Hi, I'm re-designing a yet-to-be-implemented home remote control lighting system that uses mechanical latching bistable (or impulse) relays. These particular relays will change state with a low voltage pulse on only 1 input pin. I would like to convert the system to use SSR's but I'm uncertain which to choose and how to achieve the one-wire bi-stable control. I prefer SSR's for the higher operation count. A single relay will drive a resistive load at most of 2.5A 120V AC. The control voltage can be anything below 24V AC or DC. I don't care about the retain-state-even-thru-a-power-outage feature offered by the mechanical relays. I'm using simple momentary contact pushbuttons for controls with no debouncing.
1) Do any SSR's exist that already contain some or all of the extra features I need? (debouncing, bi-stable, one-wire control)? I'd like to keep it cheap --at most $25 per relay.
2) The question #1 is no, can you point me to some simple SSR support circuits that would work?
I don't recall any SSR"s with all the functions you require.
The 4013 D flip-flop is a suitable device to use for momentary on-off toggle operation.
Some tuition material is here
Here the toggle input is provided by a 4093 astable pulse generator but this could be replaced by your momentary action switch. The SSR input would replace the output led. You can run it on 12Vdc.
Since it would be wise to co-locate the 4013 toggle close to your SSR you should have some resistive input protection on the D input (eg. 1K ohm series resistor). Your momentary switch (which might be some distance away) must have some debouncing which could be achieved by simply using a 220 - 470nF cap directly across it.
Hi, Zeb. SSRs exist which provide the function you want -- the SSAC KRPS ProgramaCube comes to mind:
If you decide to go this route, look at the data sheet to help specify your product.
The hangup, of course, is that you're asking for this functionality at about the retail price of a standard SSR. Not gonna happen.
If I might suggest something, why not just replace your switch with an alternating action pushbutton? That will give you the function you need for the price of a switch (i.e. much less than $25). You can then use a low AC or DC control voltage and replace with SSRs at your convenience.
The one nice thing about the momentary switch control is that a pulse from one switch turns it on, a pulse from a different switch across the room can turn it off --no complicated 3-way switch wiring. I could have a
OK, Zeb. If you want to use multiple inputs, an alternating pushbutton alone won't do the job.
If the SSAC device seems a little pricey, or if you just would like to try your hand at some digital logic, you might want to go with the suggestion of Mr. Herbert and use a 4013 flip-flop. I'm assuming you have a ground-referenced power source at each pushbutton here to utilize the "one wire" concept. Each switch will have to have a following diode to prevent two power supplies from being shorted to each other when two switches are simultaneously pressed (view in fixed font or M$ Notepad):
At the logic side, you're going to have to provide input protection for your logic as well as debouncing. To supplement the 12VDC power supply and the CD4013, you should also add a 74C14 hex inverter with shmitt trigger input. This will provide hysteresis to allow "snap action" for the CD4013 clock, like this:
Good to hear back from you, Zeb. First off, 4000-series CMOS is what you'd call mature -- not obsolete. I'd bet you'll still be able to find it in 7 years -- twenty, probably not. The good thing about using 12V for a power supply is that it's very common, and automatically gives you 2.4 times the noise rejection capability of the same logic at 5V for no cost. As for the rest, one question at a time:
1) You'll have to look at the SSR -- most of them are capable of being driven by 5V logic, but the current requirements differ. Some require only 3mA at 5V. Those should be directly driveable by your 74HC output (typically can source or sink at least 4mA). If your SSR requires a higher input voltage or draws more than 4mA, you might have to add the transistor again.
2) You got it -- the R-C provides a reset pulse at power on. If it truly makes no difference at all, you can omit it, and just GND the reset input like the set input.
3) Sorry -- I should have given that. The switch input should go like this (view in fixed font or M$ Notepad):
Thanks for noticing the issue -- I split the circuit up and missed the
1K pullup -- my bad. The circuit wouldn't have worked without it.
4) Schmitt trigger ICs all hapen to be inverters. Not sure why. Actually, you can make a non-inverting schmitt trigger with two CMOS inverters and a couple of resistors, like this:
You have to look at whether the flip-flop clocks on the positive-going transition or the negative-going transition. The above circuit uses a flip flop that clocks on the positive transition. That will occur about 100mS after you press the button in the circuit above. If the FF were NGT-triggered, it would be 100ms. after you released the button. Personal preference, except that people are accustomed to things happening when they push the button, not when they let it go.
Some of this circuit baffles me. I think I know 1, 2 & 3 below. But what are 4, 5, & 6 for?
1) The input to the schmitt trigger (inverter) needs to be held high because we want a high (inverted) output when the switch is pressed.
2) The 1K resistor is there because we don't want to dump the entire current of the power supply to ground when the switch is pressed.
3) When the switch is pressed the capacitor dumps its current thru the
10K resistor and to ground making the input to the trigger low. If the 10K resistor was not there the cap discharge could fry the switch contacts.
4) The diodes next to the switches?
5) The "extra" connections to VCC with the diodes?
6) The "extra" ground connection with the diode?
Hi, Zeb. 1), 2), and 3) are good. The extra diodes provide circuit protection for the logic. The extra diodes for the switches aren't necessary if you're using switches rather than another logic gate for the circuit inputs. Now the diodes from the inputs to Vcc and GND at the input to your flip flop circuit are there in case there's noise at the input which exceeds those limits. You don't want the input to the inverter to exceed the power supply evn for a microsecond -- it will fry the logic gate. The diode above the cap is there to prevent the cap from discharging through the logic gate when you turn power off.
Try building the circuit as shown, and forget the diodes at the switches if you are using switches. It should work.