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13th December 2019, 14:40 #1
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13th December 2019, 15:20 #2
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Re: how to calculate ib of PNP
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?
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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13th December 2019, 15:25 #3
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Re: how to calculate ib of PNP
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,

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13th December 2019, 15:42 #4
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Re: how to calculate ib of PNP
thank you for your answer, about I_R2 is it equal to ((5V7V_inv_d1)*R2/R1+R2)/R2??
Last edited by tronicman; 13th December 2019 at 15:48.

13th December 2019, 16:07 #5
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Re: how to calculate ib of PNP
Hi
thank you for your answer, about I_R2 is it equal to ((5V7V_inv_d1)*R2/R1+R2)/R2??
((5V7V_inv_d1)*R2/(R1+R2))/R2
then you may simplify it:
> ((5V7V_inv_d1)*R2/(R1+R2))/R2
> (5V7V_inv_d1)/(R1+R2)
(simple Ohm´s law: I = V / R)
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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