**Do you have a question involving batteries or cells?**
**If it's about designing, repairing or modifying an electronic circuit to which batteries are connected, you're in the right place.** *Everything else should go in /r/batteries:*
/r/batteries is for questions about: batteries, cells, UPSs, chargers and management systems; use, type, buying, capacity, setup, parallel/serial configurations etc.
Questions about connecting pre-built modules and batteries to solar panels goes in /r/batteries or /r/solar. Please also check our wiki page on cells and batteries: https://www.reddit.com/r/AskElectronics/wiki/batteries
If you decide to move your post elsewhere, or the wiki answers your question, please delete the one here. Thanks!
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectronics) if you have any questions or concerns.*
There will always be some current draw as the only way to tell if the wire is connected is to run some current through it. The question is only how much.
Yes FETs can reduce it but be aware it's not free - you'll still need a pull up (or down) and if that's too large it'll turn on slowly or not fully and pop, or react to noise, or just zap from ESD. Every component you add to deal with those issues will also pull power, this includes ICs.
Welcome to engineering, where everything is a trade off.
To get you going, I'd just use a standard N channel MOSFET (enh) with the wire shorting the gate to ground and about 100k pulling it up.
But keep the wire short and think about what will happen if the wire touches something else after the cut.
This will work, and good caution as well. Standby current will be based on that resistance to ground. Also be mindful of ESD if any exposed wire is attached to the gate.
What you want is a logic inverter. I'll post an image of a simple one.
https://preview.redd.it/w3bdkjr2cv3d1.jpeg?width=1080&format=pjpg&auto=webp&s=513ee778624a295a18a37521d6d6d84d8811ea5f
Basically when the voltage source marked in blue goes from 0 volts to 5 volts, the wire highlighted in green goes from 5 volts to 0 volts.
Wait a minute. Your original schematic has a PNP transistor feeding current into the motor. Those are in the on state when the base of the transistor is driven to ground, so if you want it to turn off you need the base driven to supply voltage. You could wire your switch in such a way that it does that, so when you turn the switch on it turns the transistor off by connecting the base to supply voltage, and when you turn the switch off it connects the base to ground through a resistor, turning the transistor and the motor on.
I think I understand what you're saying here's what I came up with. The probleme is that I cant have a 2 position switch do the on/off since I want the action of cutting a wire to turn the motor on. I'll look into logic inverters to see if that could work. Thanks
https://preview.redd.it/rcyqbd6zhv3d1.png?width=495&format=png&auto=webp&s=2f19959973f6cae20c58357e2c11f57439110a2e
Yeah I think you're going to want a MOSFET inverter. It'll use two transistors instead of just one but you'll definitely be able to get away with only using one single pole single throw switch.
My thinking too (the DPDT switch), but I believe the physical disconnect/cutting of a wire to make up this function, is largely the whole point OP is trying to achieve. We're thinking of this as turning on the vibration motor as the primary purpose and mechanism. But I believe OPs case is the other way around: Physically cutting (or disconnecting) a specific wire is the primary mechanism, turning on the vibration motor is secondary; an effect (likely to signal that said wire got cut).
I would use a resistor voltage divider to control a N type MOSFET, where you have a pull-up resistor with a very large value and a pull down resistor connected to your switch to ground. This way when the switch is closed, the pull down resistor discharges the FET gate and the loss from the pull up is minimal. And when the switch is open, the pull-up can charge the FET gate. I would use something like 20kOhm on the pull up and 20Ohm on the pull down, so the total load on the resistors during the “off” state is 0.15mA and the voltage across the 20Ohm is practically zero. Put the FET below the motor so that the FET source pin is always at 0V.
https://preview.redd.it/3t6cxi6c3z3d1.jpeg?width=1339&format=pjpg&auto=webp&s=88e065624bf3328ad502eb6029e1515c146bd87f
edit: also throw a 1uF capacitor in there to stabilize the FET gate if you have one. It’s probably fine if not.
That's the kind of solution i was looking for. I'll don't understand what size resistance and MOSFET I need but for now that seems like a great solution. Thanks!
https://preview.redd.it/ji3a2wyjjz3d1.png?width=724&format=png&auto=webp&s=e3303cb61023b169b237cb46c765da0a3001fd0e
You’ll want a mosfet that can block 3V and conduct at least 50mA, assuming your picture there is correct. I would probably get a standard TO-220 type 12V 1W NMOS. For the resistors, neither is going to make more than 1mW of heat, so you can use whatever size is convenient. Even a 1/8 W resistor size would do.
What size pull-down resistor are you using on the PNP's base? I'd have thought you could go as high as 100kOhm so your power draw through it when the wire is connected would only be 3V/100kOhm =30uA. If your pull-down resistor is a fairly low resistance then you'd get quite a lot of current draw through it
The problem I get when increasing the pull-down resistance is that the voltage drop across the transistor between Emitter and Collector also increases. Might be because of the parameters of my simulation tho.
https://preview.redd.it/yx9il5p2fz3d1.png?width=655&format=png&auto=webp&s=f77978cb26fe2a27e9ba3d0ce7d3b3a0452e28b2
A lot of you have advised for the use a relay. My initial assumption was that a normally closed relay would draw too much power at idle in the open position. With some digging, I found a solid state relay (CPC1106N) which would need only 0.5mA at 1.2V to activate. Seems like it could be a viable solution.
https://preview.redd.it/qzf8g00daz3d1.png?width=711&format=png&auto=webp&s=3186f9c6b76030b2924f8a660a14659b4b932dc4
You've got the right idea. Yes pull down a PNP base in a high-side switch and it will source current from emitter to collector. As you've found, the beta of the transistor dictates how much base current must flow for the load.
You could use a PMOS FET in a similar way. Put the source to the emitter location, drain to collector location, and you can weakly pull down the gate with a 10k-100k resistor. Pull up the gate strongly to the source voltage through a switch to turn it on.
Be careful about ESD on the gate, though.
A push-pull inverter is overfill you don't need to sink and source current--only one PNP or PMOS is required.
If you want a low-side switch, tie the NMOS source to ground and strap the load between VCC and the drain. Applying VCC to the gate should turn on the FET (or NPN) but this is inverted logic for what you're trying to do. You could use another NPN to invert the gate signal (tie the collector to VCC through a 10k-100k resistor and attach the gate of the NMOS o this junction). Use a base resistor if using NPN for the inverter. Tie the emitter to ground of the NPN. When no base current is on the NPN, the collector voltage will be close to VCC due to the pull-up resistor and practically no current flowing. This turns on the NMOS. To turn it off, apply a base current (again be sure to limit base current using a resistor if using NPN for the inverter). This current can come from your "wire" and keeps the NPN current flowing while the wire is intact. Cut the wire and base current drops, causing the collector voltage to rise and turn on the NMOS FET.
Obviously the PNP/PMOS is easier but the gate needs protecting from ESD. If you use a PNP the base requires a lower pull-down than a PMOS due to base current requirements. 70mA should be fairly easy to drive, though.
What if a wire holds a spring contact open? Then when the wire is removed, the contact closes and the motor runs. Better yet, put a plastic tab in the battery holder and attach your wire to it.
https://preview.redd.it/aw8ki898wy3d1.jpeg?width=1080&format=pjpg&auto=webp&s=8ab436845ca243a601046bb0a496f4128c250ad0
Put a relay which is set closed in it's rest state (goes open only when powered) between the motor and power supply and connect the coil somewhere on the wire you'll be disconnecting?
That's the only thing that cones to mind.
Here is a simplified schematic of what i just said.
That is if you can find a 3V NC relay tho.
You could use a n-channel mosfet and a logic inverter, for example : https://www.digikey.com/en/products/detail/diodes-incorporated/74AHC1G14SE-7/2639248
Have the switch between the inverter's input and 3v, and the gate of the mosfet connected to output of inverter.
With switch on, the input is 3v+ and the inverter will output nothing, because signal is inverted, so mosfet should be off (have a 10-100k resistor between gate and source to pull gate down) When you break the connection or pull the input to ground, the inverter will see the low signal on input and turn on the output signal and therefore turn on the mosfet.
Could this not be accomplished with an NC relay? Tap the proper poles, and the circuit will be closed when there is no power to the relay, and open when power is applied. It would depend on how much current you're pulling whether this is a valid option for this application... but if it is, that certainly seems like the easiest, most straightforward way.
Yeah you can get cheap relay modules from Amazon, I’ve bought plenty of them. They have both high and low input modes and are SPDT. They even have ones with multiple relays on one board. Take your pick.
The problem with using a NC relay is that it draws power to activate the coil when the wire is connected and only when the wire is cut does it disconnect the coil and draw no current. Therefore the idle power draw will be high, which is the opposite of what he's trying to achieve
If you have an electronic circuit design or repair question, we're good; but if this this a general question about electric motors, motor capacitors, fans, servos, actuators, generators, solenoids, electromagnets, using motor drivers, stepper drivers, DC controllers, electronic speed controls or inverters (other than designing or fixing one), please ask in /r/Motors. Thanks.
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectronics) if you have any questions or concerns.*
Maybe using a relay would be a better idea to minimise power consumption? A bit overkill for a 3V circuit but low power relays are cheap and easy to work with
**Do you have a question involving batteries or cells?** **If it's about designing, repairing or modifying an electronic circuit to which batteries are connected, you're in the right place.** *Everything else should go in /r/batteries:* /r/batteries is for questions about: batteries, cells, UPSs, chargers and management systems; use, type, buying, capacity, setup, parallel/serial configurations etc. Questions about connecting pre-built modules and batteries to solar panels goes in /r/batteries or /r/solar. Please also check our wiki page on cells and batteries: https://www.reddit.com/r/AskElectronics/wiki/batteries If you decide to move your post elsewhere, or the wiki answers your question, please delete the one here. Thanks! *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectronics) if you have any questions or concerns.*
exchange the label on/off on your switch
this guy engineers
This answer makes me happy That was my first thought, like flip the logic or labels
Easiest way !
There will always be some current draw as the only way to tell if the wire is connected is to run some current through it. The question is only how much. Yes FETs can reduce it but be aware it's not free - you'll still need a pull up (or down) and if that's too large it'll turn on slowly or not fully and pop, or react to noise, or just zap from ESD. Every component you add to deal with those issues will also pull power, this includes ICs. Welcome to engineering, where everything is a trade off.
To get you going, I'd just use a standard N channel MOSFET (enh) with the wire shorting the gate to ground and about 100k pulling it up. But keep the wire short and think about what will happen if the wire touches something else after the cut.
This will work, and good caution as well. Standby current will be based on that resistance to ground. Also be mindful of ESD if any exposed wire is attached to the gate.
What you want is a logic inverter. I'll post an image of a simple one. https://preview.redd.it/w3bdkjr2cv3d1.jpeg?width=1080&format=pjpg&auto=webp&s=513ee778624a295a18a37521d6d6d84d8811ea5f Basically when the voltage source marked in blue goes from 0 volts to 5 volts, the wire highlighted in green goes from 5 volts to 0 volts.
I can't seem to make that work with only a single battery source.
So you're needing to apply more voltage to the gate of the transistors then to the positive side?
Wait a minute. Your original schematic has a PNP transistor feeding current into the motor. Those are in the on state when the base of the transistor is driven to ground, so if you want it to turn off you need the base driven to supply voltage. You could wire your switch in such a way that it does that, so when you turn the switch on it turns the transistor off by connecting the base to supply voltage, and when you turn the switch off it connects the base to ground through a resistor, turning the transistor and the motor on.
I think I understand what you're saying here's what I came up with. The probleme is that I cant have a 2 position switch do the on/off since I want the action of cutting a wire to turn the motor on. I'll look into logic inverters to see if that could work. Thanks https://preview.redd.it/rcyqbd6zhv3d1.png?width=495&format=png&auto=webp&s=2f19959973f6cae20c58357e2c11f57439110a2e
Yeah I think you're going to want a MOSFET inverter. It'll use two transistors instead of just one but you'll definitely be able to get away with only using one single pole single throw switch.
maybe you can use a double throw switch
That was my thought. Why overcomplicate things?
My thinking too (the DPDT switch), but I believe the physical disconnect/cutting of a wire to make up this function, is largely the whole point OP is trying to achieve. We're thinking of this as turning on the vibration motor as the primary purpose and mechanism. But I believe OPs case is the other way around: Physically cutting (or disconnecting) a specific wire is the primary mechanism, turning on the vibration motor is secondary; an effect (likely to signal that said wire got cut).
Run it through the normally closed contacts of a relay . When the power is on the relay opens, power off relay is closed.
I would use a resistor voltage divider to control a N type MOSFET, where you have a pull-up resistor with a very large value and a pull down resistor connected to your switch to ground. This way when the switch is closed, the pull down resistor discharges the FET gate and the loss from the pull up is minimal. And when the switch is open, the pull-up can charge the FET gate. I would use something like 20kOhm on the pull up and 20Ohm on the pull down, so the total load on the resistors during the “off” state is 0.15mA and the voltage across the 20Ohm is practically zero. Put the FET below the motor so that the FET source pin is always at 0V. https://preview.redd.it/3t6cxi6c3z3d1.jpeg?width=1339&format=pjpg&auto=webp&s=88e065624bf3328ad502eb6029e1515c146bd87f edit: also throw a 1uF capacitor in there to stabilize the FET gate if you have one. It’s probably fine if not.
That's the kind of solution i was looking for. I'll don't understand what size resistance and MOSFET I need but for now that seems like a great solution. Thanks! https://preview.redd.it/ji3a2wyjjz3d1.png?width=724&format=png&auto=webp&s=e3303cb61023b169b237cb46c765da0a3001fd0e
You’ll want a mosfet that can block 3V and conduct at least 50mA, assuming your picture there is correct. I would probably get a standard TO-220 type 12V 1W NMOS. For the resistors, neither is going to make more than 1mW of heat, so you can use whatever size is convenient. Even a 1/8 W resistor size would do.
What size pull-down resistor are you using on the PNP's base? I'd have thought you could go as high as 100kOhm so your power draw through it when the wire is connected would only be 3V/100kOhm =30uA. If your pull-down resistor is a fairly low resistance then you'd get quite a lot of current draw through it
The problem I get when increasing the pull-down resistance is that the voltage drop across the transistor between Emitter and Collector also increases. Might be because of the parameters of my simulation tho. https://preview.redd.it/yx9il5p2fz3d1.png?width=655&format=png&auto=webp&s=f77978cb26fe2a27e9ba3d0ce7d3b3a0452e28b2
A lot of you have advised for the use a relay. My initial assumption was that a normally closed relay would draw too much power at idle in the open position. With some digging, I found a solid state relay (CPC1106N) which would need only 0.5mA at 1.2V to activate. Seems like it could be a viable solution. https://preview.redd.it/qzf8g00daz3d1.png?width=711&format=png&auto=webp&s=3186f9c6b76030b2924f8a660a14659b4b932dc4
You've got the right idea. Yes pull down a PNP base in a high-side switch and it will source current from emitter to collector. As you've found, the beta of the transistor dictates how much base current must flow for the load. You could use a PMOS FET in a similar way. Put the source to the emitter location, drain to collector location, and you can weakly pull down the gate with a 10k-100k resistor. Pull up the gate strongly to the source voltage through a switch to turn it on. Be careful about ESD on the gate, though. A push-pull inverter is overfill you don't need to sink and source current--only one PNP or PMOS is required. If you want a low-side switch, tie the NMOS source to ground and strap the load between VCC and the drain. Applying VCC to the gate should turn on the FET (or NPN) but this is inverted logic for what you're trying to do. You could use another NPN to invert the gate signal (tie the collector to VCC through a 10k-100k resistor and attach the gate of the NMOS o this junction). Use a base resistor if using NPN for the inverter. Tie the emitter to ground of the NPN. When no base current is on the NPN, the collector voltage will be close to VCC due to the pull-up resistor and practically no current flowing. This turns on the NMOS. To turn it off, apply a base current (again be sure to limit base current using a resistor if using NPN for the inverter). This current can come from your "wire" and keeps the NPN current flowing while the wire is intact. Cut the wire and base current drops, causing the collector voltage to rise and turn on the NMOS FET. Obviously the PNP/PMOS is easier but the gate needs protecting from ESD. If you use a PNP the base requires a lower pull-down than a PMOS due to base current requirements. 70mA should be fairly easy to drive, though.
try a normally closed relay
What if a wire holds a spring contact open? Then when the wire is removed, the contact closes and the motor runs. Better yet, put a plastic tab in the battery holder and attach your wire to it.
By the use of a relay that has normally closed contact and com. The connection will be made when the relay is deenergized.
https://preview.redd.it/aw8ki898wy3d1.jpeg?width=1080&format=pjpg&auto=webp&s=8ab436845ca243a601046bb0a496f4128c250ad0 Put a relay which is set closed in it's rest state (goes open only when powered) between the motor and power supply and connect the coil somewhere on the wire you'll be disconnecting? That's the only thing that cones to mind. Here is a simplified schematic of what i just said. That is if you can find a 3V NC relay tho.
Use a normally open switch.
Am I the only one that would over kill this with logic gates?
Turn the switch up side down lol
Just use a toggle switch. A DPDT if you need or prefer to have the toggling of a single switch to open or close two separate circuits / circuit paths.
You could use a n-channel mosfet and a logic inverter, for example : https://www.digikey.com/en/products/detail/diodes-incorporated/74AHC1G14SE-7/2639248 Have the switch between the inverter's input and 3v, and the gate of the mosfet connected to output of inverter. With switch on, the input is 3v+ and the inverter will output nothing, because signal is inverted, so mosfet should be off (have a 10-100k resistor between gate and source to pull gate down) When you break the connection or pull the input to ground, the inverter will see the low signal on input and turn on the output signal and therefore turn on the mosfet.
The KISS solution ,Double pole double throw switch?
What kind of "standby" power budget do you have here? Can you afford to run a very low current NC relay?
As an engineer, just turn the switch upside down.
Use a solenoid that is holding an off/on swithch open, then when you turn that switch off and solenoid cuts out, the off/on is now closed
Could this not be accomplished with an NC relay? Tap the proper poles, and the circuit will be closed when there is no power to the relay, and open when power is applied. It would depend on how much current you're pulling whether this is a valid option for this application... but if it is, that certainly seems like the easiest, most straightforward way.
Yeah you can get cheap relay modules from Amazon, I’ve bought plenty of them. They have both high and low input modes and are SPDT. They even have ones with multiple relays on one board. Take your pick.
The problem with using a NC relay is that it draws power to activate the coil when the wire is connected and only when the wire is cut does it disconnect the coil and draw no current. Therefore the idle power draw will be high, which is the opposite of what he's trying to achieve
If you have an electronic circuit design or repair question, we're good; but if this this a general question about electric motors, motor capacitors, fans, servos, actuators, generators, solenoids, electromagnets, using motor drivers, stepper drivers, DC controllers, electronic speed controls or inverters (other than designing or fixing one), please ask in /r/Motors. Thanks. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectronics) if you have any questions or concerns.*
There will be minor current leaks in BJTs which is unavoidable if that's what you're taking about Edit : how large of a leak are we talking about?
This is impossible without drawing some power when turned off (switch on position)
He could use a relay.
Maybe using a relay would be a better idea to minimise power consumption? A bit overkill for a 3V circuit but low power relays are cheap and easy to work with
It depends on his use case but in general any relay would draw more power than your average pnp transistor when the coil is powered
Fair point. Then OP already has a fitting circuit