Indeed, I was wondering initially how that capacitor is actually discharged, nice circuit and ideal for applications where you need to have a small on/off switch.
That reminds me, I have a whole bunch of those things in a very tiny package, SOT-23-6??? that I used for small H-bridges a few years ago. Sadly they are too tiny for me to use anymore. Wish our eyes kept up with things getting smaller and smaller.
Know what you mean. I'd prob be using SMDs, but for a benign tremor. I'd spend more time picking the damn things up than using them. Bad enough with through-holes !
I used the same with a uC to do a soft on/off latch circuit. Working very well. Even on battery powered device. All is on my instructable page for more detail. DMG6601LVT, SIL2308, DMC3021LSD-13, IRF7319TRPBF, are all pin compatible together. Thanks for sharing.
I speculate that the origin circuit loves (needs) resistive load, hates capacitive load and confused by non linear load. If we put a 10k resistor as output load, probably it will work. If we add a 100uF capacitor at the output, probably it won't work. Just my shallow speculation.
Before you added the resistor the P channel FET turned off momentarily when the switch was pressed a second time. That allowed the capacitor to charge up to whatever voltage is required to turn the N channel FET back on. The brief duration of the P channel FET being off is sometimes noticeable as the LED blinks. A much shorter button press would work. With the added resistor the capacitor can't charge up enough to turn the N channel FET back on. If you look at the original circuit with a scope you will see how the P channel FET turns off momentarily, allowing the N channel FET to turn back on. Another circuit built at random with no analysis.
Indeed. At first glance my first gut feeling was "wtf, this doesn't seem to be properly bistable - it will toggle but will feed back and go around again..."
@@bayareapianist Using the P channel FET is a bad idea. Not blaming the author of this video. The P channel FET has a higher ON resistance. Using this chip in a motor control H bridge is not a good idea either. An H bridge that mixes P channel and N channel FETs has a discontinuity at the zero crossing. Also a lack of symmetry. Try to use an H bridge like that with a PID controller and stability will be a problem. And no one wants to use P channel Fetus anyway. 4 N channel FETs, with low Ron, and driven with highside/lowside drivers is the way to go. There are lots of good integrated H bridges available.
To ascertain the reason why the original circuit didn't turn off for you, we need to analyse the circuit. When the led is on, the voltage at the gate of the N-channel is close to the output voltage, which is close to the input voltage because the gate of the P-channel is then close to ground and it is in ohmic mode with a low resistance. That assumes that the supply voltage is larger than the threshold voltage of the mosfets (1V to 2.8V). That means that the capacitor is discharged close to ground via the 100K resistor. When the switch is closed, you are then trying to dump the N-channel's gate charge into the capacitor causing the gate voltage to drop and the drain voltage to rise, but the Miller effect will tend to oppose that, with the result that the N-channel's drain does not rise high enough fast enough to turn off the P-channel before the capacitor has charged sufficiently to keep the N-channel from turning off any further. I suspect that this is likely to happen when the supply voltage is 5V or more as that gives sufficient opportunity for charging the capacitor before the N-channel gate voltage can drop below its threshold voltage. By modifying the circuit to use a resistive divider to set the N-channel's gate to only just above its threshold voltage, you increase the chance that closing the switch will turn off the N-channel before the capacitor charges up. You could test this by exploring the effect of different supply voltages on the reliability of the toggle operation. Edit: I suspect that the original circuit was using a power P-channel mosfet and a small-signal N-channel "driver" which, of course, has much smaller associated gate capacitances (the NDS9962A has Ciss of 320pF and Crss of 85pF _typical_ compared with Ciss of 60pF and Crss of 5pF _maximum_ for a 2N7000) so I would expect it to behave rather better than what you experienced.
The thing I wonder about the fix is what is the gate voltage at the n-channel when the circuit is on? It's only 1/2 the output voltage. As long as that's "well above" the threshold of the nfet, that's fine, but if it's close to the threshold, the nfet might not be fully on. Maybe it doesn't matter because the nfet doesn't carry any significant current in this circuit anyway. If it was an issue, I'd increase the value of the pull down so that the voltage divider is biased towards the high side, and the nfet gate voltage is a bit higher during "on" phase.
I'm thinking that if the circuit was connected to a larger circuit load, the pull-down resistor may not have been needed but I think its a good idea. Better safe than sorry.
It can count in binary 👍 Only up to 1 but it CAN count. If you had two of them then it could count up to 3. Three of them could count up to 7. So, someone with altogether too much time on their hands could build a computer out of them. Lol.
In order to turn on the P-channel MOSFET you should bias it by adding a resistor between source and ground, the gate must have negative voltage in respect to source. Your description about differences between P and N channel MOSFET is not accurate, it’s not about low, or high level voltage. N-channel MOSFET will start conduct when the gate is polarized with positive voltage in respect to source, and P-channel MOSFET will start to conduct when the gate is polarized with negative voltage in respect to source.
Good job. I have used the traditional npn-pnp circuit working similarly but carrying a few components more. I noted with it that in noisy environments it is required to have capacitors 10 nF to ground or so, at the bases to prevent glitches. I assume this circuit may be just as sensitive and needs to be slowed down a bit.
I have some Si4599DY chips, same pinout, slightly different specs (mine are mounted on a small breakout board). Ralph Bacon did a video using them to drive a bi-colour LED from one pin.
I got some Japanese Rohm sh8mb5, they use the same pin out. Indeed Louis Rossman showed "laptop" schematics from very well known brand and there where many mosfets.
Digikey describes the NDS9952A as "obsolete" ... none are available at Digikey, Mouser, or at JLCPCB. LCSC Has 3 pieces from alternate supplier VB Semi of Taiwan. Of course, there are plenty of other N/P pairs
This power switch is something I wanted in my arsenal, so I drew up the system in Kicad. Since I couldn't find the 9952, I searched Digikey for N/P mosfet pairs, with Rds On up to 25mO. Looking at 57 remaining options, Diodes inc DMC 3025LSD-13 shows stock of 35,000 pieces at a good price $0.66US. The one downside is the advertised RdsOn was for the N Channel, and it's the P channel that's doing the heavy lifting. That only gives 45mOhm. So for a 12V supply (I'm thinking in terms of eurorack systems) this would be 1W at 5A. Are there other characteristics I should be focusing on?
The 100k + capacitor is slow enough to avoid bounces on/off I think the problem with the original circuit is the 10k pull-up that is too strong. Replace it with 100k and you won't need the extra 10k (I hope)
I'm wondering what the quiescent current is of this circuit. I do a similar thing with a 4013 only differences I can put a cap and diode on the supply voltage and it acts like memory so the switch state remains even if the power is turned off for quite some time
A 5V chip will read anything above 2v as logic 1, and you can use a resister and a zener to drop 5v to 3.3v. Good for doing something quick that doesn't take alot of space.
@@tripplefives1402 im well aware of this i help many on a MCU server and so many tutorials will send 5v to 3.v volt logic and sometimes the sensor is tolerant and sometimes it is not. the original comment was in response to a flawed design found on the interwebs. missing a pull down resistor .
Waht do You think, can this kind of transistor be used in analog circuit, thinking of traditional audio amplifiers just for example. Of course not so much power, maybe in the driver stage or for headphone amp together with opamp?
read the datasheet: "These devices are particularly suited for low voltage applications such as notebook computer power management and other battery powered circuits where fast switching, low in-line power loss, and resistance to transients are needed." of course other applications are up to the imagination
Are matched FETs ever needed? If so, examples, please. This might be an okay way to get pretty close to matched, I would think. I can't think of any reason this would not work with BJTs. Yes? I think I will breadboard it, to see. Cheers.
5:36 “…well anyway…” is not the clearest circuit description 🤣 Great little cct though, thanks for sharing. Basically the cap is discharged all the time the output is ON, by the N MOSFET. When the switch is closed, the cap’s low voltage pulls the N mosfet’s gate low, which switches both mosfets off.
I speculate that the origin circuit loves (needs) resistive load, hates capacitive load and confused by non linear load. If we put a 10k resistor as output load, probably it will work. If we add a 100uF capacitor at the output, probably it won't work. Just my shallow speculation.
Indeed, I was wondering initially how that capacitor is actually discharged, nice circuit and ideal for applications where you need to have a small on/off switch.
That reminds me, I have a whole bunch of those things in a very tiny package, SOT-23-6??? that I used for small H-bridges a few years ago. Sadly they are too tiny for me to use anymore. Wish our eyes kept up with things getting smaller and smaller.
Know what you mean. I'd prob be using SMDs, but for a benign tremor. I'd spend more time picking the damn things up than using them. Bad enough with through-holes !
I used the same with a uC to do a soft on/off latch circuit. Working very well. Even on battery powered device. All is on my instructable page for more detail. DMG6601LVT, SIL2308, DMC3021LSD-13, IRF7319TRPBF, are all pin compatible together. Thanks for sharing.
here is his page: www.instructables.com/On-Off-Latch-Circuit-With-UC-One-Push-Button-One-P/
Yeah, also 600 amp 1200V IGBT modules are suitable to drive p-ch fet gate ROFL
I speculate that the origin circuit loves (needs) resistive load, hates capacitive load and confused by non linear load.
If we put a 10k resistor as output load, probably it will work.
If we add a 100uF capacitor at the output, probably it won't work.
Just my shallow speculation.
N and P-CH in one package is also convenient to use in H bridge motor driver.
Easiest circuit you can make with such a package is a logic inverter. Don't need any external components, just the 2 complementary CMOS transistors.
Before you added the resistor the P channel FET turned off momentarily when the switch was pressed a second time. That allowed the capacitor to charge up to whatever voltage is required to turn the N channel FET back on. The brief duration of the P channel FET being off is sometimes noticeable as the LED blinks. A much shorter button press would work. With the added resistor the capacitor can't charge up enough to turn the N channel FET back on. If you look at the original circuit with a scope you will see how the P channel FET turns off momentarily, allowing the N channel FET to turn back on. Another circuit built at random with no analysis.
Indeed. At first glance my first gut feeling was "wtf, this doesn't seem to be properly bistable - it will toggle but will feed back and go around again..."
I wouldn't use it in a medical device. I see no bistable nor hysteresis circuit here.
@@bayareapianist Using the P channel FET is a bad idea. Not blaming the author of this video. The P channel FET has a higher ON resistance. Using this chip in a motor control H bridge is not a good idea either. An H bridge that mixes P channel and N channel FETs has a discontinuity at the zero crossing. Also a lack of symmetry. Try to use an H bridge like that with a PID controller and stability will be a problem. And no one wants to use P channel Fetus anyway. 4 N channel FETs, with low Ron, and driven with highside/lowside drivers is the way to go. There are lots of good integrated H bridges available.
We used to use the IRF7389 (very similar to this one) in a tiny motor drive.
Thank for this video. Very cool schematic, i will try it!
Good chip to know about. Thanks.
To ascertain the reason why the original circuit didn't turn off for you, we need to analyse the circuit. When the led is on, the voltage at the gate of the N-channel is close to the output voltage, which is close to the input voltage because the gate of the P-channel is then close to ground and it is in ohmic mode with a low resistance. That assumes that the supply voltage is larger than the threshold voltage of the mosfets (1V to 2.8V). That means that the capacitor is discharged close to ground via the 100K resistor. When the switch is closed, you are then trying to dump the N-channel's gate charge into the capacitor causing the gate voltage to drop and the drain voltage to rise, but the Miller effect will tend to oppose that, with the result that the N-channel's drain does not rise high enough fast enough to turn off the P-channel before the capacitor has charged sufficiently to keep the N-channel from turning off any further. I suspect that this is likely to happen when the supply voltage is 5V or more as that gives sufficient opportunity for charging the capacitor before the N-channel gate voltage can drop below its threshold voltage.
By modifying the circuit to use a resistive divider to set the N-channel's gate to only just above its threshold voltage, you increase the chance that closing the switch will turn off the N-channel before the capacitor charges up. You could test this by exploring the effect of different supply voltages on the reliability of the toggle operation.
Edit: I suspect that the original circuit was using a power P-channel mosfet and a small-signal N-channel "driver" which, of course, has much smaller associated gate capacitances (the NDS9962A has Ciss of 320pF and Crss of 85pF _typical_ compared with Ciss of 60pF and Crss of 5pF _maximum_ for a 2N7000) so I would expect it to behave rather better than what you experienced.
The thing I wonder about the fix is what is the gate voltage at the n-channel when the circuit is on? It's only 1/2 the output voltage. As long as that's "well above" the threshold of the nfet, that's fine, but if it's close to the threshold, the nfet might not be fully on. Maybe it doesn't matter because the nfet doesn't carry any significant current in this circuit anyway. If it was an issue, I'd increase the value of the pull down so that the voltage divider is biased towards the high side, and the nfet gate voltage is a bit higher during "on" phase.
I wonder if it truly stays latched on indefinitely due to the eventual capacitor leakage.
I'm thinking that if the circuit was connected to a larger circuit load, the pull-down resistor may not have been needed but I think its a good idea. Better safe than sorry.
It can count in binary 👍 Only up to 1 but it CAN count. If you had two of them then it could count up to 3. Three of them could count up to 7. So, someone with altogether too much time on their hands could build a computer out of them. Lol.
In order to turn on the P-channel MOSFET you should bias it by adding a resistor between source and ground, the gate must have negative voltage in respect to source.
Your description about differences between P and N channel MOSFET is not accurate, it’s not about low, or high level voltage.
N-channel MOSFET will start conduct when the gate is polarized with positive voltage in respect to source, and P-channel MOSFET will start to conduct when the gate is polarized with negative voltage in respect to source.
There's a similar circuit using a 555 timer that does the same thing in the same manner.
RANT OF THE WEEK! RANT OF THE WEEK! RANT OF THE WEEK! :)
Good job. I have used the traditional npn-pnp circuit working similarly but carrying a few components more. I noted with it that in noisy environments it is required to have capacitors 10 nF to ground or so, at the bases to prevent glitches. I assume this circuit may be just as sensitive and needs to be slowed down a bit.
I have some Si4599DY chips, same pinout, slightly different specs (mine are mounted on a small breakout board). Ralph Bacon did a video using them to drive a bi-colour LED from one pin.
zamzam water
I got some Japanese Rohm sh8mb5, they use the same pin out.
Indeed Louis Rossman showed "laptop" schematics from very well known brand and there where many mosfets.
Interesting circuit. I think I saw something similar on EEVBlog at some point.
Digikey describes the NDS9952A as "obsolete" ... none are available at Digikey, Mouser, or at JLCPCB. LCSC Has 3 pieces from alternate supplier VB Semi of Taiwan. Of course, there are plenty of other N/P pairs
So does Arrow. And Onsemi themselves says it's obsolete.
This power switch is something I wanted in my arsenal, so I drew up the system in Kicad. Since I couldn't find the 9952, I searched Digikey for N/P mosfet pairs, with Rds On up to 25mO. Looking at 57 remaining options, Diodes inc DMC 3025LSD-13 shows stock of 35,000 pieces at a good price $0.66US. The one downside is the advertised RdsOn was for the N Channel, and it's the P channel that's doing the heavy lifting. That only gives 45mOhm. So for a 12V supply (I'm thinking in terms of eurorack systems) this would be 1W at 5A. Are there other characteristics I should be focusing on?
Seems like the original circuit had a larger capacitor and a diode somewhere. But your version is easier.
The 100k + capacitor is slow enough to avoid bounces on/off
I think the problem with the original circuit is the 10k pull-up that is too strong. Replace it with 100k and you won't need the extra 10k (I hope)
I'm wondering what the quiescent current is of this circuit. I do a similar thing with a 4013 only differences I can put a cap and diode on the supply voltage and it acts like memory so the switch state remains even if the power is turned off for quite some time
the internet were anything is possible even 5v logic to 3.3v logic
without a bi-level shift converter and it never gives up the ghost
A 5V chip will read anything above 2v as logic 1, and you can use a resister and a zener to drop 5v to 3.3v. Good for doing something quick that doesn't take alot of space.
@@tripplefives1402 im well aware of this i help many on a MCU server and so many tutorials will send 5v to 3.v volt logic and sometimes the sensor is tolerant and sometimes it is not. the original comment was in response to a flawed design found on the interwebs. missing a pull down resistor .
@@joejane9977 in the original the load is the pulldown.
@@tripplefives1402 you missed the point its ok
Waht do You think, can this kind of transistor be used in analog circuit, thinking of traditional audio amplifiers just for example. Of course not so much power, maybe in the driver stage or for headphone amp together with opamp?
read the datasheet: "These devices are particularly suited for low voltage applications such as notebook computer power management and other battery powered circuits where fast switching, low in-line power loss, and resistance to transients are needed."
of course other applications are up to the imagination
Are matched FETs ever needed? If so, examples, please. This might be an okay way to get pretty close to matched, I would think. I can't think of any reason this would not work with BJTs. Yes? I think I will breadboard it, to see. Cheers.
Need them for analog stuff. Mosfets have less resistance and consume less power even when working in the linear region.
?? Not clear how that turns off.
5:36 “…well anyway…” is not the clearest circuit description 🤣
Great little cct though, thanks for sharing.
Basically the cap is discharged all the time the output is ON, by the N MOSFET. When the switch is closed, the cap’s low voltage pulls the N mosfet’s gate low, which switches both mosfets off.
@@VividSolutions duh, thx
Can i use this circuit to drive a relay or TO-220 power FET for switching higher current?
I speculate that the origin circuit loves (needs) resistive load, hates capacitive load and confused by non linear load.
If we put a 10k resistor as output load, probably it will work.
If we add a 100uF capacitor at the output, probably it won't work.
Just my shallow speculation.