hello, i wanna know the objective of the attached schematic ,

 

From what I was it was a switch. The capacitor is to make it faster.

 

what do u mean by faster ?
the base current start with high value , then it will decrease.
i was asked about the maximum frequency of switching of transistor?

 

I believe the applicable term for this capacitor is that it is a "speed-up" capacitor. PRS has characterized the purpose of the capacitor nicely. The idea is that the impedance of the capacitor being an inverse function of the frequency of the signal applied to it. The net impedance of the capacitor and the resistor across which it is placed decreases as the frequency of the input signal increases. The result is that the input signal drives the transistor harder at higher frequencies.

hgmjr

 

It's guesswork with no component values, but the capacitor can be a high pass filter around the one base resistor, leading to faster turn-on with increasing frequency.

 

it was a question:
base voltage has a maximum value of 10v, peak base current is 1.5mA, steady state base current is 1 mA,

draw base voltage and current?
choose the values of C1 R1 R2?
what is the maximum permissible switching frequency of the transistor?

 

If this is a class assignment it should be relocated into the Homework Section. Is this the case?

hgmjr

 

it was a question in last year exam , i try to solve it and understand the objective of this configuration?

 

I think it is one of this 'stupid' exams questions.

Assume a Vbe of 0.7V.

To have a maximum current of 1.5mA and assuming the capacitor as short circuit when the switching happens:

R2=(10-0.7)/1.5mA=6.2K

To have a 1mA at steady state R1 has to be:

R1=(10-0.7)/1mA-R2=3.1K

To find C we need more info?

Is the datasheet of the transistor given.