"Reset" circuits. - Page 1
Forum section: electronics / beginners

author: paulca
author_rank: Super Contributor
timestamp: June 11, 2025, 10:43:39 am
content: It's amazing how complicated such a "simple" little circuit gets when you open pandora's box.I got this one of Stackover flow via a bot. I studied it for a while, decided it covered all three of my requirements. The fact it was doing something else as well, I left for later to figure out and breadboard with the scope.Having done initial tests and after staring at it a while, I think it's OTT. I think it's trying to cover too small,negligible corner cases that I don't really care about and could possibly lose at least one component.Requirement one. Allow RESET to rise in a controlled fashion so that it passes the "HIGH" level "ONCE" on power up.Requirement two. Additionally. Provide a few microseconds for VCC to stabilise board wide, caps to charge et. al. CPU to reach stable reset state.My analysis.Initial: VCC is "OFF" - assume it's close enough to GND to call it 0.When VCC bounces like a child on sugar when the power switch snaps closed, RC circuit of (R2 and C5) causes a slow rising level which should give plenty of time for the bounces of a switch and the CPU state. Around a ms to 3.5v I think my scope measured the rise around 900us to reach the "HIGH" level from the datasheet.I think it passes the requirements with the caveat.... it depends on your clock speed and the values of R2 and C5 such that the CPU is held in a RESET LOW state for a "TBC" number of close cycles to properly brings it's internal state to consistency. Failure to do so causes "false boots" where the thing launches into fetch exectute while there is still garbage in the PC.Requirement three: Debounce the RESET Switch.The switch connects C5 via the 100R to ground. So it will have a discharge time from 5V down to about 1V of roughly 100us. Long enough to debounce a tactile micro-switch.Circuit analysis.When the switch is closed the 100R discharges C5 to ground. C5 is still being charged via the 1k, but is being discharged by the 100R. RESET will stabilise at the VCC/1100*100 or < 1V but not 0, as R1 and R2 form a voltage divider with RESET the mid point.Nothing has involved the diode yet.This is the "other things" it's doing, or attempting to do. I believe it is intending on getting rid of the charge of C5 when VCC drops to (near)0V.When VCC is high, the diode will not back flow and is "effectively not there".When VCC drops while C5 is charged, assuming no diode, C5 would immediately attempt to power VCC via the 1k. This will discharge it back to the CMOS low in about 1ms - assuming there IS actually power draw there - remember kids, ALWAYS put a "load" on your circuits under test as the results are VERY different here if you don't.Is 1ms too slow? Well, too slow for what exactly?Unless we attempt to turn it back on again within that ms? Thats a bit of an "out there" case, no?The only thing I can see D1 doing is "flooring" C5 to 0.6V as fast as possible by giving it a low resistance path to power the circuit through VCC. How long depends on how much other reservoir capacitance there is.What corner case is it trying to cover off?

author: Ian.M
author_rank: Super Contributor
timestamp: June 11, 2025, 02:10:36 pm
content: Note that the MCU or CPU will probably come out of reset at some mid-rail voltage approx halfway between the reset input's logic thresholds, so you need to calculate the worst case pullup resistor and capacitor values to keep reset below its logicLOWthreshold for long enough.The 100 ohm resistor is to damp the 1uF capacitor ringing with the switch track and wiring inductance when the switch shorts it. Without it, the RESET line coud ring to -5V with a current spike of several amps, which if connected to a typical MCU or CPU reset input could easily take it negative enough, hard enough with respect to the substrate to cause charge injection resulting in catastrophic latchup. However IMHO its in the wrong place as it compromises the logic low margin - it should be in series with the pullup at its bottom end, feeding the capacitor and RESET, with the switch to ground from the junction of the two resistors.The diode improves the chances it will reset cleanly if there is a short power dropout. Without it the risk of a hang or an unexpected execution state vastly increases. However if the power is notably glitchy, it may not be sufficient and you may need a proper supervisor IC.

author: paulca
author_rank: Super Contributor
timestamp: June 11, 2025, 02:34:52 pm
content: Quote from: Ian.M on June 11, 2025, 02:10:36 pmI see that would "Ground" the mid point of the two resistors removing the voltage divider, everything "south" of the RESET is then 0V (steady state).

author: jwet
author_rank: Frequent Contributor
timestamp: June 11, 2025, 05:30:33 pm
content: Maxim and many others make thousand of flavors of tiny little supervisors that cost about a dime in qty that do a nice job. It is a surprisingly complicated problem and supervisors are worth the cost in my opinion. They're likely cheaper than the RC diode lashups too.Some of the benefits-1. They trip at a fixed voltage fairly precisely over temp and time.2. Timing is accurate enough especially as a minimum value and is stable.3. They initialize themselves so even chattering Vcc generate a full timeout each time. They also have hysteresis so once triggered complete a cycle.4. They don't store energy that has no where to go on power down.5. Their output is defined as Vcc disappears.A couple to look at Max809,Max811- the first is just a three terminal basic device and the second has a button input including pullup R's. Suppiy current is 12 uA IIRC.These parts are 20 years old and have been copied to death by Asia.

author: Ian.M
author_rank: Super Contributor
timestamp: June 11, 2025, 05:39:06 pm
content: +1.However for O.P's simple circuit, one of the other jobs the diode does is give (4.) the energy stored in the capacitor somewhere to go on power-down. It prevents Reset from going more than one diode drop above Vcc as power is lost so protecting the CPU/MCU, so don't omit it.N.B. the diode also prevents HV programming of MCUs that take Reset above Vcc to enter programming mode. This is especially problematic for legacy Microchip PICs.You can get supervisor ICs with open drain HV programming tolerant outputs which avoid this issue.

author: BrianHG
author_rank: Super Contributor
timestamp: June 11, 2025, 08:15:13 pm
content: I prefer to add a NPN 2N3904, with the reset cap tied to the VCC and base with series resistor and discharge diode. I tune and get a rapid full digital swing output transition at 0.1sec after VCC fully stabilizes.

author: Benta
author_rank: Super Contributor
timestamp: June 11, 2025, 08:27:40 pm
content: I just use a MAX803.

author: PGPG
author_rank: Frequent Contributor
timestamp: June 12, 2025, 10:18:08 am
content: In 90s I was using some reset ICs, but later uC get integrated brown-out detectors and are self-sufficient (at least uC we use).