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Run timer with reset

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New Member
I'm looking to make a run timer with a reset. I have a thermostatically controlled refrigerator on my boat, and I want to be able to "force" it to run for 30 minutes when I have extra power available (engine, shore power, etc). I also want to be able to cancel it. What I envision is a single push button, that when pressed engages a relay (and an indicator lamp) and starts the timer. The timer would run for 30 minutes and then shut down. I'd also want to be able to press the same button, and have it shut down.

I can find many appropriate circuits using a 555 as a timer (but I think it sounds like I can't really get 30 minutes from that...), and I can also find latching flip flops to turn a single push button switch into an on/off circuit -- but how to do both? I'm sure that there are lots of mechanical solutions, but I'd really like to make a semiconductor solution as a learning exercise.




Well-Known Member
Most Helpful Member
Look into a CMOS 4060 timer chip, followed by a transistor to switch the relay.
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Well-Known Member
Try a 555 first. The datasheet says it's capable of producing delays of usec to hours. I think capacitor leakage current is cited as the main reason for problems with long delays. Doesn't hurt to give it a go, just have a few different capacitors available if you need to change the timing.


New Member
I appreciate the comments on the time duration, but my primary concern is the basic circuit. I don't know how, and can't find any with much Googling, to make a single-switch to initiate time delay and abort.



Well-Known Member
Hope you don't mind the sketch, the left 2 gates form a toggle FF, while the right is the pulse limiter. Output will have to be buffered with something to drive your fridge obviously.

Play around with the 220u & 10M to get the timing you want


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New Member

COOL! I think I can even follow it. Let me see if I have it straight:

* NAND 1 inputs are high (is this a natural event? No matter anyway, since the circuit will be "always on."), and it outputs low.
* NAND 2 then outputs high, latching NAND 1
* The output of 220u is high (and discharged), NAND 3 sees 1/1, and outputs low.

* You press S1
* The 100k/1u are discharged, and take NAND 1 low (the R/C also provide debounce, I think)
* NAND 1 outputs high (and the R/C begin to charge)
* NAND 2 outputs low
* NAND 2 keeps NAND 1 inputs low, as a latch
* The 220u is discharged, and so takes 1 leg of NAND 3 low, and NAND 3 outputs high. A transistor (or opamp) drives a relay.
* The 220u starts to charge through the 10M
* When the 220u gets high enough, NAND 3 sees 1/1, and goes low, shutting off.
* To abort, a second press of S1 takes inputs of NAND 1 high (from 1u, which is charged)
* NAND 1 goes low, driving output of NAND 2 high (which latches NAND 1)
* Input of 220u + charge state of 220u is high (and discharges through D1 -- Schottky?)
* NAND 3 sees 1/1 and outputs low.

It all makes sense!

One remaining question:

* When the unit naturally times out, the FF is still in the "activated" mode, with NAND 1 output high and NAND 2 output low. A press of S1 will "reset" the unit, requiring TWO presses to get it going.

Could this be fixed by splitting NAND 1 inputs, one as shown and one from NAND 3 inverted through NAND 4? I think the "start" and "abort" results will be the same, but when it times out, NAND 3 will output low, NAND 4 will go high, and NAND 1 will see 1/0, so it will output high and the rest is the same as an abort.

Have I read all this right?

Thanks again!



Well-Known Member
I'm impressed that you put the effort in to try to understand the circuit!

I think you got it pretty much. I'll try to explain in my words (which I now realise are a bit convoluted, sorry, I hope it can still help though):

Nx = NANDx (where x is 1, 2, 3)
o/p = output, i/p = input

Whenever N1 input is low:
N1: inputs are low
N2: inputs high (due to N1 o/p), o/p is low
N3: i/p low, o/p high (while 220u charges through 10M)

when 220u charges to 2/3vcc, N3 output goes low

Whenever N1 input is high:
N1: inputs are high
N2: inputs low (due to N1 o/p), o/p is high
N3: i/p high, o/p low

220u discharges quickly through diode & 10M

When the switch is off:
1u cap (dis)charges to N1 o/p voltage through 100k

When the switch is on (pressed):
N1 i/p voltage is pulled to 1u cap voltage - N2 o/p becomes the same value and latches both N1 & N2 (stable state). The 1u cap remains at the same state due to the 10k from N2 output. The cap state only changes when the switch is released.

Re your question about giving it a known state on start-up, just connect a 100n capacitor from vcc to N1 input. --> this is to start up in the 'OFF' state (i.e. N3 output low)
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