# How To Hold Power ON For Some Time Without Using A Microcontroller or 555?

This is my first question here, so I'm hoping for help.

I'd like to take a Momentary event and use it to turn on a device (LED, for starters to model the concept) for a minute (give or take) at a steady Current & then go off. I'll need to provide ~3v at ~.350 to ~1A (.350 most likely, 1A not-so-much) from a low-voltage DC Source.

Imagine walking up to your door, stepping on the mat (with a pressure switch under it), and having the porch light come on for a minute without using your hands and you'll get the concept. With a minimum component count and all (other than wiring) in as small a package as possible. EDIT: I'm trying to build a proof-of-concept prototype as described, but the end result will be human-wearable.

I've been playing with Monostable Multivibrator ideas ("one shot" being irresistible), but seem to have hit some saturation point in my soggy melon & can't get a steady current while ON.

There's more to this project, but this specific part is scrambling my brain & I fear I'm down a rabbit hole & need expert help digging back out.

Currents, voltages & time are approximations, but my primary concerns are NOT using a microcontroller or 555, holding the Output steady during the entire "unstable state", using no or minimum quiescent current during the "stable state" (not that a one-shot is the right answer!) and keeping the bounding box as small as possible as well as keeping the cost at a minimum.

No need for complexity, just a simple trigger-and-hold-for-a-minute latch. EDIT: I'd like to emphasize the need for brutal simplicity. And that a quick momentary bump would hold non-TTL current to the load for about a minute.

I hope some creative electronics genius reads this and can suggest some possible circuit(s) to try.

Thank you for reading, and more-so for answering with a simple circuit or a link to a page with one on it.

J

• you could just be doing something wrong with the 555 timer.. a One-shot should be all you need, there must be something going on there. Otherwise a bit of discrete logic in the form of fast ON slow OFF capacitor timing with an RC timer, comparator, MOSFET or other transistor to do the switching of the load. It's not really worth the discrete component complexity Aug 11, 2015 at 4:15
• I'm not using a 555. The point is to NOT use one, if possible. Aug 11, 2015 at 5:00
• Not sure what could be simpler than a 555. Maybe your goal is steampunk? Use relays to latch it on, and have a thermal switch over the light bulb that heats up after a minute. Aug 11, 2015 at 18:20
• I can understand not wanting to use a microcontroller, because of the maintainance cost associated (writing code, keeping it so one can understand the circuit); but why no 555? Of course the old (NE555) draws lots of current, but the CMOS versions are much less power-hungry! Aug 12, 2015 at 7:30
• Perhaps sensitive relay and capacitor. Jul 19, 2017 at 21:08

my primary concerns are NOT using a microcontroller or 555,

Okay. Not a real engineering goal but maybe an interesting goal for learning from. You'll find lots of monostable circuits online if you just google image search for them. It's also still possible to design digital logic from single gate building blocks.

holding the Output steady during the entire "unstable state"

You can probably solve this by using a buffer between the logic circuit's output and the load. The buffer might be a simple as a single MOSFET.

using no or minimum quiescent current during the "stable state"

Almost certainly the best way to do this is using a microcontroller instead of a monostable. Many microcontrollers can enter a very low power sleep state, and then respond when they receive an interrupt (for example due to a pressure switch being activated).

for a minute (give or take)

It's generally tricky to get a monostable to keep time accurately over such a long period. Again, the low power, low cost solution is likely a microcontroller. You could also design a digital circuit using discrete logic (start with a counter), but it will almost definitely consume more quiescent power than the microcontroller solution.

A general tip: When you're designing something for an application you've never tried before, first make a working design --- just one that has the basic required functionality --- then once you know how that performs start thinking about how to optimize it for speed or power or whatever's important in that application.

• Thank you for your answer. I'm warming up to the microcontroller idea, as I've found a couple of Very Small packages that might be made to fit. If it's bigger than the metal part of a USB 'A' connector, fitment will be a problem. And I have made a basic model, it's just too big. And the current to the load dwindles away as the RC tank winds down. Basically I'm in a rabbit hole & can't find my way out, Google or no. Aug 11, 2015 at 5:03
• For a uC appropriate for this job, the package sizes go all the way down to "too small to hand-solder". Same for discrete logic gates, but you'd need more of them to get the job done. Aug 11, 2015 at 5:07
• Thanks. (I just found out you can't hit <Enter> in these comments.) I'd like to know more of these small uC's you mentioned, especially if they'll supply ~1A DC? Aug 11, 2015 at 5:23
• Won't supply 1 A DC. A buffer MOSFET, or a buffer MOSFET driving a bigger buffer MOSFET, ought to solve that problem. As for small parts, the first one I checked at Microchip had a 3x3 mm package, and the first MSP430 I checked at TI had a 4x4 mm package. Aug 11, 2015 at 5:26

How long do you want the light to stay on for? Are you willing to tolerate a slow decay?

You can do a simple delay timer using nothing more than a MOSFET and a RC network on the gate. But there are a few caveats and one subtlety.

You want to use a MOSFET with a low threshold voltage, then charge the timing capacitor on the gate to a relatively high voltage. This will keep the MOSFET fully enhanced for the majority of time, then the light will decay to full off some time later.

The first caveat is that the MOSFET must be both large enough and with enough heatsink so that its temperature does not get too high while the MOSFET is in the linear region. Your power level is fairly low, so this shouldn't be a problem.

The MOSFET need to have a low threshold voltage. Head over to Digikey and search for "Trench FET". These have very low threshold voltage and are ideal for this application.

You need to charge the timing capacitor to a fairly high voltage - 12 Vdc is ideal. You mention that you have a 3 Vdc supply for the LED. Because the timing current is so low, you could put a standard 9V battery in series with the 3 Vdc supply to get 12 Vdc. The battery will last for several years but does need to be replaced when the timer period starts to get too short.

Simple sketch follows:

simulate this circuit – Schematic created using CircuitLab

The FDP8880FS-ND costs about a dollar and has a gate threshold near 3V. It should be fully enhanced at about 4 Vdc or so.

When the pressure switch on the mat closes, it charges up C1. The FET turns on hard and the light goes on. C1 remains charged so long as the switch is closed.

When the switch opens, the capacitor begins to discharge. However, the FET will remain saturated until the voltage drops down to somewhere between 3 to 4 Vdc. The light will then slowly decay until it is fully off.

C1 must be a quality capacitor. Ceramic or Film is good, electrolytic is bad. Increase the value to give you the time delay that you want.

You want faster decay time, so the addition of M2 and R3 should do that for you.

This is untested but it should work. In operation, M2 serves to rapidly discharge C1 whenever it is turned ON. It is turned ON whenever M1 is not fully enhanced.

Circuit operation is as follows:

Assume that the circuit is OFF. M1 is OFF, as is the load. M2 is fully enhanced and places a 10k resistor across C1.

When the switch closes, R2 charges C1 to the full 12 Vdc from the two series-connected batteries. R2 is so much smaller than R3 that R3 has almost no effect on M1's gate voltage.

M1 turns on hard and the gate voltage on M2 drops to about zero. M2 turns OFF and remains that was so long as M1 is fully enhanced.

When the charge on C1 decays to the point where M1 starts to turn OFF, M2 turns ON and rapidly discharges C1.

One final advantage of modifying the circuit this way: MOSFET M1 no longer needs to spend significant time in the linear region. It can now become a physically much smaller package not needing any heatsink.

• I think that's a step in the right direction. Thank you. The problem I have is, I'm severely limited in battery capacity as well as space. Aug 11, 2015 at 17:59
• The decaying current through the load is one of my problems -- the LED gets dimmer gradually over time, which isn't really helping. If it had some "clipping" waveform -- where it stuck at some level -- say 350mA -- and then ramped down, that would be okay. Aug 11, 2015 at 18:05
• You can do pretty much anything you want if you keep adding components. I'll modify my schematic with an untested change that should do what you want: it will make the decay time MUCH faster when the lamp starts to turn off. Aug 12, 2015 at 2:34
• I see what you're doing there, Dwayne, and I like it. The 2nd MOSFET shuts off the first w/o making me slog through the ever-dwindling current as the 1st goes out of its Linear mode into Saturation... Aug 13, 2015 at 0:43
• Why 2V? MOSFET M1 sees a gate voltage that starts off around 12V, decaying to somewhere between 2-4V. The 2N7000 FETs that I use regularly seem to start turning on somewhere above 1V. Note that M2 doesn't have to become fully enhanced: the goal is to make R1 look much smaller than 10M. That should happen quite nicely. Aug 13, 2015 at 5:27

Try this one. you can adjust the "VAR1" to change on time. To further increase/decrease change capacitor (C1) value.

I have uploaded Proteus simulation file.

• how does this work? How does C1 discharge? Aug 12, 2015 at 2:11
• push the button Aug 12, 2015 at 4:16
• push the button, then the capacitor get charged and V+ > V-, U1 will out Vcc which makes Q1 to turn on. It will stays as it is until you release the button. After you release button, V+ will be the capacitor charged voltage and it will discharge through U1. Output of U1 stays until V+ > V- Aug 12, 2015 at 4:29
• I think he just forgot to add a resistor parallel to C1... But that's not a low-power design, at least due to the resistor divider. Aug 12, 2015 at 7:26
• you can replace resistor divider with zener diode to give reference voltage. Aug 12, 2015 at 10:25

maybe something crazy like this:

simulate this circuit – Schematic created using CircuitLab

R3 is a potentiometer which you can tune how sensitive the pressure is required to trigger the logic change from the Force Sensistive Resistor (FSR) becoming lower resistance when pressure is applied (person steps on front mat).

An inverting buffer with schmitt trigger is used to turn that signal into digital goodness. The digital signal will go "HIGH" when someone steps on the pressure pad. The high signal goes into the S input of the "CHARGE" SR latch. The R input is connected over at the DISCHARGE SR Latch which should be off at this point.

The output of CHARGE goes high (Q = HIGH), and NOT Q will be LOW. LOW turns ON the P-Channel MOSFET shown as M1, which is normally held OFF when NOT Q is force high, or during a tri-state condition the resistor R6 is there for reliability.

When M1 turns on, current from the 3V supply flows directly into a 100uF capacitor. This node/rail is being monitored for a logic HIGH by the "RDY" (meaning ready) and "NRDY" digital buffers. When the capacitor node is READY, the DISCHARGE SR latch jumps into action, disabling the CHARGE input latch by putting it's R input high, and then turning on the low-side N Channel MOSFET M2. M2 allows the capacitor C1 to drain through a known resistor R8, forming an RC network with a time constant of ~1 second.

The charge state will go from ~2.3V (when the signal to RDY was valid) down to ~1V to be logic LOW, making the DISCHARGE SR latch reset and stop, and release the R from the CHARGE input SR Latch, allowing a new pressure step cycle to begin.

Honestly I whipped this up in about 15 minutes and there are some real-world issues such as possible race conditions, un-equal schmitt trigger thresholds all over the place, and the need for probably at least 2 more RC-networks to slow down signals on purpose and avoid these race conditions, especially for RDY and NRDY.

both MOSFETs would need to be very low voltage logic level MOSFETs, i'd go for 1V Vgs_th values, and probably a 12->20V VdsMax.

All of this should work without an issue at 3V. Especially if you use CMOS logic chips and use MOSFETs as suggested. Using PNP and NPN BJTs instead of the FETs could be done, but you'd lose way too much voltage and too much current just to saturate them to reduce the forward voltage drop. All of these CMOS devices will have ultra low leakage/quiescent current too.

In terms of solution size, you can do surface mount 0402 size passives, get a dual SR Latch IC, a P and N complementary MOSFET package which is probably a tiny SOT23-6, and a single hex inverter buffer + Schmitt trigger output style logic chip (such as the Texas Instruments SN74LV14ADR), and the "non inverting" buffer is just two inverters chained in series, which is great because then you use more of the available circuits in that package.

If anyone could add to this circuit or expand on the theory to make it more robust, please feel free to comment!

• My first thought is, I'd love to know where to find a MOSFET with 1V Vgs_th!! That would pass more than mA current, I mean... (There's a limit as to how many components I can string together and still stay in the housing I have planned. And that's assuming I can even find a 3V battery that will source it!!) Aug 13, 2015 at 0:47
• Hey @Jimbo, 1-2V Vgs is not difficult, there are many shapes and sizes of logic-level MOSFETs. There is a huge difference in Drain-Source current from 2V to 3V, as in, 2V Vgs_th max will pass at least 250uA, 3V will pass many amps. Depends on the FET of course, but that's normal - A very small window of voltage after the threshold. And how did you intend to power the system then, if you weren't already going to need a 3V battery? You said you will need to provide ~3V to the load, so I thought why not power the system with 3V too. I havent shown in, but you'll need another FET. Let me add that Aug 13, 2015 at 3:05
• @Jimbo here's a tiny little P-channel FET, which would be fine for the output (high current) MOSFET I just added to the schematic, its got a 1V threshold, as soon as you go above 1.5V at the gate it can do 5+ Amps digikey.ca/product-detail/en/SI1401EDH-T1-GE3/… Aug 13, 2015 at 3:19