I want to find current through Zener diode. correct answer is 10.3 mA.
I tried to solve it this way.
i found Ib( base current)
Ib = Vcc-Vbe-Vz/10 so 20-10-0.7/10 = 0.93 mA.
My answer is wrong. Can you correct me?
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2 Answers
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Are you calculating the base current or Zener current? The base voltage will be 10V, so the emitter voltage would be 9.3V. The emitter current would be 9.3/10 = 0.93A. Base current would be 9.3mA. The voltage across the base resistor is 10V, so the current is 20mA. The Zener current is Ir - Ib = 20 - 9.3 = 10.7mA
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\$\begingroup\$ thanks. can you tell me how you calculate voltage across base resistor?? \$\endgroup\$– BeginnerCommented Mar 10, 2017 at 13:44
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\$\begingroup\$ Since the zener holds the base at 10V, the drop across the resistor is 20-10 = 10V. \$\endgroup\$– AngeloQCommented Mar 10, 2017 at 13:45
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\$\begingroup\$ Zener current is Ir - Ib, not Ib - Ir. Obviously just a typo :) \$\endgroup\$– ColinCommented Mar 10, 2017 at 14:00
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it is not that simple.
Two things going on here:
- the zener could dominate the b-e junction + resistor: this would be a case where the beta is sufficiently high. essentially the base current would be so small that the voltage drop on the 0.5k resistor isn't enough to shut off the zener.
In this case, 9.7v to follow through the 10R resistor, -> Ie = 1a, and Ib = 10ma. that current will cause a voltage drop over the 0.5K resistor of 5v. insufficient to shut down the zener.
So the zener will be conducting at 10v, and the total current going through the zener is 10v / 0.5K = 20ma. 10ma of that goes through the transistor so the remaining 10ma through the zener.
you don't have the following case but you could.
2) the base current is so large that it could shut off the zener. See the 0.5K resistor is really 5K. going through the same math and you will conclude to that the voltage drop off the 5K resistor now is 5K * 10ma = 50v -> invalidating the assumption that led us to this calculation. This means that the zener would not be conducting reversely. Instead, the b-e junction + 10R leg dominates. that means the current through the zener is 0ma.
the key is to make an assumption, follow through the calculation and then see if that answer is consistent with your assumption.
without doing that, you really don't know if the answer is valid or not.
edit: it is not hard to calculate the beta at which the circuit switches from one scenario to another. that's when Ib * 0.5K = 10v = 0.7v + Ie * 10. or Ib = 20ma, and Ie = ~1amp. or beta = 50.
that is: With a beta > 50, the zener will conduct; and with a beta < 50, the zener will not.
