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constant to momentary output info


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dualsport 
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Joined: September 27, 2005
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Posted: April 05, 2008 at 4:27 PM / IP Logged  
They're very handy as a general purpose switch; I accidentally blew out a relay driver chip on an alarm module, and I just wired in a transistor, using the signal from input to the blown driver to control it. Worked just as good as new.
Another use I had was to use it to control a wireless doorbell transmitter, driven from the normal household doorbell. I wanted to have the portable wireless doorbell controlled by the one button, so the transistor isolated the AC voltage from the transmitter.
It doesn't draw any current on the input, so it can be switched without any loading on the input signal. It's just a handy thing to have available for all sorts of applications, so I figured it would be worth the whole ten bucks.
steelroe 
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Location: Ireland
Posted: April 05, 2008 at 4:45 PM / IP Logged  

Forgive all the greenhorn questions here but it is a long time since i had anything to do with electronics.

How does this circuit work?

Where and for how long is the output for?

R1 is a 1 mega ohm resistor correct?

steelroe 
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Joined: April 03, 2008
Location: Ireland
Posted: April 05, 2008 at 4:50 PM / IP Logged  
So i am going to need 2 of these circuits to open and close the mirrors
dualsport 
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Posted: April 05, 2008 at 5:30 PM / IP Logged  
Actually, the second circuit is set up for for a constant to pulse output, I'll make a few changes to tailor it for what you'll need. I'll draw it up and post it up in a bit-
dualsport 
Silver - Posts: 983
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Joined: September 27, 2005
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Posted: April 05, 2008 at 6:27 PM / IP Logged  
constant to momentary output info - Page 2 -- posted image.
When everything is sitting there before the trigger comes in, the capacitor is discharged.
When the voltage between the gate (pin 2) and the source (pin 3) of the transistor is low, it's off, and basically an open circuit between the drain (pin 1) and pin 3. The relay is not turned on as a result.
When your trigger comes in, the capacitor charges up to the input voltage through the diode D3, and the transistor turns on.
When on, the transistor connects the relay coil to ground, turning on the relay for the start of the output pulse.
With no more input from your trigger, the cap discharges through the R1 resistor, until the voltage drops below the turn on point of the transistor, and the relay opens up again.
You can use a larger capacitor or resistor to get a longer pulse time, and vice versa.
steelroe 
Member - Posts: 17
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Joined: April 03, 2008
Location: Ireland
Posted: April 05, 2008 at 6:37 PM / IP Logged  

Look sgood, makes more sense when explained by someone who knows what they are talking about.  What duration of output can I expect from the circuit you posted.

dualsport 
Silver - Posts: 983
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Joined: September 27, 2005
Location: United States
Posted: April 05, 2008 at 7:15 PM / IP Logged  
It should be in the range of 5 seconds; but just adjust the resistance to make it longer or shorter as required. You don't have to change both at the same time, just go up or down on either to change the timing.
You don't have to use the same values shown, if you happen to have some spare parts handy, just pick combinations that will give you roughly the time you want when you multiply R x C.
e.g., If you use a 500k resistor, it would drop the pulse time in half; if you use a 2M resistor, and it'd double the time.
Keep in mind the time of the pulse starts after your input pulse turns back off, so you need to add your input pulse time to the timing. If your input pulse was 2 seconds long, it just extends the pulse by the designated time.   If you put a constant (+) input to it, the output would go on and stay on until you remove the input.
The diode across the relay coil is needed to suppress the voltage spike that occurs when the relay is turned off; it would go well over the maximum voltage that the transistor can handle and destroy it otherwise.
The diode on the input allows the input pulse to pass in only one direction, in case your signal goes to ground on standby. If you didn't have the diode there, the charge on the cap would drain off back through the input instead of the resistor, affecting the timing of the pulse.
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