Electrical – Some questions about a transistor latch

You know that driving enough base current will fall the transistor in saturation region.

But recall that for an NPN transistor, if \$V_C < V_B > V_E\$ --or simply, if \$V_B\$ is greater than both \$V_C\$ and \$V_E\$-- then the transistor will be in saturation region.

Let's test this in your circuit:

\$V_{C_{Q2}} = V_{CC} \$

Even if Q1 is saturated, \$V_{C_{Q1}}\$ will be between \$0\$ and \$V_{CC}\$. Because, as you can see, there's a voltage divider formed by R1, R2 and C-E resistance of Q1.

\$V_{B_{Q2}}=V_{C_{Q1}}\$, thus

\$0 < V_{B_{Q2}} < V_{CC} \$ which means \$V_{B_{Q2}} < V_{C_{Q2}}\$.

\$V_{B_{Q2}}\$ will never be greater than \$V_{C_{Q2}}\$ so Q2 cannot be in saturation region.

Here are several things that are happening:

1. When the switch is open, the base is floating and so Vbe is not well defined. There can be a small charge/voltage remaining from when the switch was last closed, or from capacitance to the surroundings. This will tend to turn the transistor on even though the switch is open.

2. The collector-base junction is like a reverse biased diode: leakage will occur from collector to base (\$I_{CB}\$) and (since the base floats) this current must go to the emitter, appearing like a base-emitter current (\$I_{BE}=I_{CB}\$). This gets amplified as a collector-emitter current of roughly \$I_{CE} = h_{FE}I_{CB}\$. This will be worse at higher temperatures.

To reduce 1 and 2, try adding a resistor from base to gnd (e.g. 10 kOhm).

3. The voltmeter you are using affects the circuit as it appears like a resistance in parallel. If possible, measure the voltages across the LED and transistor simultaneously with two voltmeters - the values should then at least add up (though the circuit behaviour will still be modified by introducing shunt currents).

4. The LED has an exponential voltage-current relationship, whereas the resistor's is linear. A small leakage current will cause a bigger voltage drop in the LED than in the resistor.

E.g. 1 uA would give an unmeasurable 0.1 mV across the resistor but perhaps 1 V across the LED.

For a similar problem, see this: Voltage drop across LED in open circuit

\$ \$

Also note: in bright light an LED can generate its own voltage, which would confuse the situation, but I don't think that is happening here (you could easily check though).