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here two diodes are connected in series, one is forward biased and the other is reverse biased, are connected across 5V. thermal voltage is given as 26mV. the voltage drop across each diode is asked.

enter image description here since one of the diode is reversed biased the current through the circuit will be reverse saturation current of the diode. so by applying shocley's equation we can find the voltage across each diode. but that doesn't add upto 5V

is there any mistakes in my method

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  • \$\begingroup\$ Hint: Most of the voltage drop will be across the reverse biased diode. \$\endgroup\$ – JIm Dearden Oct 23 '16 at 14:49
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The voltages must add up to 5V, that is a constraint.

The current through each diode must be the same- they're in series.

If you use Shockley's equation for each diode and apply those constraints you should be able to solve it.

Note that you can be sure that the voltage across the reverse biased diode is between -4.4V and -5V so the current should be pretty easy to approximate if you have a realistic idea of what Is should be.

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  • \$\begingroup\$ I applied shockely's equation \$\endgroup\$ – Ampatishan Arun Oct 23 '16 at 14:48
  • \$\begingroup\$ But for both diodes the equations appears to be same for me \$\endgroup\$ – Ampatishan Arun Oct 23 '16 at 14:48
  • \$\begingroup\$ The sign is different on the voltage \$\endgroup\$ – Spehro Pefhany Oct 23 '16 at 14:54
  • \$\begingroup\$ Only on voltage or on both voltage and diode current \$\endgroup\$ – Ampatishan Arun Oct 23 '16 at 14:56
  • \$\begingroup\$ Yes, on both, that's important. \$\endgroup\$ – Spehro Pefhany Oct 23 '16 at 15:00
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You can accurately calculate the Voltage drop in each diode from theoretical Shockley equation if you know all the parameters of which diodes are used.

This is not how I would compute the answer in an exam, but rather as a design engineer, choose a diode and see the exact results.

The better way is to choose a common diode and read the datasheet curves to compute the effective voltage drop at the estimated current where that slope represents the incremental or effective series resistance (ESR) with a small offset voltage which we know is 26mV @20'C at zero current for a silicon diode. But we can neglect this for practical reasons.

The most common old small signal silicon diode was the 1N4148. We know one diode will be in reverse league mode and hence have most if not all the voltage drop, and will be much higher impedance than your 10 MOhm DMM so you would not be able to measure its impedance or voltage drop using this method, but we can calculate its voltage drop directly from Figure 2 & 3. You may be inclined to extrapolate the curves since the applied voltage is only 5V and diode leakage drops with voltage like a linear resistor. You may find fixed leakage currents at a fixed voltage, but better datasheets such as this show the actual behaviour over a wide range of currents. But again for practical reasons when you get down to nanocurrents, you can round off.

So why not measure the typical Resistance of reverse leakage @20'C as shown for typical diode datasheet.

Then do the same for the other diode using this effective current limit in micro or nanoamps to determine the forward voltage. This too you may find "off the chart edge" so we can round off the values to say, "3 significant figures" and know exactly what these values are.

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enter image description here

So again we know the current is same thru both, but one is forward biased while the other is reverse biased so let's call it Vf and Vr, where Vf+Vr=5V in this question.

Although it is an academic question ,

  • What is Vf ?

  • What is Vr ?

    • such that total adds up to 5V for this example.

ref : datasheet and distributor pricing

http://www.digikey.com/product-detail/en/fairchild-semiconductor/1N4148TR/1N4148FSTR-ND/458811

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