Discussion in 'General Electronics Chat' started by TAKYMOUNIR, Nov 9, 2014.
What the current according to the vinput
I'll bite. Sounds like homework. What do you think the current is?
What'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!
this is not homework this ciruit i use it in my work
yes the input is poistive
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?
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?
I 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 ua
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
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?
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 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
and if vin is negative Q1 will never turn on?
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.
thanks so much and what is vd? is this v anode of diode or what?
Vd = V_anode - V_cathode = diode forward voltage drop