB = 0.7 and want VEC > 0.4 you get the requirement of VC < 0.3

VC is directly related to IC through the resistor:

VC = -5 + 10k*IC < 0.3
This gives IC < 0.53 mA

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The text is making a rough estimate of the collector current value that would indicate the onset of saturation. The estimate is that VB is at roughly 0V - which incidentally it isn't - and that the onset of saturation occurs when VC and VB are equal. VC would be zero volts when the Ic is 0.5mA. So roughly speaking, saturation would begin at Ic = 0.5mA.

13. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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The condition for saturation in the book for a pnp is for Vcb >= 0.4V

14. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Which is quite reasonable.

15. ### Ghar Active Member

Mar 8, 2010
655
73
Yeah you're right, I remembered wrong.
It's just a rule of thumb however, so it doesn't matter what it is exactly.

16. ### ihaveaquestion Thread Starter Active Member

May 1, 2009
314
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No need to apologize... as an unrelated question, how does the beta value B affect BJT circuits, or more specifically, the three terminal currents?

Thanks

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Consider the relationships .....

Ic=βIb
Ie=(1+β)Ib

Have you checked the forum notebooks?

e.g.

http://www.allaboutcircuits.com/vol_3/chpt_4/4.html