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Consider the circuit below. Assume that SW1 is open. What voltage drop is observed across D1? What voltage drop is observed across Q1 (from collector to emmitter)? Why?

Since the base current is zero, we know that Q1 is in cutoff, and there will be no current through D1, R1, or Q1. So Ohm's Law tells us that there is no voltage across the resistor: the voltage drop across D1 and Q1 must add up to V1 = 5V. But without any current flowing, none of the I-V characteristic equations for either the diode or transistor would be applicable. So how can we reason about the results?

schematic

simulate this circuit – Schematic created using CircuitLab

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The simplest models of a diode and transistor can't answer this question. You need to use a more detailed model of your devices to understand what will happen.

In particular, the transistor will allow a small leakage current to flow, even in cut-off mode. From the Fairchiled 2N3904 datasheet:

enter image description here

Of course your circuit can't apply more than 5 V to the collector, so the leakage will be somewhat lower, maybe 10 nA.

10 nA or even 50 nA through the diode will produce almost no voltage drop. 10-50 nA through the 100 ohm resistor will similarly produce only a few microvolts of drop. Thus the bulk of the voltage drop (~5 V) will appear across the transistor's collector and emitter.

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  • \$\begingroup\$ To expand on this, shouldn't we expect to see 5V across the collector-base junction, and 0V across the emitter-base junction? If the EBJ were forward biased, carriers would diffuse across the depletion region. Since there is little current in the circuit, most of those carriers would stay in the BJT. This means they would recombine, widening the depletion region and increasing the barrier voltage. So a forward-biased PN junction that is not conducting is not possible (or at least, unstable). In contrast, a reverse biased PN junction (the CJB) conducting pico or nano amperes is quite normal. \$\endgroup\$ – odougs Jul 30 '16 at 2:03
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Consider the circuit below. Assume that SW1 is open. What voltage drop is observed across D1?

0 V. There is no current through it so voltage drop is zero. If there was any voltage drop across it the diode would start to conduct a tiny current.

What voltage drop is observed across Q1 (from collector to emmitter)? Why?

5 V. Because it is open circuit.

Since the base current is zero, we know that Q1 is in cutoff, and there will be no current through D1, R1, or Q1. So Ohm's Law tells us that there is no voltage across the resistor:

Correct.

the voltage drop across D1 and Q1 must add up to V1 = 5V.

Correct.

But without any current flowing, none of the I-V characteristic equations for either the diode or transistor would be applicable. So how can we reason about the results?

Q1 is an open circuit. D1 isn't and its I-V equations still stand. Look at it the other way around: if no current is flowing through the LED then the voltage across it must be zero.

Imagine we switch on the 5 V to your circuit very quickly and let's say that due to some stray capacitance inside the LED and transistor that the collector instantaneously jumps to 2.5 V. What will happen? Q1 is still open so no current flows there. The LED is forward biased so current will flow there. It will be a very short pulse. OK, lets work it out very roughly.

enter image description here

_Figure 1. 2N3904 \$ C_{CBO} \$ and \$ C_{EBO} \$ are quoted as 4 and 18 pF.

I can't find a figure for a typical collector-emitter capacitance value so lets go with the higher of the two from Figure 1, 18 pF. Your R1 is 100 Ω. We can calculate the discharge time constant \$ \tau = RC = 100 \times 18p = 1.8~ns \$. This is very crude but the general idea is that in a real circuit the capacitance would be discharging as the circuit powered up. In any measurements you took you would see Q1's collector voltage rise up with and equal to the supply voltage.

Remember: even though the LED won't light the I-V curve shows that the slightest increase in forward voltage above zero will result in a (very small) current.

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  • \$\begingroup\$ I think I understand the concept now, but I find it confusing to say that Q1 is "open circuit" while D1 is not, since they are in series. I now think it may be better to consider Q1 as two back-to-back PN junctions. Then by applying the I-V equations for PN junctions, it becomes clear that the bulk of the 5V must appear across the collector-base junction, because only a reverse biased PN junction can support that much voltage while conducting only pico or nano amperes. \$\endgroup\$ – odougs Jul 30 '16 at 2:11

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