[SOLVED] how to calculate ib of PNP

[tronicman]

hello guys,
i would be thankful if you help me to find how to calculate ib of this PNP ??

 

[KlausST]

Hi,

hard to say without values. I guess 300 uA.

No - joke aside.

do you really want us to go give you a mathematical solution for all cases you can think of?
* DC, what range?
* AC, what frequency, amplitude
* What values of the parts?

****
some approches for DC cases: (for DC cases you may omit C1)
* when V_IN is about 5V7 --> no current flow through D1, R2, R1 --> Ib is about zero.

* when V_IN < (5V7 - D1_forward_voltage) ... (= when D1 becomes conductive) calculate: V_R1 = (5V7 - V_in - V_d1) * R1 / (R1 + R2)
* now you have some new cases:
.. * when V_R1 < V_BE_th: then ib is about zero
.. * When V_R1 > V_BE_th: then consider V_R1 is about V_BE. Calculate I_R1. calculate anew I_R2, the difference between both is Ib.

It all depends a lot on device types, temperature and so on.

****
Why don´t you do just an LtSpice simulation?

Klaus

 

[tronicman]

hello klaus,
sorry for the messing informations, for Vin it is varies from 0v to 2,5v and for R1=3320ohm and R2=510ohm, thank you for your answer,

 

[tronicman]

thank you for your answer, about I_R2 is it equal to ((5V7-V_in-v_d1)*R2/R1+R2)/R2??

 

[KlausST]

Hi

thank you for your answer, about I_R2 is it equal to ((5V7-V_in-v_d1)*R2/R1+R2)/R2??

you missed the brackets:
((5V7-V_in-v_d1)*R2/(R1+R2))/R2
then you may simplify it:
--> ((5V7-V_in-v_d1)*R2/(R1+R2))/R2
--> (5V7-V_in-v_d1)/(R1+R2)
(simple Ohm´s law: I = V / R)

Klaus