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I'm trying to find the collector, base and emitter currents of the transistor in the following simple topology:

schematic

simulate this circuit – Schematic created using CircuitLab

In order for the transistor to be on, the emitter needs to be at a higher potential than the base. Therefore, since Ue=0, Ub=-0.7. This implies that there is a higher voltage drop across the top 100Κ resistor(10.7 Volts vs 9.3) meaning that the current through the top 100K resistor is greater than that of the lower 100K resistor. Therefore, the math demands that the base current goes towards the base.

However, in pnp transistors the base current should flow away from the base, towards the left in this case. The equations here demand that it goes to the right. Is it possible? Or does this mean that the base current is zero?

Update: It is implied that |Vbe|=0.7 volts if the transistor is on. I'm trying to prove that the base current is zero. I assumed that the transistor is on and Vbe=0.7 as given. However, as mentioned above, this led to a base current flowing towards the base. Is this enough proof that Ib=0 and the transistor is off?

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  • \$\begingroup\$ If you reverse-bias the BE junction, you will get leakage currents at first but then soon (with enough reverse-bias) you will get avalanche behavior (sometimes just called "zenering.") Small signal BJTs, as a broad rule, shouldn't be subjected to more than about four volts of reverse-bias. One or two volts would be better, though. In your schematic, which I may be reading right, should present 0 V through 50 k to the base of the PNP. Given the emitter is grounded, this would seem to mean no reverse-bias to me. Why do you think differently? \$\endgroup\$
    – jonk
    Commented Jan 12, 2020 at 19:11
  • \$\begingroup\$ @jonk First of all, I'm not so used to transistor circuits so it is quite likely that I've made some mistake but yes you read the schematic right. We can replace the two transistors with what you described, a 50K resistor connected to the base and ground. However, it is given that |Vbe|=0.7 Volts and therefore the base is at -0.7 Volts. That means, a current will flow from ground through the resistor and towards the base. Is this direction correct for a pnp transistor? Checking the solutions I have, the result for the base current should be zero. I can see that Vb<Ve but still... \$\endgroup\$ Commented Jan 12, 2020 at 19:21
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    \$\begingroup\$ It cannot be given that the base to emitter voltage, under these circumstances, has a magnitude of 700 mV. It's simply not possible. The transistor will be, for all intents and purposes, off as there will be no base recombination current. There is a specification for BJTs which may allow some leakage current from emitter to collector, even in this circumstance. But it won't be much so long as the collector-emitter breakdown voltage isn't exceeded. \$\endgroup\$
    – jonk
    Commented Jan 12, 2020 at 19:22
  • \$\begingroup\$ @jonk I updated the question. Vbe is 0.7 only if the transistor is on. So, I'm supposed to assume that it's on with this Vbe and check the current results to see if my assumption was correct. You mention that the transistor is indeed off. How is it explained? Is the explanation I gave adequate? \$\endgroup\$ Commented Jan 12, 2020 at 19:31
  • \$\begingroup\$ It's really easy to prove the base current is zero. You have a Thevenin voltage source of 0 V and resistance 50 k tied to the base. The emitter is tied to ground. So the total circuit has 0 V at both ends of a series pair consisting of a resistor and a diode. There is no potential difference across the series pair which would motivate a current through the series pair. So the current must be zero. To have a current you must have an electromotive force. Period. You don't have one here. End of story. \$\endgroup\$
    – jonk
    Commented Jan 12, 2020 at 19:43

2 Answers 2

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General Forward-Biased Solution

Here's the equivalent schematic you need to analyze:

schematic

simulate this circuit – Schematic created using CircuitLab

The solution of the KVL equation is:

$$\mid \:I_\text{B} \mid\:=\:\frac{V_T}{R_{_\text{TH}}}\cdot\operatorname{LambertW}\left[\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}}{V_T}\cdot e^{^\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}-V_{_\text{TH}}}{V_T}}\right]-I_{_\text{SAT}}$$

The above equation can be used for any forward-biased situation. So if the practical values of the resistors in the resistor-divider aren't perfect, you can make measurements and compute the appropriate Thevenin voltage and resistance for the divider as a voltage source to the base and plug those in, above.

If \$V_{\text{TH}}=0\:\text{V}\$ then set \$u=\frac{I_{_\text{SAT}}\cdot R_{_\text{TH}}}{V_T}\$ and then \$\operatorname{LambertW}\left[u\cdot e^{^u}\right]=u\$ and so \$\mid \:I_\text{B} \mid\:=\:0\:\text{A}\$.

But I don't expect you to be able to develop the above equation. And I am pretty sure your problem does not require it of you.

Summary

Your circuit presents \$0\:\text{V}\$ across \$R_{_\text{TH}}\$ and the base-emitter diode of \$Q_1\$. Since in this ideal case there is no voltage gradient there cannot be any current between the two sources (which are both \$0\:\text{V}\$ at either end of the series pair.) The base current is zero.

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  • \$\begingroup\$ +1, very good answer. I'd thought roughly the same argument so I then wondered: can you also show that without Thévenin's theorem? One thought I had would be going through the possible regions and show that such hypothesis lead to absurd results. Can you think of a faster way? \$\endgroup\$
    – edmz
    Commented Jan 12, 2020 at 22:20
  • \$\begingroup\$ @edmz You'd get to the same place. When there is no difference in the electric field between two points, there cannot be a continuous flow of charge from one to the other. Symmetries (which yield conservation laws) would forbid it, regardless of what stands between or what path you take. \$\endgroup\$
    – jonk
    Commented Jan 12, 2020 at 22:27
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First, let's get rid of the transistor, and note that the unloaded base voltage is zero:

schematic

simulate this circuit – Schematic created using CircuitLab

Now let's connect the transistor, and note that the V.BE voltage is zero as well. We assume that the BE diode does not produce any thermal currents.

schematic

simulate this circuit

Since V.BE=0, the base current is nil, and the approximate emitter-collector current is zero. Thus the collector load resistor has no voltage drop - both its terminals are at the same potential:

schematic

simulate this circuit

Now you may ask: OK, that's an idealized circuit analyzed in a simplistic fashion. How about a real circuit?

We can replace the R1+R2 divider with its Thévenin equivalent, and use a high-valued shunt resistor to measure the collector current:

schematic

simulate this circuit

According to the CircuitLab simulator, the voltage across the 10MOhm resistor is about 5uV, so the current is 5uV/10MOhm=0.5pA.

The choice of the voltmeter equivalent series resistance is not arbitrary: that's the input resistance of a most digital multimeters. Thus, the multimeter alone acts like a parallel combination of RVM1 and VM1.

I've picked a random 2N3906 from my bin-of-transistors, and connected it with base shorted to emitter, and used a 9V battery as the voltage source. The measured voltage - and thus current - is on the same order as the simulation predicts (10uA, give-or-take), and is very sensitive to external disturbances.

After connecting the base resistor, there's no significant change in the collector current, although the current was at the edge of what I could measure without shielding.

Due to the below-picoampere currents involved, low-leakage measurement technique is essential. No probe wires should touch, even though they are insulated - PVC is not the best insulator for such measurements. Silicone or Teflon (PTFE) is much better. Even body movements can affect the readout, and the entire circuit up to the voltmeter and supply source should be shielded from air currents.

Ideally, the circuit should be in a metal box acting as a shield, plugged directly into the multimeter. The transistor and any other insulating surfaces in the box, e.g. the banana plug standoff, should be washed in IPA followed by a deionized water rinse. It doesn't take much conductance due to contamination to skew the results here.

I might re-do the measurements later after sticking the circuit into just such a box.

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