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I'm trying to build a circuit that will stay turned on once triggered, and ordinary SR latch seems like a perfect solution. However, I cannot find an efficient way to turn SR Latch to deterministic state on power on. What could be the simplest yet efficient solution? Only one requirement - I want to build it entirely from basic components, without using any ICs, for a learning purpose. As I understand, I just need to generate some short pulse when circuit is powered on, and I can achieve it by connecting a capacitor in series to S or R input of the latch, but then after I turn circuit off it stays charged. Of course I can use bleed resistor and some kind of a filter before input from the capacitor to latch S or R (schmitt trigger?) to cut off small current that comes thru bleed resistor, but I feel that there are better solutions for this particular task.

I will appreciate if anyone can give me a direction to learn instead of the final solution, except searching for it directly (already tried without success)

I do not think that this question is a duplicate of Starting a latch in a known state since I asked for non-IC solutions, but if moderators disagree, please feel free to close current one.

I apologize for my bad English and a newbie question, thanks in advance.

devnull
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Georgy Farniev
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  • No ICs, right? So just discrete BJTs and FETs and passive components? Correct? – jonk Aug 24 '21 at 05:38
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    only passive components and BJTs – Georgy Farniev Aug 24 '21 at 05:42
  • So you intend a switch to turn things on and some kind of "power on reset" circuit that guarantees an initial condition of OFF before the switch is used. But once the switch is applied, the circuit stays ON, forever. (Until the circuit is disconnected somehow from power.) Is that about it? – jonk Aug 24 '21 at 05:54
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    Yes, that's correct. – Georgy Farniev Aug 24 '21 at 06:10
  • So you are going to want a circuit that does something [like this](https://i.stack.imgur.com/NcVXc.png) for the power-on duration -- it tracks the power-on voltage as it rises until it can come into managed control, at which point it pulls LO for a set period in order to reset the circuit. This is not hard but takes some thought. This LO-going reset pulse can be used to reset the rest of the circuit that monitors the switch. I have a few ideas. But perhaps someone else will do a better job of helping you on this. Let's see how it goes. – jonk Aug 24 '21 at 06:22
  • Currently, I did some pulse circuit using capacitor and resistor, but not sure if it will work in real life. I will construct an SR latch and give it a try. – Georgy Farniev Aug 24 '21 at 06:24
  • You will want carefully designed ***hysteresis*** for the circuit. It's not something to leave to an RC by itself. Seriously, an RC is insufficient. I'll +1 the question. – jonk Aug 24 '21 at 06:26
  • Don't forget to add a diode to discharge the capacitor on power-down. – the busybee Aug 24 '21 at 06:50
  • Can you please give me an example on how to connect a diode to discharge the capacitor? Thanks. Also, I built an SR as here and concept with resistor and capacitor works very well, given that I will not switch device on and off very fast, faster than capacitor discharge time after it's off. – Georgy Farniev Aug 24 '21 at 08:36
  • You may find this question and answer informative. https://electronics.stackexchange.com/questions/350978/bistable-multivibrator-using-discrete-transistors – Peter Smith Aug 24 '21 at 10:14
  • Here you can find a solution with just one transistor: https://electronics.stackexchange.com/questions/566737/initializing-an-nor-sr-latch/566745#566745 –  Aug 24 '21 at 17:05

3 Answers3

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You could use a small RC pair to provide a signal that stays high or low for a short time after power on:

schematic

simulate this circuit – Schematic created using CircuitLab

D1 solves your problem of the capacitor remaining charged after power off, but probably isn't necessary, as the input protection diodes on most CMOS device inputs would perform the same function.

As to how to use this with an SR latch, here's a hint (since you don't want the full solution): You don't have to use a 2-input gates.

Simon Fitch
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Don't dismiss a bleed resistor. It is simple, reliable, and does not depend on any other circuit components to work.

A diode across the timing resistor will discharge the timing capacitor only if there is a low enough resistance through other circuit components when the power is removed. From the circuit's point of view, there are two ways power can be removed. 1. The positive rail goes open circuit. 2. The positive rail remains connected to the power source, and that source presents a low resistance to GND when off. A discharge diode is not guaranteed to work in case #1.

For a simple 2-transistor bistable latch circuit, the pulse width at the Reset input can be pretty short, on the order of few milliseconds; let's say 10 ms. If a bleeder resistor across the timing capacitor is 10x the timing resistor value, the discharge period to less than 1% charge (5 time constants) will be approx. 50 ms. This should be short enough for even a fast power-off-power-on cycle.

AnalogKid
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Here's a circuit utilizing an electromagnetic relay.

enter image description here

vu2nan
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