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Circuit design for review

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Kookie

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Hi, I am about to make this circuit and I wanted to submit it for scrutiny before I even breadboard it. (The values for "x" have not been determined yet.) Will it work?

http://mywpages.comcast.net/Kookie/Motorcir.jpg

Its been a very long time since I did this sort of thing and I'm not sure if I designed it correctly.

The purpose of the circuit is frivolous, so I wont go into that, but basically its replacing an indicator light (plug in LED) with this circuit that will light the LED plus a couple more and turn a motor.

Here are my questions about the circuit:

1) Will it work in theory?

2) I have R1 and LED1 on the emitter side of the transistor in the optoisolator. Is that ok?

3) I want to make the pair of LEDs (LED23) slowly turn off. Dim to off. To do this, I'm thinking C1 can achieve this. Am I correct?

4) Is there a voltage drop of 0.7 between the collector and emitter of the transistors, or is that just between the base and emitter? (Can't remember :oops: )

5) I want to design this as small as possible, with as little parts as possible. Here's what I have to put it in: http://mywpages.comcast.net/kookie/motolayout2.jpg
Its very cramped, but I should be able to put it all in. Depending on the size of C1, if that will work at all. Would there be a better design?

So, what do you think? I really appreciate your time and expertise with any help and input you can give me. Thank you very much. :D
 

Kookie

New Member
P.S. One more question. What is D3 for? (The diode across the motor.) I must admit, I had seen that in other circuits and I copied it without understanding it. I'd assume its some protection for the circuit for when the motor stops and the fields collapse. Is that correct? I'm not sure how that works.
 

Klaus

New Member
1. No, have a good look at T2 and see what turns it on? Nothing, as it needs at least 0.7V positive potential on the base compared to the emitter. Hint, you're NOT feeding AC to the capacitor.
You need some sort of DC bias to the base junction.
Why not try each section of the circuit individually on a bread board? That way you learn by doing rather than by somebody telling you.
2. yes
3. R3 and C1 will give you a time constant, depending on their values (t in seconds = R (Ohms)x C (Farads). To get a visible effect you need at least 1/2 second or longer, but first sort out the charging / discharging of C1.
4. Collector - emitter voltage drop depends on what state the base is. ).7V or thereabouts (depending on type) when fully on, Vcc when off.

You will have problems with the motor if the pulses are short.
Motors have inertia and take time to accelerate or stop. If you want to use fractional revolutions you need a stepper motor and its associated driving circuitry.

The diode is to stop the back EMF of the motor, it is also used across relay coils for the same reason.

Klaus
 

Kookie

New Member
Ah, thank you Klaus for answering my questions. Like I said its been a long time for me since I've done this sort of thing. Believe it or not, but I was good at this back when I was a kid, :lol: . I'd like to see it work on paper fist before I start letting the smoke out of everything and get my concepts straight again. I've forgotten more than I like. (Yeah, like simple stuff, ack, caps block DC, pass AC. That one slipped my mind there and I was thinking of only charging and discharging the cap, doh.)

As far as the motor start / stop drift I'm not concerned with. It wont be a problem.

I will have to rethink how I can do this. Hmmm, I'm tired and will resubmit something tomorrow.

One last thing here, you said: The diode is to stop the back EMF of the motor, it is also used across relay coils for the same reason.

Ok, but how does that actually work? What's going on there? So, when the potential across the coil stops, the magnetic field collapses, and you get a little burst of high potential. What does the diode do? I'm not getting this, ack.

Thanks again!
 

Kookie

New Member
Eh, crap, I can't think of a way to do this. I took that cap out, don't know what I was thinking there. I'm not sure how I'd make those LEDs dim out. Can't think of anything.
 

Klaus

New Member
Sorry, still no go. you have to provide a path for C1 to discharge and D5 prevents that. So, C1 will charge at the first pulse and stay thus charged.
You appear to like diodes, D2 does nothing and is not required.

Remember that if you charge one side of a capacitor to some positive potential its other side will assume the opposite potential. You must provide a path for the capacitor charge to equalise if you want a repeated charging function.

D4, R3 is going to defeat your R/C part which got me wondering why you drew them there.
Did you want to dim from bright to OFF or from OFF to bright or both??

You got the back EMF part right, the diode shorts it out so it cannot do any harm elsewhere in the circuit.

I still think that learning on paper is more fraught with problems than learning by doing. Electronic parts are cheap anyway and the odd puff of smoke reinforces the lesson to take care and double check before you apply power to it.

Klaus
 
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