Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange

New answers tagged

6

That is a kind of motion detector. It looks very much like the SQ-SEN-200 from SignalQuest. Datasheet Picture from the datasheet: They are meant to be used in devices that should sleep when not in use, but wake up and do something when moved. Like a GPS tracker. You want to record where your cat goes, but don't need 18 hours a day of unchanging GPS ...


0

There is no general answer for this. You will have to check the datasheets of the components for both solutions and pick the one that has lowest battery drain.


1

You're not driving the FET with a high enough voltage. Remember, the threshold voltage \$V_{th}\$ is the voltage at which the FET just barely begins to allow current through--not the voltage at which it's fully on. If you look on page 4 of the datasheet: you'll see that the on-state resistance is specified at a \$V_{GS}\$ of 10V, twice what you have. And ...


0

There a re two ways to address your problems: Put a high power diode across the switched 24V supply Put a small wattage Zener diode and diode from Drain to ground You also need to address the V(GS) for the device you used. That alone may be killing your device. simulate this circuit – Schematic created using CircuitLab


2

You blew your PMOS. Semiconductors usually fail short. You are cutting it real close with your PMOS's max Vgs (25V max). Any ringing or spike and you blow your PMOS gate. My money is on that. It's so close that 24V zeners or TVS diodes to clamp the gate voltage would not help. Use a 12V or 15V clamping diode instead. simulate this circuit – ...


1

It's probably inductive kickback when you switch the PMOS off. Put a diode in there; cathode to PMOS drain, anode to ground (so it's reverse-biased when the PMOS is on). When you turn off the PMOS the two inductors will pull current from ground until they're discharged.


2

simulate this circuit – Schematic created using CircuitLab Figure 1. OP's proposal redrawn. With SW1 in position shown BAT2+ will be shorted to U1's BAT- which according to the TP4056 datasheet is supposed to be connected to ground so the arrangement is likely to have a very unhappy ending. simulate this circuit Figure 2. A circuit with less ...


3

If the switch was on the neutral wire, then toggling it would not be noticeable when measuring the wires in the junction box. The hot wire would constantly be live. Considering you're getting power on one of the wires that can be switched on or off suggests your switch is on the live side, as it should be.


-3

you can add hysteresis like this simulate this circuit – Schematic created using CircuitLab


3

R6 and C2 form a 33.8 kHz lowpass filter that reduces high frequency noise and slows the rise and fall times of the output squarewave. Because R2 is the collector load, the actual filter corner frequencies are 33.8 kHz for the trailing edge, and 3.08 kHz for the leading edge. You can see this in your sim image, where the output square wave has three steps ...


3

Figure 0. The schematic (subsequently removed) on which this answer is based. simulate this circuit – Schematic created using CircuitLab Figure 1. A typical polarity reversal switch. Your circuit is difficult to understand due to a poor schematic and poor labeling. It's also a bit dodgy as all the switches are independent and short circuits can be ...


3

From your schematic, the current flowing from PB0 to T1's base-emitter junction is \$I_B = (5V - 0.6V) / 10\Omega = 440mA\$, which is extremely high for a MCU-pin's drive capability. Thus PB0 may limit the output (I'm not sure, but I hope). To saturate the transistor, a base current of \$I_{Bs} = \frac{I_C}{ (\beta_{min}/10)}\$ is sufficient (I cannot ...


1

There's been some detailed analysis of similar problems - for example here Summary - RF energy gets into the switch IC and causes random ON/OFF. The solution is to wire inductors or resistors in series with the touch connection to the lamp metal housing to isolate the control from a hostile EMC environment. See the link for details, but the values are ...


1

Since I can not comment I will post my comment as an answer. I have a very similar sight (same?) and I have tried many things to 'fix' it. Finally one day I discovered that at the same time I pulled my back door closed (well, sort of slammed it) the light turned ON. Then the same action would also turn it Off. I knew both the lamp and the back door light ...


2

Some relays are provided with a holding mechanism comprising a small coil in series with the contacts; this coil is on a small electromagnet that acts on a small armature on the moving contact assembly to hold the contacts tightly closed once they have established the flow of trip-coil current. This coil is called a "seal-in" or "holding" coil. It ...


4

A "seal in" contact is just another normally open contact on a relay. The name "seal in" comes from the application of the contact. Most electronics people aren't likely to have heard of the terminology of "seal in". The term is used in industrial controls (which now days use relays and PLC). Choose a relay that is suitable for your needs. Edit 1 : ...


0

Yes, you could replace it with momentary switches but it's not simple. That is a 3 Pole 3 Throw switch. For a push button you would need at least three 3 pole Single Throw switches, one per input, preferably ones that are ganged together (so pressing one disconnects the other). And if its momentary you would have to hold the button in for it to work. Or you ...


0

The type of switch you are looking for is called a "momentary" switch. They are available in different forms including a 'toggle' (or bat handled) switch, slide switches and even rotary switches (less common). They are available in different sizes too. You may find that a DPDT (double pole, double throw) momentary (spring loaded) off-on-off may be able to ...


5

You can use a buffer per LED like this. In this diagram, R1 through R3 are pullup resistors. Closing any of the switches will cause the buffer directly connected to it to go to 0, which turns low all the buffers below it. 4050 has 6 buffers. You will need 2 of them for 9 LEDs. This solution does only needs a voltage to power the 4050 (3V to 20V for ...


1

An alternative approach is to use a LM3914 to drive the LEDs, with an external 10-resistor ladder powered from the reference voltage. Then the rotary switch simply selects a voltage from the ladder which will light the required number of LEDs. This is just an outline; for example, the topmost resistor of the ladder would be selected to set the step voltages ...


2

Sort of, this kind of switch is commonly used as a shutter release for digital cameras. They are not the old style toggle type but they work more like tactile momentary type and rely on two different pressure levels to operate. A half press is used to activate the auto-focus and then fully pressing it will take the photo. Questions about where to buy ...


2

I wouldn't suggest this unless you're eager to climb the learning curve for FPGAs (including buying a programming pod and dealing with a SMT part with lots of pins), but you could use a Lattice LCMXO2 series with internal flash and oscillator. Circuit would look like this (plus some power supply connections, a programming connector and bypass caps): ...


1

Similar to the micro controller method, another way is to use a OP amp ICs. The positive inputs are all connected together and they connect to a potentiometer that produces varying voltage, instead of a switch. The negative connections connect to a series of resistors to give each one a different Voltage. As the knob is turned, the lights turn on one by one. ...


14

Oldfart and Mattman944 give very similar answers involving complex diode networks. If brightness variation is acceptable a simple diode ladder is sufficient. Red LED's typically have 2V voltage drop and diodes typically have 0.6V voltage drop, so the combined effect of the diode voltage drops in a ladder is can be significant. With a 9V battery and the ...


15

One way to achieve your progressive LED lightup as you turn the rotary switch is to use a current sink on the common of the switch and then wire the LEDs across the selector switch terminals as shown below. The constant current sink shown is a low cost way to get a 20mA sink for the LEDs so that there is no brightness variation as the number of lit LEDs ...


2

Yet another example of trying to use an N-channel MOSFET as a high-side switch. It simply doesn't work. Use P-channel MOSFETs and pull the gates to ground to turn them on.


4

It might seem like overkill but it would be fewer parts, and possibly less expensive, than some other solutions to use a micro-controller. Many Ardunio boards have over 9 digital output pins - you could drive one LED with each of nine pins. By having the switch pick different points of a voltage divider and feeding it to one analog pin, you could determine ...


14

If you’re not wedded to the specific switch you have, get a “progressive shorting rotary switch” to replace it. That works just like your drawing.


64

Using a regulated current source to light them, wire the LEDs in series and short out the segment which you want to be dark. simulate this circuit – Schematic created using CircuitLab You can possibly use a buck-boost converter to make the 30V if you don't already have a suitable voltage. Here's a simple way to build one using a LM2596S module: ...


4

If you can afford another 0.5V drop you can use a massive array of diodes. Here is an example with three LEDS which requires 6 diodes. (Sorry for the SW, SW2.., circuit lab does not have a rotatory switch symbol) simulate this circuit – Schematic created using CircuitLab


9

Here is a low tech solution that requires a lot of parts. Only 4 positions shown, you need 45 diodes for 9 positions. Sunyskyguy has a clever solution if you have a high voltage available.


1

The circuit below will provide DC bias for inputs to the switch, and remove DC bias on the outputs. simulate this circuit – Schematic created using CircuitLab "in" is any input to the switch. "out" is any output from the switch. The input bias circuit has a corner frequency below 1 Hz. It will pass frequencies above that. The output bias will ...


1

As you don't wish to use a relay, you could possibly use a CD4066, CMOS Quad Bilateral Switch (datasheet): Although as suggested in Finbarr's comment, as you might want to use a double pole switch, you would want to use all four switches within the IC package. See image below taken from How to Build a 4066 Quad Bilateral Switch Circuit Just modify the ...


0

The appropriate component here would be a relay – either in its electromagnetic form, or in the form of an analog switch. There's dedicated bus switch ICs.


0

There are numerous reasons to choose one type of switching over the other. If your circuit / load can tolerate the ground currents created when switching the load.. generally low side switching is easier and cheaper. If your circuit can not tolerate this (too much disturbance in the ground plane of the more sensitive / lower voltage processor/logic) .. ...


Top 50 recent answers are included