Relay pulser circuit - Page 1
Forum section: electronics / projects

author: pertinax
author_rank: Newbie
timestamp: July 28, 2025, 09:40:10 am
content: Hello, I have drawn this little circuit to allow a relay to turn on momentarily. The intended application is as part of a starting circuit for a single phase fan blower motor (a common application for the season)The circuit shall be connected in series with a 230 V relay coil and the AC input. When AC is first applied, the 555 is powered through the relay coil, by the voltage drop across the zener diode, D1. The timer can then short the MOSFET gate using its DISCHARGE output after a specified time, turning off the relay. Since the circuit is powered through the relay coil, it is protected by an additional impedance from the mains. No fuse is needed. Initial power is supplied from the drive current of the coil. When the relay is turned off, supply voltage for the 555 is upheld using R3. The specific relay I have looked at will have a coil current of around 3 mA and a specified "must release" coil current 15% of this, i.e., 450 µA. This also works out.It seems to work in LTspice. However, I am concerned that the discharge output from the 555 will not go higher than VCC (by a diode drop or so). Does anyone know if this is the case for the CMOS 555?Thanks!

author: Zero999
author_rank: Super Contributor
timestamp: July 28, 2025, 08:26:14 pm
content: You might get more replies if you post the .asc file, along with any models and symbols.Note there's no need to make your on symbol for the 7555. Click on component, , DIP8. Place the component and set its value to the subcircuit name of the model, which can also be pasted into the .asc file.I haven't tested a 7555 to see if the discharge pin voltage can be higher than the supply voltage. I suspect there's an ESD protection diode between pin 7 and pin 8. If it can exceed the supply voltage, then it will be limited to the maximum voltage rating of the IC. Adding a transistor to form a cascode, with the internal transistor on the discharge pin, will enable it to switch higher voltages.

author: pertinax
author_rank: Newbie
timestamp: July 29, 2025, 07:57:25 am
content: Thanks for helpful suggestions. I didn't know about the general 8-pin dip symbol, looks more tidy indeed! I thought that I had included a zip with the simulation files. I will try again.Here is an updated schematic with your cascode idea. Also, I added a second transistor in order to sink the gate current into the VCC node, so that it can be used to maintain VCC also when the main MOSFET is turned off. This way, I don't need a second high voltage pullup resistor.

author: Zero999
author_rank: Super Contributor
timestamp: July 29, 2025, 09:36:09 pm
content: Silly question: why not use an off the shelf timer relay?Is it alright if there's a short delay, before it turns on, say 200ms? I have an idea about using a delayed pulse to turn on the MOSFET, after the power supply capacitor has charged enough. This will reduce the power dissipation in the 7555 circuit. I don't have time to post the circuit now, but here's how to get a delayed pulse.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 29, 2025, 11:15:52 pm
content: Quote from: pertinax on July 29, 2025, 07:57:25 amWith all those parts, do you still even need a 555 ?QuoteThat already has hysteresis, so your simplest circuit can be a capacitor with a bleed resistor on the bridge pins.Or you can get more precise timing with just 2 MOSFETS.With 2 MOSFETS the ON-OFF transfer can be regenerative, so you get a nice snap-action on the drive.M1 is a low threshold small MOSFET and M3 is a Mains voltage rated part.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 29, 2025, 11:37:09 pm
content: QuoteHere is that simplest possible delay circuit. Works with the holding current of the relay.Very simple, but the recovery time is a trade off of R7 and min relay current. Will be a few seconds, so much slower recovery than 2 MOSFETS.

author: pertinax
author_rank: Newbie
timestamp: July 30, 2025, 09:03:24 pm
content: Quote from: Zero999 on July 29, 2025, 09:36:09 pmThats not a silly question. I guess a bought part is the straight forward solution, but think about all the interesting circuits and follow up ideas which have resulted.Yes. It would be alright with a short delay before the pulse. I also thought about the idea of delaying the pulse in order to have time to charge the supply capacitor. This doesn't seem to be a concern with the updated design from my last post, though. Since the pulse occurs directly and in fact helps to charge the supply rail.Quote from: PCB.Wiz on July 29, 2025, 11:15:52 pmWell, I guess the 555's existence can be debated.I didn't directly find a simple circuit which suited my requirements without it. Also, since its main purpose is timing, when using the 555 within its specifications, the datasheet parameters and equations can be used to calculate pulse times etc. This gives the circuit some "anchoring" and predictability, so to speak.I like your circuit with only transistors. This is what I was looking for originally. I guess, though, that the actual timing will depend on the M1 threshold voltage. So it may need to be tweaked.Quote from: PCB.Wiz on July 29, 2025, 11:37:09 pmNice! The recovery time shouldn't be a problem. As long as the fan doesn't have time to spin down. It should take more than a couple of seconds. One concern is that the capacitor needs to handle some voltage and will be physically somewhat large.I have drawn up a small board with the 555 solution. It may be a bit overcomplicated, but my hope is that it will work out of the box without tweaking. It contains the timer circuit, the relay and connections for a 3.3 µF run capacitor as well as the motor slow speed winding and neutral. This should allow it to be easily integrated into my existing cooling fan.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 30, 2025, 09:15:50 pm
content: Quote from: pertinax on July 30, 2025, 09:03:24 pmYes, during charge phase, M3 clamps and M1 gate charges, so the charge current is related to Vth(M3)-Vth(M1)