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Timing circuit using LM339

dahappychappy

Apr 10, 2022
3
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Hi,
I am trying to mod my de-soldering station by adding in a small tank to hold a vacuum, and a solenoid is actuated after a certain amount of time allowing the gun to have a decent burst of negative pressure instead of the slower build up. After 1.5 seconds or so the solenoid will open and hopefully suck all the solder out so the bad parts can be removed.
Anyway, I have designed this circuit using one of the LM339 comparators as a schmitt trigger, then using another comparator on the chip to invert the signal to use on a transistor to trigger the coil. I don't understand what is happening with the circuit. The inverting input on U2B is going below the reference voltage and the output doesn't change until the trigger is released. The trigger is normally open as per the diagram.

Thank you for your assistance.
 

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Alec_t

Jul 7, 2015
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U2A needs a pull-up resistor on its output.
Welcome to AAC!
 

crutschow

May 7, 2021
857
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The LM339 has an open transistor collector output so needs a pull-up resistor on its output, as Alec_t noted.
 

dahappychappy

Apr 10, 2022
3
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Apr 10, 2022
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Hi,
Thank you for that :) I have have tried it on my circuit with a 10k resistor and it didn't work. I could have a wiring problem for all I know. Now I am trying to just simulate the circuit in MicroCap as it is free. It is saying it looks not too bad. A 10Meg seems to work for U2A's pull up. Maybe I need to drive the transistors harder so they are in saturation as I am getting a 4volt drop across it when on.
Thank you
 

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Last edited:

crutschow

May 7, 2021
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You have a feedback load on U2A of 15kΩ so a 10meg pull-up resistor will only give a small output voltage.
You need a smaller pull-up resistor to get a significant voltage out.
Putting random component values into a circuit seldom gives good results.
 

dahappychappy

Apr 10, 2022
3
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Apr 10, 2022
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That is true. I increased the size of the feedback resistor on U2A so I could use a smaller pull up resistor. I reworked the circuit and now it is working fully. I don't understand how you figured out the feedback load above.
I think the lack of pull up was the initial problem. It couldn't drive the next comparator properly. I also found I was charging the first capacitor to half of the supply voltage, then I was measuring it against a reference of half of the supply voltage. I also found I had soldered the first BJT in backwards at some point. I have changed the circuit so many times at this point it's crazy.
Got it working now.
Thanks for your assistance.
 

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crutschow

May 7, 2021
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I don't understand how you figured out the feedback load above.
The feedback load is R5 (10k) in series with the parallel value of R3 and R4 (also each 10k).
Thus the feedback resistance is 10k + 5k or 15kΩ.
 
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