hello
i have a question about this circuit: http://www.israup.net/images/84dc5ddecc1b5ca9d43efabd6e557db2.png
Rb=470, R1=R2=1k (near the inductor) and L=2mH, Vcc=5V
the input is 0 and 5 in a square wave. the inductor responds when a current is changing. the response when the input voltage goes from 0 to 5 is much smaller then the response for 5 to 0. here is the pulses on Vx: http://www.israup.net/images/b0370a738c1c03e2eb7dd3f45465a053.png
you can see in the graph that for 0->5 there is a low peak of about 2.4V and for input 5->0 the peak is 9.6. my first question is why is that? as I understand, when the transistor is in cutoff there is no current Ic so the VCC see the two resistors in the first moment so the voltage in Vx will be (Vcc-0.2)*R/2R. when the input goes from 5->0 there is a current through the inductor so the inductor acts like short circuit and we do not count the parallel resistor so the Vx will be (Vcc-I(L)*R)-0.2
is there something wrong with this logic? I mean, i know i got the right values, I'm asking if my thought process is ok.
the second question is how to calculate the time constant? for some reason im getting the values 4.4usec (0->5) and 2,2usec(5->0). is there any difference between calculating the time constant for the inductor (L/R) and the voltage peak in Vx?
thank you for your help
i have a question about this circuit: http://www.israup.net/images/84dc5ddecc1b5ca9d43efabd6e557db2.png
Rb=470, R1=R2=1k (near the inductor) and L=2mH, Vcc=5V
the input is 0 and 5 in a square wave. the inductor responds when a current is changing. the response when the input voltage goes from 0 to 5 is much smaller then the response for 5 to 0. here is the pulses on Vx: http://www.israup.net/images/b0370a738c1c03e2eb7dd3f45465a053.png
you can see in the graph that for 0->5 there is a low peak of about 2.4V and for input 5->0 the peak is 9.6. my first question is why is that? as I understand, when the transistor is in cutoff there is no current Ic so the VCC see the two resistors in the first moment so the voltage in Vx will be (Vcc-0.2)*R/2R. when the input goes from 5->0 there is a current through the inductor so the inductor acts like short circuit and we do not count the parallel resistor so the Vx will be (Vcc-I(L)*R)-0.2
is there something wrong with this logic? I mean, i know i got the right values, I'm asking if my thought process is ok.
the second question is how to calculate the time constant? for some reason im getting the values 4.4usec (0->5) and 2,2usec(5->0). is there any difference between calculating the time constant for the inductor (L/R) and the voltage peak in Vx?
thank you for your help