Simple Power On Reset Circuit - Page 1
Forum section: electronics / projects

author: Matt Coates
author_rank: Contributor
timestamp: July 14, 2025, 10:17:40 am
content: I'm looking for suggestions for a simple circuit that will generate a 0.5s pulse, once, as soon as 5v power is applied to the circuit.The aim is for a low cost and simple solution. There a bunch of neat POR ICs, but they all monitor rail voltages which isn't helpful. This isn't being used to reset a MCU, its a pure timing application as part of a clock circuit.TI has a simple circuit using a RC + Inverter here -https://www.ti.com/lit/po/scet013/scet013.pdf?ts=1752170217746- but for a 0.5s to 1s pulse the capacitor will be large and may take a while to discharge between power cycles.Does anyone have any ideas or alternitive options? Open to the idea of a 555, but you still need to use the same RC circuit to feed the trigger pin, so you may as well use the inverter as its a lower part count option.Thanks!

author: Zero999
author_rank: Super Contributor
timestamp: July 14, 2025, 10:31:39 am
content: Quote from: Matt Coates on July 14, 2025, 10:17:40 amThat will work, just make the resistor value very high, say 10M for 1s. CMOS gates have a very low leakage current. The capacitor will discharge very quickly, via the internal ESD protection diodes, when the power is removed.A 555 will work, but will requite a larger capacitor because it has BJT inputs which a have a high bias current.Note any application using an RC circuit will be innacurate, 20% tolerance being typical. A crystal oscillator and counter will be more precise, but bear in mind it takes time for the osicllator to start, although that's less of afactor for such a long delay.

author: Paradroid
author_rank: Contributor
timestamp: July 14, 2025, 10:34:57 am
content: Instead of making the capacitor huge, you can also increase the resistance to slow down charging of the cap. Also, note that TI's app note clearly states that it must be a Schmitt-trigger inverter, not just any inverter.If you're looking for a lower component count solution, the best idea I'd have is one of those 3-cent MCUs in tiny packages. Dave had that "3-cent blinky" video a while back, featuring a Padauk MCU, which has a proprietary MCU core. However, there are Chinese manufacturers that make MCUs with PIC-compatible instruction set that are extremely cheap - at least in bulk.If this is a hobby thing, you should stick to the TI app note - can't get any simpler than that. For mass-production, a pre-programmed MCU might increase reliability and decrease assembly cost.Jens

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 14, 2025, 12:00:36 pm
content: Relying on the ESD diodes to discharge your capacitor probably works, but it's not a neat design. You can also put a (schottky) diode over the resistor, so it attempts to power the whole circuit, and that usually drains it quickly (but with these <100nA uC's these days it may still take a while.)You can make something around a thyristor that shorts the capacitor (or the whole power supply) when it detects the supply voltage is dropping. But this is probably more complicated then is justified.Thyristor circuits that short out a power supply is quite common. (See "crowbar"). In your case you don't want it to blow a fuse, so add a small series resistor.

author: Zero999
author_rank: Super Contributor
timestamp: July 14, 2025, 01:05:59 pm
content: Quote from: Doctorandus_P on July 14, 2025, 12:00:36 pmThe problem with a Schottky diode is it has a very high leakage current, compared to an internal ESD protection diode. A 100nF capacitor charged to 5V, is only 1.25µJ, which won't damage the ESD protection diode.QuoteHow do you stop the circuit from triggering every time the power is applied?The circuit needs to be activated, when the input votage drops below the output voltage. I wouldn't recommend anything latching. Here's a simple circuit using a BJT, which turns on when the output voltage exceeds the input voltage, by a diode drop. The downside is the output voltage is lower than the input voltage, by a Schottky diode drop. No doub D1 could be replaced by an idea diode circuit, but it's more complex.

author: Ian.M
author_rank: Super Contributor
timestamp: July 14, 2025, 01:32:59 pm
content: An unconventional approach to generating a long one-shot active low reset pulse without large RC time constants is to use a CD4060 CMOS 14-Bit Binary Counter and Oscillator. Give it a RC oscillator of slightly over 16KHz, and its Q14 output will toggle at approx 1Hz. Add a diode from Q14 to its CLOCK in pin, cathode to CLOCK and that will stop the oscillator as soon as Q14 goes high. Add a simple RC POR circuit to its RESET pin to guarentee it powers up with all outputs low, and you'll get a nice 0.5 second low pulse on Q14, starting from powerup, then once it goes high, no further activity. Not bad for a single IC, three resistors, two capacitors and a diode!If you need active high reset its less attractive . . .

author: Peabody
author_rank: Super Contributor
timestamp: July 14, 2025, 02:09:04 pm
content: It seems there's a lot we don't know. How fast does the 5V come up? Will this be happening often? Does it have to be a square wave? How much tolerance on the 0.5s? What's the input impedance of whatever the pulse is driving, and what is its Vih?Anyway, depending on those answers, you might not need the inverter. A series capacitor would immediately transmit the rise in voltage, and then the right discharge resistor would bring it back down at more or less the right time.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: July 14, 2025, 02:49:46 pm
content: Quote from: Matt Coates on July 14, 2025, 10:17:40 amIf you're looking for something that can give youconsistentlyandprecisely0.5s (within a certain reasonable expectation, of course), then that disqualifies the common POR/voltage supervisor ICs anyway. They don't give any kind of precision with the output timing - it's very approximate, and varies with temperature and input voltage. For example, I have personally found that clone MAX810s give a longer output pulse with higher VCC voltage; for example, the pulse length is longer with 5V versus 3.3V.

author: Matt Coates
author_rank: Contributor
timestamp: July 14, 2025, 04:14:33 pm
content: Thanks for all the responses!So the output pulse will be driving the enable input of a high side swtich which is being used to drive an indicator. As soon as 24V power is applied a DC-DC will generate a 5V rail and this will (amongst other things) power/trigger the pulse generation circuit. The purpose of the pulse is just to illuminate some indicators and a buzzer - any duration between 0.5s and 1s would be sufficient.From the sounds of things the RC and inverter approach with a diode across the resistor should work well. The reverse leakage should be around 0.1uA, so if using a 470k and 2.2uF RC that will be charging up at 10uA giving only a 1% error due to the diode. The ESD diode drop is nominally 0.5V for the SN74LVC1G17 so if using a BAT54 it should conduct well before that and stop the bulk of the current passing through the internal diodes when the capacitor discharges.I do like the sounds of the oscillator and counter though as it would be more consistent in the duration. The time isn't too critical but it would be bad if we got 0.3s on some units and 0.9s on others - I will investigate further!

author: Terry Bites
author_rank: Super Contributor
timestamp: July 14, 2025, 06:35:48 pm
content: NCP302, NCP303 from OnSemi.Run on gnat farts, a single fart may well be enough though.Add a 470n cap and thats job done.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: July 14, 2025, 07:07:35 pm
content: I guess something like the attached circuit could work.As soon as the the capacitor starts discharging with more then 600mV/1M = 60uA Q1 opens and triggers the thyristor. But I I'm not sure the capacitor can even deliver enough current to trigger the thyristor.A 100nF ceramic capacitor can still deliver a high peak current, but ESD diodes are made for high peak currents. But just putting a series resistor of a few hundred Ohm in between the capacitor and the logic gate ensures there are no high peak currents, and still limit the discharge time to less then a ms. (Of course depending on all the other bulk and decoupling capacitors on the power supply rails).From the RC constant you can expect a timing deviation of less then 10% , but the trouble is with the voltage at which the logic gate triggers. This has some variability that influences the timing, and this treshold is not well specified (It may be better for a schmitt trigger).Idea:Remove the thyristor, replace the BJT with a P-channel fet with low threshold voltage, with it's drain to GND.(Or just connect the collector of the BJT to GND). I think BJT's tend to have a lower leakage current then FET's, but it's all also temperature dependent, so testing is not trivial.The circuit (and variations) are easy to build on a breadboard. If you add an opamp with FET inputs to buffer the capacitor voltage, you can also measure it easily with your oscilloscope.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 14, 2025, 07:21:32 pm
content: Quote from: Matt Coates on July 14, 2025, 10:17:40 amWhy isn’t that helpful?What you describeissupply monitoring.Does precision matter ?Most POR parts have nominal timing, but you can get external cap versions or OV/UV parts can have external RC for more precise timing.Monostable parts like 1G123 have Schmitt triggers and precise RC timing, with discharge at end of time, which gives good recovery.addit:Quote from: Matt Coates on July 14, 2025, 04:14:33 pmSo long as you avoid the truly terrible capacitors, you should be able to get 5-10% type spreads from a schmitt and RC.Monostables will be more precise, and there are CMOS timers like HEF4541 that allow NPO caps to be used, for even better precisionQuote from: Matt Coates on July 14, 2025, 04:14:33 pmIf the enable is well defined and has hysteresis, you could apply the RC directly.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: July 14, 2025, 07:28:39 pm
content: Quote from: Matt Coates on July 14, 2025, 04:14:33 pmKeep in mind that many logic parts have only one diode to GND, to allow hot plug use.A cap to Vcc can use that diode to discharge.

author: Zero999
author_rank: Super Contributor
timestamp: July 14, 2025, 08:21:13 pm
content: Quote from: Doctorandus_P on July 14, 2025, 07:07:35 pmThat circuit will only work if the supply voltage falls quickly enough to enable the voltage drop across R1 to exceed 0.6V and trigger the SCR. Imagine the power supply voltage falling slowly due to a large capacitance on the output, the voltage across R1 will never reach 0.6V. Another potential issue is if the power is reapplied, before the current through the SCR falls below the holding current, causing it to remain latched on and short out the power.

author: Solder_Junkie
author_rank: Frequent Contributor
timestamp: July 15, 2025, 09:07:53 am
content: A reset on powering up equipment is quite a common requirement. The attached circuit is an add-on for an Elecraft K4 amateur radio transceiver that grounds a pin on power being applied to the radio, pretty simple stuff that may be useful.SJ

author: youngda9
author_rank: Regular Contributor
timestamp: July 15, 2025, 01:34:39 pm
content: Put a diode across the resistor in the OP suggested circuit to discharge the capacitor quickly when power is cycled.

author: Peabody
author_rank: Super Contributor
timestamp: July 15, 2025, 01:42:18 pm
content: I'm still unclear as to how much current the pulse has to supply.