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PN Junctions
19 questions By Tony R. Kuphaldt
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Question 16 of 19
In order to simplify analysis of circuits containing PN junctions, a “standard” forward voltage drop is assumed for any conducting junction, the exact figure depending on the type of semiconductor material the junction is made of.
How much voltage is assumed to be dropped across a conducting silicon PN junction? How much voltage is assumed for a forward-biased germanium PN junction? Identify some factors that cause the real forward voltage drop of a PN junction to deviate from its “standard” figure.
Reveal answerSilicon = 0.7 volts ; Germanium = 0.3 volts.
Temperature, current, and doping concentration all affect the forward voltage drop of a PN junction.
Notes:I’ve seen too many students gain the false impression that silicon PN junctions always drop 0.7 volts, no matter what the conditions. This “fact” is emphasized so strongly in many textbooks that students usually don’t think to ask when they measure a diode’s forward voltage drop and find it to be considerably different than 0.7 volts! It is very important that students realize this figure is an approximation only, used for the sake of (greatly) simplifying junction semiconductor circuit analysis.
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Question 17 of 19
∫f(x) dx Calculus alert!
A forward-biased PN semiconductor junction does not possess a “resistance” in the same manner as a resistor or a length of wire. Any attempt at applying Ohm’s Law to a diode, then, is doomed from the start.
This is not to say that we cannot assign a dynamic value of resistance to a PN junction, though. The fundamental definition of resistance comes from Ohm’s Law, and it is expressed in derivative form as such:
$$R= \frac{dV}{dI}
The fundamental equation relating current and voltage together for a PN junction is Shockley’s diode equation:
$$I = I_S(e^{\frac{qV}{NkT}} -1)$$
At room temperature (approximately 21 degrees C, or 294 degrees K), the thermal voltage of a PN junction is about 25 millivolts. Substituting 1 for the nonideality coefficient, we may simply the diode equation as such:
$$I = I_S(e^{\frac{V}{0.025}} -1) \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ I = I_S (e^{40V} - 1)$$
Differentiate this equation with respect to V, so as to determine \(\frac{dI}{dV}\), and then reciprocate to find a mathematical definition for dynamic resistance \((\frac{dV}{dI})\) of a PN junction. Hints: saturation current (IS) is a very small constant for most diodes, and the final equation should express dynamic resistance in terms of thermal voltage (25 mV) and diode current (I).
Reveal answer$$r \approx \frac{25mV}{I}$$
Notes:The result of this derivation is important in the analysis of certain transistor amplifiers, where the dynamic resistance of the base-emitter PN junction is significant to bias and gain approximations. I show the solution steps for you here because it is a neat application of differentiation (and substitution) to solve a real-world problem:
$$I=I_S(e^{40V}-1)$$
$$\frac{dI}{dV}=I_S(40e^{40V}-0)$$
$$\frac{dI}{dV}=40I_Se^{40V}$$
Now, we manipulate the original equation to obtain a definition for IS e40 V in terms of current, for the sake of substitution:
$$I=I_S(e^{40V}-1)$$
$$I=I_Se^{40V}-I_S$$
$$I+I_S=I_Se^{40V}$$
Substituting this expression into the derivative:
$$\frac{dI}{dV}=40(I+I_S)$$
Reciprocating to get voltage over current (the proper form for resistance):
$$\frac{dV}{dI}= \frac{0.025}{I+I_S}$$
Now we may get rid of the saturation current term, because it is negligibly small:
$$\frac{dV}{dI} \approx \frac{0.025}{I}$$
$$r \approx \frac{25mV}{I}$$
The constant of 25 millivolts is not set in stone, by any means. Its value varies with temperature, and is sometimes given as 26 millivolts or even 30 millivolts.
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Question 18 of 19
Measure the forward voltage drop of a silicon rectifying diode, such as a model 1N4001. How close is the measured forward voltage drop to the “ideal” figure usually assumed for silicon PN junctions? What happens when you increase the temperature of the diode by holding on to it with your fingers? What happens when you decrease the temperature of the diode by touching an ice cube to it?
Reveal answerDid you really think I was going to give away the answer here, and spoil the fun of setting up an experiment?
Notes:Diodes are quite temperature-sensitive, so this experiment will be very easy to conduct. You may not have ice available in your classroom, but that’s okay. Your students should realize that experiments such as this are perfectly fair to perform at home, where they probably do have access to ice.
