A simple circuit to simulate a momentary switch

davea0511

Junior Member
Oct 23, 2003
9
0
0
Here's my problem...

The power switch on my LCD monitor is a momentary switch, and if I disconnect then reconnect the power to the monitor, I have to manually push the momentary switch again to turn on the monitor, which is a pain because the monitor is behind a control panel. So I want the monitor to turn on automatically whenever I plug in the power.

Any ideas? AFter much consternation I came up with a super simple circuit. I animated the function of it here: http://brainstep.com/circuit2.html

Obviously I'm not a EE (EE's: please keep the laughter down to a dull roar), so... will someone tell me if am I completely whacked out - will this even work and if not why? I've seen some possibilities using a 555 timer circuit, but my idea seems so much easier (less soldering) - I could even probably do it all on a long terminal block and cover it with heatshrink tubing.

Thanks in advance for ANY feedback!
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
It's twisted, but I can see what you're aiming at. If anything, you should move the capacitor between the gate and ground instead of gate and input. When the input jumps high, the voltage may couple across the capacitor and so there is a good chance that the gate will shut off right away. To get a good slow RC delay, you want to ground the other end of the capacitor. That aside, I still have no idea if this crazy idea will work.
 

blahblah99

Platinum Member
Oct 10, 2000
2,689
0
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You can do that with a comparator w/ large hysteresis, and an RC time constant at the input.

IE, an opamp like an LM393 with some feedback (for hysteresis), and an RC with a time constant of a second.

That'll solve your problem.
 

xSauronx

Lifer
Jul 14, 2000
19,586
4
81
Originally posted by: davea0511
Here's my problem...

The power switch on my LCD monitor is a momentary switch, and if I disconnect then reconnect the power to the monitor, I have to manually push the momentary switch again to turn on the monitor, which is a pain because the monitor is behind a control panel.

me being the lazy...err...efficient bastard i am; id be more inclined to cut a small hole in the cabinet over the button and just push it. guess thats out of the question?
 

davea0511

Junior Member
Oct 23, 2003
9
0
0
Originally posted by: TuxDave
It's twisted, but I can see what you're aiming at. If anything, you should move the capacitor between the gate and ground instead of gate and input. When the input jumps high, the voltage may couple across the capacitor and so there is a good chance that the gate will shut off right away. To get a good slow RC delay, you want to ground the other end of the capacitor. That aside, I still have no idea if this crazy idea will work.

Thanks for the insight! I think I get what you mean, but the way I addressed your feedback is a little different than what you suggested. Have a look:

http://brainstep.com/circuit2.html

The ground in this circuit gives the capacitor something from which to couple a negative charge when the input first goes high. Am I still whistling dixie or should I give it a shot?
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
Originally posted by: davea0511
Originally posted by: TuxDave
It's twisted, but I can see what you're aiming at. If anything, you should move the capacitor between the gate and ground instead of gate and input. When the input jumps high, the voltage may couple across the capacitor and so there is a good chance that the gate will shut off right away. To get a good slow RC delay, you want to ground the other end of the capacitor. That aside, I still have no idea if this crazy idea will work.

Thanks for the insight! I think I get what you mean, but the way I addressed your feedback is a little different than what you suggested. Have a look:

http://brainstep.com/circuit2.html

The ground in this circuit gives the capacitor something from which to couple a negative charge when the input first goes high. Am I still whistling dixie or should I give it a shot?

Well, how about when I say it this way. When you turn anything on abruptly, capacitors essentially act like short circuits for a brief moment. So if your input jumps rapidly from 0 to 3.3V, the gate will jump to a value SOONER than what you were expecting. You should put the capacitor between the gate and ground.
 

JTWill

Senior member
Feb 2, 2005
327
0
0
Are you thinking of an SCR or a FET?
What happens if you hold down the power switch? Hold down the switch and unplug it , plug it back in if it powers up and stays on take a toothpick and jam the switch on all the time.
I'd look at the schematic of the monitor and bias it on all the time.
 

davea0511

Junior Member
Oct 23, 2003
9
0
0
Well, how about when I say it this way. When you turn anything on abruptly, capacitors essentially act like short circuits for a brief moment. So if your input jumps rapidly from 0 to 3.3V, the gate will jump to a value SOONER than what you were expecting. You should put the capacitor between the gate and ground.[/quote]

I see. Something like this then:

http://brainstep.com/circuit4.html

I like that. It removes all the funny effects that the capacitor might have on the transistor (speaking strictly in "I don't have a clue what I'm doing" talk).

Will I also need a resistor between the capacitor and ground?

Also, if a capacitor momentarily shorts out when voltage immediately goes from 0 to 5V couldn't I simply replace this whole circuit with just a capacitor?
 

davea0511

Junior Member
Oct 23, 2003
9
0
0
jtwill, elecrzy, xsauronx -

This is part of a kiosk so other people are going to be plugging the kiosk into the power outlet (like a janitor might do when he's vacuuming) - that's one of the reasons I want it to automatically turn on when they plug it in. My Motherboard has a jumper that changes the momentary-on switch to an always-on, but unfortunately the LCD monitor doesn't so that's why I need a circuit to automatically toggle the momentary switch when power is applied.
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
Originally posted by: davea0511
Well, how about when I say it this way. When you turn anything on abruptly, capacitors essentially act like short circuits for a brief moment. So if your input jumps rapidly from 0 to 3.3V, the gate will jump to a value SOONER than what you were expecting. You should put the capacitor between the gate and ground.

I see. Something like this then:

http://brainstep.com/circuit4.html

I like that. It removes all the funny effects that the capacitor might have on the transistor (speaking strictly in "I don't have a clue what I'm doing" talk).

Will I also need a resistor between the capacitor and ground?

Also, if a capacitor momentarily shorts out when voltage immediately goes from 0 to 5V couldn't I simply replace this whole circuit with just a capacitor?[/quote]

That may not work. It depends on how much current will the 'turning on' process require. If it requires some good amount of steady current, then the capacitor may quickly lose its charge and cause it not to turn on. I think your solution as you have it has the best chance of working. Just double check that your transistor can support up to 5V.
 

RaynorWolfcastle

Diamond Member
Feb 8, 2001
8,968
16
81
What you just described is a one shot oscillator, which in and of itself is a simple circuit that you can build with a 555 timer and a few minutes (you can add a power transistor as a quick and dirty output stage if you need to source a lot of current). The bigger issue however is whether you're going to be taking apart the display to do this... Also, who knows if you can get specs on what kind of voltage/current that switch expects as an input...

The circuit itself is simple, the rest is a little more complicated... I'm sure you could hack it though.
 

davea0511

Junior Member
Oct 23, 2003
9
0
0
Suprisingly, I got both circuits I linked to above to work as expected, but there was a problem: the capacitor kept the signal high for a short time period after I disconnected the power. That is unacceptable because in the event of a power glitch the circuit wouldn't do anything to turn the monitor back on (the monitor will shut off during a momentary loss of power, but the charged capacitor keeps my circuit from turning the monitor back on).

So there goes my pride... I'm now doing the 555 circuit. I have no doubt that will work... I just thought it would be cool to do it with a few individual discrete components that I could all throw on a terminal block and cover with heat-shrink tubing).

ps- I also tried a comparator chip (4 comparators), hooking up the inputs to outputs to try to simulate a momentary switch with the built-in propagation delay - but no worki, the propapation delay was probably too short.
 

TuxDave

Lifer
Oct 8, 2002
10,572
3
71
Originally posted by: davea0511
Suprisingly, I got both circuits I linked to above to work as expected, but there was a problem: the capacitor kept the signal high for a short time period after I disconnected the power. That is unacceptable because in the event of a power glitch the circuit wouldn't do anything to turn the monitor back on (the monitor will shut off during a momentary loss of power, but the charged capacitor keeps my circuit from turning the monitor back on).

So there goes my pride... I'm now doing the 555 circuit. I have no doubt that will work... I just thought it would be cool to do it with a few individual discrete components that I could all throw on a terminal block and cover with heat-shrink tubing).

ps- I also tried a comparator chip (4 comparators), hooking up the inputs to outputs to try to simulate a momentary switch with the built-in propagation delay - but no worki, the propapation delay was probably too short.

haha.. ah well, nice effort.