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I am trying to create an electronic circuit, that is sort of a "not?" When the switch is pushed, nothing happens. When the switch is released, after a momentary press, then the Led would light. This will drive a simple light, and finally the entire process would be reset with a second button. The reset can happen at the momentary press of switch #2.

A friend of mine at work seemed to think this would be a good application for a 556 timer chip, which I gather has two 555 timer circuits in it, which are somehow linked, and the output of one of them controls the start of the second timer. I'm not exactly sure how you would apply that to my project, but I suppose it makes sense, the first button would trigger the timer circuit, for as long as button #1 is pressed, and then AFTER that button press, trigger the second timer, which would go on forever until a signal from Switch #2.


I assembled and created this circuit. I don't currently have a switch that does what #2 does, but with a jumper wire, I can simulate it. It doesn't work. :(

My friend from work seems to think it was in my description - But I re-read the description and it's pretty straight forward.

  • LOAD TO BECOME ACTIVE WHEN SWITCH #1 IS RELEASED.
  • SWITCH #2 TURNS OFF THE POWER TO THE LOAD.

Doesn't SOUND that complicated to me, although I wasn't able to figure out a way to make it work. In my mind, logically, sure, you have a latch that's activated by the release of the first button, and is de-latched by the second. Easy peasy!

Tons of Thanks in advance guys! (and/or gals)

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  • \$\begingroup\$ Would logic gates be fine? It involves wasting a lot of logic gates unless you use SMD. \$\endgroup\$ – Bradman175 Jan 6 '17 at 7:00
  • \$\begingroup\$ Sounds like \$SW_1\$ is grounded, with a pull-up, and that its rising edge triggers a state change of a flip-flop and that \$SW_2\$ is just attached to the CLEAR input. But perhaps I misunderstand. \$\endgroup\$ – jonk Jan 6 '17 at 8:52
  • \$\begingroup\$ If you're using a jumper wire to simulate SW2, you'll probably need to use a much larger capacitor, on the order of 100 uF or more. The problem with the wire is that too much time elapses between the breaking of the NO contact and the making of the NC contact, which allows the capacitor charge to leak away before it can be used. In a real switch, those two events are just a few milliseconds apart. \$\endgroup\$ – Dave Tweed Jan 11 '17 at 13:08
  • \$\begingroup\$ I replaced the wire with the recommended switch. I sort of realized that, it's almost more akin to a selector than a momentary switch. Anyway, it's all hooked up properly, I even noticed the labels NO/NC even matched the drawing! Is there no way to post a picture here? facebook.com/… \$\endgroup\$ – BrianDP Jan 11 '17 at 14:05
  • \$\begingroup\$ @BrianDP: Yes, you can post a picture by editing your original question. But I see you created a second account -- did you lose the password for your original account? If so, you should get in touch with a staff member (contact us link at the very bottom of the page) and have the two accounts merged. \$\endgroup\$ – Dave Tweed Jan 11 '17 at 15:40
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Bradman175 is on the right track, but perhaps a diagram that shows what he's talking about would help:

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Not sure if it will allow me to add pictures, but I picked up a switch that IS a momentary NO/C like depicted. And, it seems like it should work! When you hit SW2, the capacitor is charged. Then, when you release, it comes back and kicks the Q1/Q2 feedback loop to turn itself on. Finally, Switch 1 Momentarily grounds the (base?) of Q2 stopping the feedback loop. Looks like EXACTLY what I'm trying to do. I have pictures of my breadboard, maybe if you guys saw it you would say "Doh! You can't X, Y, Z...Do it A, B C instead." \$\endgroup\$ – BrianDP Jan 11 '17 at 8:03
  • \$\begingroup\$ I am Pretty sure I've put it together the way you've described, and the way the drawing looks. It seems like it ought to preform as designed! To reiterate- Just so I know what this thing is doing (or supposed to be doing) and please correct me if I have misunderstood any of these components, or their functions. When you click SW2, C1 becomes charged, and when the button is released, it sets the feedback loops of the 3906 / 3904 into motion. This continues, thus feeding the load, until SW1 is pressed, which momentarily breaks the power to Q1/Q2 stopping everything. \$\endgroup\$ – BrianDP Jan 11 '17 at 8:20
  • \$\begingroup\$ Hope this links correctly.. facebook.com/… \$\endgroup\$ – BrianDP Jan 11 '17 at 14:01
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Here is one of many possible solutions. This one is very straight forward and uses 1/2 of two edge-triggered D-flip-flops inside a 74HC74 IC.

Basically, whenever there is a rising edge on its CLK (clock) input the flip-flop transfers whatever is at the D (data) input to its Q output. Here, D is tied to VDD so it is always logic 1.

When START button is pressed, nothing will happen because CLK only responds to a positive (rising) edge. When START is released, R1 pulls CLK high and the flip-flop transfers the 1 at the D input to its Q output. Additional presses on START do nothing since Q is already high.

When STOP is pressed, /CLR is pulled low and clears the flip-flop so Q goes low. /CLR input is negative logic (denoted by the bubble/circle on the input).

enter image description here

Note that the LED is wired to /Q. /Q is the complement of Q. But the LED is powered from VDD so it still lights up when Q goes high. The reason this is a bit better configuration is that most CMOS ICs can drive low better than high.

Also, if the user is pressing the START and STOP at same time, the STOP always overrides the START button and holding STOP down will cause it to ignore START signals. I think this is what you would want if it was controlling some kindof machinery. You always have to design your circuits to take unexpected inputs from users. Hope that helps, -Vince

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You basically want to use an arrangement of transistors along with a circuit called the silicon controlled switch. You want a SPDT button that charges a capacitor. When released, the capacitor would go where the on button is to latch the SCS on. Then you can use the off button to turn the SCS off. The motor in the diagram is the load.

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  • \$\begingroup\$ Is that he same as an SCR? I have some of those!! They seem like trensistors though... \$\endgroup\$ – BrianDP Jan 7 '17 at 16:30
  • \$\begingroup\$ @BrianDP Very similar. However it's not recommended to pull the gate down to ground in order to turn it off, as it could create high currents for a short period of time. However, for some reason my SCR can still be turned off if connected to a switch and a resistor in series which is safe, however it's not how it's supposed to operate, so it may not always work. You probably want to find a special "breed" of SCRs called GTO thyristor, but these are almost impossible to find, especially in small sizes. \$\endgroup\$ – Bradman175 Jan 7 '17 at 23:07

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