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

 

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.

 

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...

 

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.

 

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.

Here's what you said that's currently still up..

 

Here's what you said that's currently still up..

Yes, Vec = Ve - Vc
Vc becoming more negative increases Vec.

 

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.

 

Yes, Vec = Ve - Vc
Vc becoming more negative increases Vec.

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.

 

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

 

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.

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.

 

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

 

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

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.

 

The condition for saturation in the book for a pnp is for Vcb >= 0.4V

 

The condition for saturation in the book for a pnp is for Vcb >= 0.4V

Which is quite reasonable.

 

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

 

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

 

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

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