What the current according to the vinput
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this is not homework this ciruit i use it in my workI'll bite. Sounds like homework. What do you think the current is?
yes the input is poistiveWhat's that diode doing on Vinput? Have you drawn it correctly?
Is your input voltage positive or negative?
If it is correct and positive then your homework will be easier!
if the input is 0 vdc ,this mean that diode is forward biasing and emitter of Q1 WILL BE ZERO VOLTS AND IF THE INPUT IS HIGHER THAN(1.89+.7) ,THE DIODE WILL BE REVERSEBIASED AND VB OF Q1 WILL BE 1.89 VDC ,SO IS THIS TRUE?I'll bite. Sounds like homework. What do you think the current is?
SO WHEN VIN =0 VDC THE CURRENT IN 50 OHM IS ABOUT 416 MICRO AMP AND WHEN THE DIODE IS REVERSE BIASED THE CURRENT ABOUT 449 MICRO AMP? SO IS THIS TRUE?if the input is 0 vdc ,this mean that diode is forward biasing and emitter of Q1 WILL BE ZERO VOLTS AND IF THE INPUT IS HIGHER THAN(1.89+.7) ,THE DIODE WILL BE REVERSEBIASED AND VB OF Q1 WILL BE 1.89 VDC ,SO IS THIS TRUE?
r4I dont see a 50 ohm resistor on the supplied circuit.
R4I dont see a 50 ohm resistor on the supplied circuit.
R4 CONNECTED TO COLLECTOR OF Q2
Hi Crutshow can you see if the current at vin=0 is 416ua and at vin>2.6 the current is about 449 uaI'll bite. Sounds like homework. What do you think the current is?
No, for sure. For Vin = 0V Q1 is Cut off and Q2 is equal toHi Crutshow can you see if the current at vin=0 is 416ua and at vin>2.6 the current is about 449 ua
so E of Q1 will be .8v and B of Q1 will be about 1.5 v and what the voltage of the cathode of diode will be?No, for sure. For Vin = 0V Q1 is Cut off and Q2 is equal to
Ic2 = ((15 * 3/30) - 0.7 + 15)/18 = 0.8777mA. And Q1 will start to turn on if Vin > (15 * 3/(30) - 0.7) = 0.8V and Ic2 current will start to drop from 0.877mA to 0A
and the diode will be out of circuit when vin >(1.89v+.7) and when Q1 will turn on E will be about 15 v that is why Q2 will be offso E of Q1 will be .8v and B of Q1 will be about 1.5 v and what the voltage of the cathode of diode will be?
and also when vin =zero this mean that diode forward biased and most of the current will go through the diode so Q1 will be off?
and if vin is negative Q1 will never turn on?and the diode will be out of circuit when vin >(1.89v+.7) and when Q1 will turn on E will be about 15 v that is why Q2 will be off
thanks so much and what is vd? is this v anode of diode or what?Without the diode Q1 base voltage is equal to :
Vb1 = 15 * 3.9/(27 + 3.9) = 1.89V. And if we assume that diode will be OFF if Vd =0.5V. For Vin > 1.89V - 0.5V = 1.39V diode is OFF.
And yes, for negative voltage diode is ON so Q1 is OFF.