Diodes, electronics, Current question?

J

Janet

New member
How can I determine the peak forward current through each diode?

I know the answers are A. 7.86mA B. 8.5mA C. 18.8mA D. 19.4mA

When I work these problems. I'm not getting the right answers to match. Need help solving these.

Also how would I solve the output voltage for these?

Here is a picture of the circuits.

http://www.flickr.com/photos/22774022@N06/3981573381/
 
A. Current starts when the input exceeds 12.6 volts to overcome the diode and battery. At 30 volts peak, you will have 30–12.6 volts drop across the 2.2k, which gives you I = 17.4/2.2 = 7.91 ma (using 0.7 instead of 0.6 for the diode gives you 7.86b ma)

B. Current is in the other direction. At the peak, you have 30+12 volts, –0.7 or 41.3 volts across 2.2k = 18.8 ma

C. you can do the next two.

.
 
A. Current starts when the input exceeds 12.6 volts to overcome the diode and battery. At 30 volts peak, you will have 30–12.6 volts drop across the 2.2k, which gives you I = 17.4/2.2 = 7.91 ma (using 0.7 instead of 0.6 for the diode gives you 7.86b ma)

B. Current is in the other direction. At the peak, you have 30+12 volts, –0.7 or 41.3 volts across 2.2k = 18.8 ma

C. you can do the next two.

.
 
Answers:
A. Vout = 0.6 + 12 = 12.6 volts
I peak = (30 - 12.6)/ 2.2K = 7.9 ma
B. V out = 12 - 0.6 = 11.4 volts
i peak = (30 - 11.4)/2.2k = 8.45 ma
C. V out = -12 +0.6 = -11.4 v.
I peak = (30 - (-11.4))/2.2k = 18.8 ma
D. V out = -12-0.6 = -12.6 v
I peak = (30 - (-12.6))/2.2k = 19.36 ma

You need to understand the concept of a diode. If you find it difficulty to understand,hook it up and measure it.
 
A. the 12V opposes the forward current so

30V - 12V = 18V

the diode drops .7 volts

18V - .7V = 17.3V

I = V / R
I = 17.3V / 2200 ohms
I = 7.86mA

Hope that points you in the right direction
 
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