Effect of a transistor on a voltage divider? (or vice versa)

Question

I'm learning about transistors and am trying to make a basic circuit that uses a photoresistor as a sensor to turn an LED on and off depending on light conditions (the 150kOhm resistor is where the photoresistor will be). I'm still trying to learn how to choose the appropriate resistor values, but messing around in Multisim I noticed that the voltage divider in the circuit below does not behave as I expected.

Here's the circuit with the top resistor in the voltage divider being 1kOhm:

According to Multisim, the voltage across each of the resistors is about the same, when almost the entire supply voltage should be across the bottom resistor since it's much larger than the top.

Here's the circuit with the 1k resistor changed to a 10k:

If I take the transistor out of the circuit, then the voltage divider behaves as it should.

Why isn't the voltage divider working properly when the transistor is attached to it?

Answer 1

The formula for the voltage at the center of a voltage divider depends on the assumption that the current through R2 is exactly equal to the current through R4.

In this case, that's not true. There's an alternate path for current to flow, through R5 and the Q1 b-e junction.

In fact, because of it's high value, R4 is effectively irrelevant in this design. Current mostly flows through R2, R5, and the Q1 b-e.

And the results are about what you'd expect. The voltage across the Q1 b-e is about 0.7 V, and R2 and R5 are about equal.

As for the 2nd version, by reducing the current through the resistor divider that you're trying to use, you've increased the impact of "leakage" through R5 and Q1.

Answer 2

Vbe varies from 0.6 to 0.7V at low currents, so essentially you have a 50% divider in cct #1 with two 1k resistors above this threshold .

Using 0.65V for Vbe as a guestimate, the 50% drop of 4.35 = 2.175V above this threshold or 2.825. Your simulating says 2.869 , so thats pretty close. The 100k is negligble compared to 1k. You can also treat the Vbe junction at threshold voltage and ESR being the slope of VI.

In the 2nd cct. You have a 1/11 ratio to Vbe so 0.11111 of the drop is 0.111*4.35 =0.483 then add Vbe of 0.65V to get 1.133 which is close to your simulation.. Using exact Vbe from NPN specs will give more precise values, but tolerances need to be considered. Again the 150k can be neglected considering Vbe variance.

Treat all diodes as fixed voltages with variable resistance in series and you will be able to do this in your head, once you understand the Rbe and RdsON and Vce(sat) values of various parts is pretty low.

Answer 3

If you move R3 to the emitter, then your base will have (roughly) the value you think it should have. That is because the emitter will be allowed to go above ground when current is running through the circuit, and the resistance looking into base will be beta*R3. If beta*R3 is much higher than R4, then you can treat the R2 R4 voltage divider as unloaded.

It might be easier to use the output of the Photoresistor resistor voltage divider as the input to a comparator or a 7414, if you want a binary off/on result.