PNP Transistor Biasing resistor calculation

[KerimF]

The typing error was just R1 had to be R2 on the last 3 lines.
R2 = ( 5 - 0.7 ) / 1.189 = 4.3 / 1.189 = 3.62 right?

 

[mailus]

YES,but

R1 = 0.7 / 0.4 = 1.75 Kohm , [calculated_4]

so I_r1= Vbe_q1/R1

I_r1=0.7/1.8K only (not 0.7/1.8)

I_r1=388uA

I_r1 = 0.7 / 1.8 = 0.389 mA , [calculated_5]

so
I_r2 = Ib + I_r1 , [equation_3]
I_r2 = 0.8 + 0.389uA = 0.800389 mA , [calculated_6]

 

[KerimF]

Oh I see... you are confused by the Kohm

0.7 / 1.8 = 0.388 Amp
0.7 / 1.8K = 0.388 mA
0.7 / 1.8Meg = 0.388 uA

 

[mailus]

I am so Sorry.............

little bit confused.......
388uA=0.388mA

- - - Updated - - -

Can you help me how to point out this hfe value?

In this case (saturated Q1), the ratio Ic/Ib could be estimated about 20.
G_i = Ic / Ib = 20 , [estimated_4]

 

[KerimF]

Can you help me how to point out this hfe value?

I will try

Let us see what happens if the current Ib is relatively big, say 100mA, and hfe is 200 (as read on the datasheet).
What could be Ic in this case?
Ic = hfe * Ib = 200 * 100 = 20,000 mA = 20 A

In our simple circuit, here, this Ic value of 20A cannot be reached. So what is the limit of Ic in a real circuit supplied by 5V and the collector load is 1Kohm?

 

[mailus]

but in my circuit i only control 20mA. that is Ic=20mA.

so Ib=Ic/hfe;

Ib= 20 mA /200=100uA only;

then why we can't take hfe;

 

[KerimF]

Ok it seems my question on post #25 wasn't clear enough so you couldn't answer it.

if Rc is 1K (connected between Vcc and an NPN collector) and since the emitter (for this NPN transistor) is connected to ground, the current in Rc cannot exceed:
Ic_max = Vcc / Rc = 5 / 1K = 5 mA (in this case Vce = 0, actually it could be made close to zero but not zero)

If you get the point, you see that even for Ib = 100 mA, Ic (in this example) cannot exceed 5 mA and the ratio Ic/Ib = 5/100 = 1/20 ONLY
But even if we decrease Ib to 5 mA, Vce will still be close to zero hence Ic = 5mA and Ic/Ib = 1 only.

As you see, when the transistor is over-driven (or saturated, because Ib is relatively high), Beta is no more a characteristic of the transistor but it is defined by the external circuit of the transistor. In this case, Vce is always close to zero and Ic is almost constant and equals to Vcc/Rc.


Now let us assume Ib small, say 0.01 mA. If beta is 100 then Ic should be:
Ic = 100 * 0.01 = 1 mA

This current is less than the previous limit of 5 mA (Ic_max). This means Vce is no more close to 0V (ground):

Vce = Vcc - Ic * Rc
Vce = 5 - 1mA*1K
Vce = 4V

Let us keep Ib as 10uA (0.01mA), and increase Rc from 1K to 10K. In this case:
Vce = 5 - 1mA*10K
Vce = - 5 V

But this result is unreal because Vce cannot be less than 0 volt, therefore:
Vce = 0 = 5 - Ic * 10K
Ic = 0.5 mA

As you see, Ic now is independent of Beta.

I am not a good teacher and I find hard to show you the difference between Beta of the linear region (when Vce > 1V) and the ratio Ic/Ib of the saturation region (Vce < 1V). Unfortunately the transition between the two regions (linear to saturation) is not abrupt (and may not start at 1V, sometimes less or higher than 1V) this is why you saw me assuming Ic/Ib = 20 for Vce = 200mV (these values are based on my personal observations and are just a start up ones, also they differ for small, medium and power transistors).

I hope someone else will help you get this from another angle.