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I'm trying to teach myself some basic electronics and I'm stuck on understanding what's going on with this simple NPN transistor circuit that is using a diode for bias.

enter image description here

I've simulated the circuit with SPICE and I get a base voltage of about 0.61 V and a collector voltage of about 0.85 V. I also built the circuit and get very similar values on my breadboard.

It seems that the transistor is conducting, otherwise the collector voltage would be 9 V, but if it's conducting that means that there's a current through the base and thus the diode, in which case the drop across the diode should be about 0.7 V, but instead I'm getting about 0.24 V.

Something seems to be wrong/missing from my mental model of either how the diode or the transistor (or everything!) works. I'd be very grateful if someone can help me fix that.

Changing R1 to provide different diode currents and voltage drops (measured from C-B):

1      364 mA  1.14 V
10     13.4 mA 0.75 V
100    521 uA  0.58 V
1k     45.2 uA 0.46 V
10k    4.87 uA 0.35 V
100k   568 nA  0.24 V
1MEG   73 nA   0.14 V
10MEG  10 nA   0.06 V
100MEG 1.65 nA 0.02 V
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    \$\begingroup\$ The diode ensures that the collector is a diode drop above the base. The BJT is fully out of saturation, now. And no, I've no idea why it was done as you don't provide any context. The diode voltage is low because there's not much current in it. Aren't you aware of the Shockley diode equation? \$\endgroup\$
    – jonk
    Jun 7, 2020 at 20:15
  • \$\begingroup\$ Are you sure the diode really is a 1N4148 (Silicon junction) and not a Schottky diode of some kind? \$\endgroup\$ Jun 7, 2020 at 20:38
  • \$\begingroup\$ @BrianDrummond, the diode in my test circuit is labelled ST4148 if I'm reading it correctly. I assumed this meant that it's an 1N4148 manufactured by ST Electronics. Perhaps that's not true? \$\endgroup\$
    – Rob Duncan
    Jun 7, 2020 at 21:58
  • \$\begingroup\$ @jonk, thanks for directing me to the Shockley equation. My basic diode model was that there is negligible diode current until it crosses the threshold. \$\endgroup\$
    – Rob Duncan
    Jun 7, 2020 at 22:04

2 Answers 2

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The Shockley diode equation, expressed so that you get the diode voltage as a function of diode current, looks like:

$$V_D=\eta\,V_T\,\ln\left(1+\frac{I_D}{I_\text{SAT}}\right)$$

Aside from the thermal temperature, \$V_T\$, the two key parameters defining the dioide's behavior are the emission co-efficient, \$\eta\$, and the saturation current, \$I_\text{SAT}\$. It turns out that \$I_\text{SAT}\$ is also highly temperature-dependent. But assuming that the temperature is constant, at \$T=27^\circ\text{C}\$, we can take \$V_T\approx 26\:\text{mV}\$ and just worry about the specific values of \$\eta\$ and \$I_\text{SAT}\$. The model I have in LTspice provides \$\eta=1.752\$ and \$I_\text{SAT}=2.52\:\text{nA}\$.

I've also examined the default \$\beta\$ for the 2N3904 model in LTspice. It's \$\beta=300\$.

Here's a simulation that varies the load resistance over the wide range you specified (\$1\:\Omega\$ to \$100\:\text{M}\Omega\$) and displays a few interesting details:

enter image description here

Please note that I am using a .STEP card to modify the load resistance value, automatically. This way I don't have to do separate runs and write down answers one at a time. Instead, I can just plot useful information.

I've chosen to plot three different values. The green line is the diode voltage. The dark blue line is the diode's current (also the BJT's base current.) And the red line is the BJT's computed \$\beta\$ value.

Before we dig in too deeply, let's check out a few hand-calculations. First off, from the above equation, for every \$10\:\text{X}\$ change in diode current I can expect to see \$1.752\cdot 26\:\text{mV}\cdot \ln\left(10\right)\approx 105 \:\text{mV}\$ change in the diode voltage. That should be the approximate slope of the dark blue line. We can also compute an arbitrary value for the diode voltage. Let's say we want to work this out for \$I_D=1\:\mu\text{A}\$. I'd compute \$1.752\cdot 26\:\text{mV}\cdot \ln\left(1+\frac{1\:\mu\text{A}}{2.52\:\text{nA}}\right)\approx 273 \:\text{mV}\$. Now, on the right side find the tick for "1e-006A" and move left until it intersects the dark blue line. Now, go directly downward from that intersection until you find the green line. Note that this is about \$270\:\text{mV}\$. Very close to prediction.

Now, if you examine the curves, you'll find some interesting details. The red line should be flat at \$\beta=300\$, but it isn't. This is because the BJT experiences current-crowding and Ohmic resistance issues that, among other reasons, complicate the actual \$\beta\$. It doesn't reach a flat \$\beta\approx 300\$ until the load is about \$1\:\text{k}\Omega\$ (which corresponds to about \$I_C\approx 8.3\:\text{mA}\$.) If you examine the OnSemi datasheet for the 2N3904, then you will see this:

enter image description here

Which shows you that the \$\beta\$ does start to decline at about that collector current. So this is about what should be expected.

With lower \$\beta\$ (moving leftward on the above chart), the diode will experience rapidly growing (relative to the collector's current) currents on its own. So you'd expect the diode to show a change in slope for its diode voltage as you go from about \$1\:\text{k}\Omega\$ to about \$100\:\Omega\$. And, in fact, you do see that change roughly in that region. But the diode appears to be remarkably flat, from just a little between \$100\:\Omega\$ and \$1\:\text{k}\Omega\$ to a little between \$1\:\text{M}\Omega\$ and \$10\:\text{M}\Omega\$. That's more than six orders of magnitude! Maybe even seven! Not bad.

When the currents get very tiny in the diode, then other new effects take place. These include the formation of emitter-base surface channels; the recombination of surface carriers, and the recombination of carriers in the emitter-base space-charge layer. So once again, it's not "flat" as these new effects begin to dominate in the very low current regime.

I'll leave you to inspect the chart. Hopefully, you'll leave from that experience a little more comfortable with the whole situation.

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  • \$\begingroup\$ thank you very much for taking the time to provide such a detailed and informative answer. I am puzzled by this: "With lower 𝛽 (moving leftward on the above chart), the diode will experience rapidly growing currents on its own". How is the diode current affected by the transistor current gain? \$\endgroup\$
    – Rob Duncan
    Jun 8, 2020 at 16:05
  • \$\begingroup\$ @RobDuncan \$\beta=\frac{I_C}{I_B}\$, by definition. If the transistor's \$\beta\$ declines for the reasons stated (or more), then either \$I_C\$ declines with respect to \$I_B\$, or \$I_B\$ increases with respect to \$I_C\$, or both. In short, since only small voltage changes are involved and the resistor current stays mostly the same, more current is required to be diverted towards the base to sustain the resistor current, which means more current in the diode. But I mean that "with respect to." But I don't mean that on an absolute scale. Just a relative one. \$\endgroup\$
    – jonk
    Jun 8, 2020 at 16:46
  • \$\begingroup\$ got it! Thank you very much for all this. \$\endgroup\$
    – Rob Duncan
    Jun 8, 2020 at 20:39
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The circuit is low-current.

With 100K ohm in collector, at most the collector current is 9v/100,000 and if we consider a 100,000 ohm resistor as 10uA/volt, then we have 90uA at most thru the collector.

Assume beta of 100X (quite probable for this current level), and the base current will be 0.9uA.

Consider your diode likely has 0.6v at 1mA.

That drops by 0.058 volts for each 10X drop in diode current.

At 1uA, expect 0.6v - 3 * 0.058 ==== (0.6 - 0.2) == 0.4v across the diode.

However that 0.058 volts per 10:1 current change is ONLY TRUE if the diode junction is Abrupt Junction. Which is only a theoretical model. (see "jonk" answer for more realistic number for 10:1 effect)

So use the simulator to test your SPICE model of the diode:

  • at 100mA (0.1 amp)

  • at 10mA

  • at 1mA

  • at 0.1mA

  • at 0.01mA

  • at 1uA

  • at 0.1uA

  • at 0.01uA

  • at 1nanoAmp

and please add these results to your question, so we all can learn.

And thank you.

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  • \$\begingroup\$ I changed the resistor in the circuit above to provide different currents (I hope I've understood your directions correctly). I get the following: \$\endgroup\$
    – Rob Duncan
    Jun 7, 2020 at 21:25
  • \$\begingroup\$ I get the following diode current and voltage drops: (364 mA, 1.14V), (13.4 mA, 0.75V), (521 uA, 0.58 V), (45.2 uA, 0.46V), (4.87 uA, 0.35V), (568 nA, 0.24V), (73 nA, 0.14V), (10 nA, 0.06V), (1.65 nA, 0.02 V). \$\endgroup\$
    – Rob Duncan
    Jun 7, 2020 at 21:34
  • \$\begingroup\$ Sorry to be picky, but could you edit your question with those results? And make it clear where the drops are happening -- because 1.14V across the diode seems wrong. \$\endgroup\$
    – TimWescott
    Jun 7, 2020 at 21:37

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