Proving that the transistor in this simple schematic is off

Question

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

^{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?

Answer 1

General Forward-Biased Solution

Here's the equivalent schematic you need to analyze:

^{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.

Answer 2

First, let's get rid of the transistor, and note that the unloaded base voltage is zero:

^{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.

^{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:

^{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:

^{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.