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a timing circuit

joseph barbuzza

Feb 16, 2015
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KrisBlueNZ i hope you got the pictures. Please let me know if they are of any good. I appreciate any help i can get. Thank you
 

KrisBlueNZ

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Sorry Joseph, but neither those photos, nor that construction method, is acceptable. You need to build the circuit up on a piece of stripboard like this: http://www.digikey.com/product-detail/en/8022/V2018-ND/565947

While you're doing that, you can remove the 100 kΩ resistor - it doesn't do anything in that circuit. You can also remove the 9V battery, and connect the 330Ω and 3.3 kΩ resistors to the positive side of the first 12V battery (the battery whose negative side connects to the MOSFET's source) instead.

Google how to make a stripboard layout for information on how to translate your diagram to stripboard and how to lay the components out.

Once you've built it on stripboard, and all the components are close together and clearly identifiable and nothing will short onto anything else, I will be able to tell you where to measure, if it still doesn't work.

BTW the MOSFET is static-sensitive so you should put a paper clip across its leads when you've removed it from the old circuit, and take the paper clip off when it's installed in the stripboard.
 

joseph barbuzza

Feb 16, 2015
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I can't do a breadboard so i made due with what i have. I hope it's acceptable
 

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KrisBlueNZ

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OK, that's reasonable. I've marked up the six "nodes" in the circuit.

epoint 272724 clear layout marked up.jpg
Connect your 9V battery (it doesn't matter about the other battery) and measure these DC voltages, and post the results:

Cover the phototransistor e.g. with the cover from the end of a pen.
Test #1. A to Z. This just checks the voltage of your 9V battery.
Test #2. A to B. This is across the 330Ω resistor.
Test #3. B to Z. This is the voltage across the LED. If it's an infra-red LED, this voltage should be about 1~1.5V.
Test #4. C to Z. Since the phototransistor is covered, this voltage should be nearly equal to the 9V battery voltage.
Remove the cover from the phototransistor and point the LED directly at it.
Test #5. C to Z. Since the phototransistor is exposed to bright light, it should pull the voltage at C down to less than 2V.
 

KrisBlueNZ

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OK, the LED is either damaged/faulty, or connected the wrong way round. Try reversing it. When it's connected the right way round, the voltage from B to Z will be about 1~2V.

It's possible that the LED has been damaged by being connected backwards. So even if you get the right voltage when you reverse the LED, it may not work properly, and you may need to buy another LED. If you do, make a note of the right way to connect it! It should have a slightly flattened part on the "skirt" which normally indicates the cathode lead (the side that connects to Z), as shown in this diagram from http://www.garyfallidou.org/en_led_polarity.html

led_polarity_en.jpg
 

KrisBlueNZ

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Joseph PMed me and said he has reversed the LED and is getting 1.24V across it.

OK Joseph, can you repeat the measurements I listed in post #28.
 

KrisBlueNZ

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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.
 
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joseph barbuzza

Feb 16, 2015
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  • Ok i had replaced the LED for a new one. In truth every bit of the circuit is new, checked and verified before soldering into circuit. I did notice as I'm checking the voltage that as i moved the LED slightly the voltage did change, increased. Is it possible i need a brighter LED and if so what do i ask for
 

KrisBlueNZ

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1. The LED may have been damaged when it was connected backwards, so if you haven't replaced it already, replace it.

2. If the voltage from C to Z still doesn't go below 2V with the LED shining on the phototransistor, try reversing the photoransistor to see whether point C goes any lower with the LED shining on it.

3. If not, put the phototransistor back the way it was.

4. Try increasing the 3.3 kΩ resistor to 10 kΩ and see whether C will go below 2V.

5. If C still won't go below 2V, increase the resistor to 33 kΩ.

6. If that still doesn't work, try a new phototransistor.
 

joseph barbuzza

Feb 16, 2015
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Ok i had an idea from all your help and removed the 100k resistor from the circuit and replaced it with the 3.3k resistor just to see. Now the mosfet is switching. When covered 24 v uncovered 1.24 v.
 

joseph barbuzza

Feb 16, 2015
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But i have more question, in one of your posts you said i coud remove the 9v battery and connect it to the first 12v battery. If im working with a 24v power supply. Simple positive and negative lead. Is that still possible. Would like to get rid of a separate 9v source
 

KrisBlueNZ

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I'm not happy with that fix. Changing the 100 kΩ resistor to 1 kΩ will reduce the voltage at C, but the voltage at C should swing over a wide range. It should be at least 7V when the phototransistor is covered up, and it should be less than 2V when the LED is pointing directly at the phototransistor. If the signal doesn't swing over that range, the MOSFET won't switch cleanly and it may overheat.

Yes, you can get rid of the separate 9V source if your 24V power source is made from two 12V batteries connected in series. The +9V supply doesn't need to be exactly +9V; it can be +12V. But it can't be 24V. That is too much voltage for the gate of the MOSFET and it will damage it.

You see that the negative side of the first 12V battery, AND the negative side of the 9V battery, are both connected to point Z in the circuit. So, if your 24V supply is made from two separate 12V batteries, you can use the positive side of the first one as the positive supply for the circuit that runs from the 9V battery.
 

joseph barbuzza

Feb 16, 2015
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I'm not sure i understand. I replaced the 3.3k ohm resistor with the 100k ohm resistor that i removed from the circuit and the circuit is working. The same 100k that you said does nothing for the circuit. Also thank you for clarifying the 12 to 24v thought for replacing the 9v
 

KrisBlueNZ

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Oh, OK. I get it. So you took the 100kΩ resistor that was connected from C to Z, which I said wasn't doing anything, and you removed it from that part of the circuit. Then you replaced the 3.3 kΩ resistor with the 100 kΩ resistor. Now the circuit works OK.

That's fine. As long as you get at least 7V from C to Z when the phototransistor is covered up, and less than 2V from C to Z when it's uncovered and the LED is shining on it, then everything is OK.
 
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