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Can you please help me with this exercise?

In each circuit, calculate the current flowing through the 10kΩ resistor and calculate the voltage drop across the germanium diode. Please assume that a germanium diode has a forward-biased junction potential of 0.3V.

schematic diagram

Circuit (a)

Supply Voltage = 5V, R= 10kΩ, Junction potential = 0.3V, I =? VDrop=?

VD = (5V – 0.3V) = 4.7V
I = VD/R = 4.7V / 10kΩ = 0.47mA

Circuit (b)

Supply Voltage = 10V, R= 10kΩ, Junction potential = 0.3V, I =? VDrop =?

VD = (10V – 0.3V) = 9.7V
I = VD/R = 9.7V / 10kΩ = 0.97mA

Can you please review my solution?

I am not sure about the voltage drop, which in my calculation is indicated as VD. Also the second circuit, which is reverse biased, I'm not sure about the calculation I made.

Thanks in advance.

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    \$\begingroup\$ Look more closely at the two drawings. And remember what the basic function of a diode is. \$\endgroup\$
    – Hearth
    Jan 29, 2019 at 19:14
  • \$\begingroup\$ I had a science teacher back in grade school with the last name Dibenedetto. But back to your question. It's also important to note that you should consider drawing an equivalent circuit of a diode as well. \$\endgroup\$
    – user103380
    Jan 29, 2019 at 19:23

2 Answers 2

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You claim your second diode is reverse-biased. You can only make this assertion if it doesn't conduct current, and not because its terminals are switched.

Note that for forward-biased diodes, the forward-biased junction potential is the voltage drop from P to N, meaning that the voltage drop from N to P is the negative of this. If you look at your calculations, then you will see that you used the same polarity for both!

EDIT:

Please note that when you're assuming that the diode is forward-biased, the resultant current you calculate through the diode (from P to N) should be positive; if its negative, then your initial assumption that the diode is forward-biased is incorrect, and therefore it is reverse-biased (and hence doesn't conduct current). It was mentioned in the comments that OP may not know about this, so I provided clarification.

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    \$\begingroup\$ This answer implies that reverse-biased diodes generate power, which is false. \$\endgroup\$
    – Hearth
    Jan 30, 2019 at 0:14
  • \$\begingroup\$ I don't understand. Where have I said that it generates power? If the current through the diode is negative, then the initial assumption that it's forward-biased is incorrect. \$\endgroup\$
    – Metric
    Jan 30, 2019 at 0:22
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    \$\begingroup\$ You didn't say it, but you implied it, probably unintentionally, by saying that the asker should use the negative of the given potential drop without explaining why that's unphysical (since the asker seems unfamiliar with the basics of electronics, this may give them the wrong idea). \$\endgroup\$
    – Hearth
    Jan 30, 2019 at 0:26
  • \$\begingroup\$ Edited for further clarification. \$\endgroup\$
    – Metric
    Jan 30, 2019 at 0:31
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Bad terminology and bad circuit understanding.

In case (a) the diode is forward biased. The diode has voltage 0,3V as said. The rest of the battery voltage is over the 10 kOhm resistor and the circuit current is 0,47 milliamperes.

In case (b) you cannot assume the voltage over the diode, you must find it by reasoning. I guess the teacher wants you to recall Shockley's I versus V law for PN diodes. That formula says I=Is(exp(V/Vt)-1) where Is is the reverse saturation current which depends on diode materials and dimensions. Vt is the thermal voltage kT/q but that's about 0,026 volts at 300 kelvins. You must solve Is from 0,47mA = Is(exp(0,3/0,026)-1)

Then you should find in case b such negative V that the caused I causes the remaining diode voltage (after subtracting from 10 volts the resistor drop I * 10kOhm) to be just = the same V. That can be iterated only numerically. But you should see that Is is so small compared to 10V/10kOhm that you can well assume the circuit current I = Is. Then subtract from 10V drop Is * 10kOhm and that's your diode voltage in the reverse case. It's so near 10V that with usual low cost multimeter it's impossible to measure directly how much less it is than 10V.

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