OK, Joseph reversed the LED and has PMed me some new measurements. Let's keep the info on this thread now Joseph.
1. A-Z 8.61v
2. A-B 7.36v
3. B-Z 1.24v
4. C-Z 7.49v
5. Uncovered, C-Z 4.36
OK, those voltages are all good except #5.
The way that circuit works is that the 3.3 kΩ resistor between A and C "pulls" point C up to the battery positive voltage, unless the phototransistor (which is connected between C and Z) pulls point C downwards (towards Z). The MOSFET' responds to the voltage between its gate and its source, which are connected to C and Z respectively.
Measurement #4 shows the voltage at C (relative to Z, which is the "0V" rail of the circuit) at 7.49V. This voltage comes from the battery, via the 3.3 kΩ resistor. The phototransistor has no (or very little) effect on this voltage, because no light is falling on it. This high positive voltage at C (relative to Z) makes the MOSFET conduct, and will energise the load (in the drain circuit).
When light falls on the phototransistor, that causes it to conduct, and it should pull the voltage at C down to less than 2V, perhaps less than 1V depending on how hard it is conducting. This is measurement #5 with the phototransistor uncovered.
But measurement #5 is only 4.36V. The phototransistor is not pulling the voltage down low enough, and the MOSFET will not turn OFF properly. The phototransistor needs to pull the voltage below about 2V to ensure the MOSFET turns OFF completely.
There are several possible reasons why the circuit isn't doing what it should.
- The LED may not be emitting much infra-red light. This could be because it was damaged when it was connected backwards before. The only way to find out would be to buy a new LED and try that.
- The phototransistor may not be receiving enough light, or the amount of light may not be enough to make it pull point C down below about 2V. Try moving the LED closer to the phototransistor. You should see the voltage between C and Z decrease as you move it closer and aim it more accurately. If you can't get it below about 2V then you can try increasing the resistance of the 3.3 kΩ resistor.
- The 3.3 kΩ resistor may need to be weaker (a higher resistance value). When light falls on the phototransistor, it "pulls" the voltage at C down towards Z. To do this, it has to fight with the current coming from A, via the 3.3 kΩ resistor. Increasing the 3.3 kΩ resistor, to 10 kΩ for example, will weaken the effect of the resistor and may allow the phototransistor to win the tug-of-war game that it's playing. If 10 kΩ doesn't allow the phototransistor to pull C below about 2V, try 33 kΩ. If that still doesn't work, there's probably another problem.
- The phototransistor may be an unsuitable type, or it may be damaged. You could try replacing it.
- It might be possible that the phototransistor is connected backwards. If you're going to get new components anyway, try reversing the phototransistor and see whether this gives you a voltage at C that's over 7V with the phototransistor covered, and under 2V with it uncovered. Doing this might damage the phototransistor but if you're going to replace it anyway, you might as well try it.