# Tag Info

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This is basically a link-only answer, so it may well be downvoted and/or removed. Here is a set of specifications for makeshift systems that might be acceptable for emergency short term use, according to the UK government. Note the mandatory safety devices, overpressure and a whole bunch of other things will cause more damage (including instant death) than ...

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You may use magnetic latch relays which require only a short pulse to set or reset. Here's a schematic, using a diode matrix, 3 push button switches and 3 magnetic latch relays. Freewheeling diodes for the relay coils and LED indicators are not shown. 'NO' contacts of the relays K1, K2 & K3 are to be used to select the taps. How it works: The 'S' (...

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You cannot manipulate the gate control of the LM74610 such that the mosfet will not conduct, because the body diode of the mosfet will always be conducting (in case of correct polarity). You need the apply a second mosfet to switch on the circuit: either like you suggested or e.g. using mosfets back-to-back and e.g. the LTC4359 as alternative for the ...

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Is it possible to share the positive wire from the breaker box with both light switches and go back to the breaker box with a shared negative wire? Yes you can do that, as long as cabling is thick enough to handle currents from all of your lights. Will this work or is there a better alternative? Circuit looks perfectly fine.

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I think it's from the Panasonic EVQQW series, see the datasheet page 3. However, I can't find out what the numbers mean

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You seemingly have thought to switch AC in a clever way: For +cycles of the mains AC one IGBT is active switch and another works passively through its protection diode. For negative cycles the roles of the module halves are changed. Both IGBTs seem to get the same drive signal (Vge = +15V for conduction and -8V for OFF-state). It's a puzzle why this doesn't ...

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Your problem is your assumption that you can switch AC currents with an IGBT directly. Your IGBT keeps turning on and off because of the positive and negative voltage swings on the gate compared to the emitter. The AC voltage influence also depends on its relation to your gate driver's potential. Did you get the idea for this circuit yourself, or did you ...

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There are plenty of reasons that can cause IGBTs blowing. Here are just some suspect points regarding IGBTs. A) Your 2 IGBT connection look suspicious. PS-Collectro-Emitter-Emitter-Collector-Load? B) Driver output resistors according DS - top one 3 times bigger than bottom one. In your circuit they are both 20R. C) Wires that goes to Gate should be as ...

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I had an idea of using a analog switch SPDT IC to switch the pins, i could only find one where it can more than 5v TS12A12511 This circuit is not finished though since the voltage divider for enabling/disabling the switches is proving quire hard, below 5v the IN pin must read <0.8v and above 6v the IN pin must read >5v EDIT: I think i found the way to ...

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You could use a 4053 multiplexer with a comparator and voltage reference for selecting. Be sure to carefully analyze this idea to see if it makes sense in all possible situations. The reference has to work from the lowest possible voltage with one cell, so you'll have to divide the battery voltage down. A better solution might be to add a couple of jumper ...

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A couple of comments/suggestions. First, have you considered using a solid-state relay (SSR) to turn ON|OFF the +12 VDC power? From your description it seems an SSR is a good option. Second, I recommend you use a comparator with hysteresis for your "low battery" detector circuit (Reference1, Reference2, Reference3). Note that a battery's voltage changes ...

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If you connect the drain of Q2 to the source of Q1, then Q1 will be disabled when the gate of Q2 is pulled low. (assuming they are N-MOSFET's). If this is your intent, then S2A must be at logic zero to disable the relay. (It's not very clear what else you wish to disconnect.) Also I advise you to increase the resistors between the gates and ground to ...

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So it'll be another characteristic I'm not taking into account. Could you please help me? Well, it's the right characteristic you are considering but you need to dig into the detail of the what you say here: - IPD60R180P7S (Coss = 19pF) The truth is, if you compare apples with apples, the $C_{OSS}$ of this device is more like 250 pF. The apples are ...

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The CD4016 is a CMOS quad bilateral switch. This would work if your signals are between 0 and about 15 V. Figure 1. Switching a CONTROL pin high connects the associated IN/OUT pins. Check the series on-resistance of the switch. It may be a couple of hundred ohms and this may affect your circuit. I seem to remember that the CD4066 had a lower on-resistance.

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Your opto output doesn't make sense. All it does is short gate-source without ever applying a voltage to gate-source. IGBTs cannot conduct in reverse, which would mean either D1 or D2 would be carrying the full current of a particular polarity plus any associated forward conduction loss. Why are you using IGBTs instead of MOSFETs at such low voltages? ...

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Your S1 is almost certainly not an closing switch, simulate this circuit – Schematic created using CircuitLab but a momentary pushbutton, simulate this circuit considering it's used to cycle through modes. This sounds very much like Throw away existing circuit, replace S1 by toggling button, be done. Alternatively, you could also Throw ...

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You could power the switch part between +2.5V and -2.5V. Part only supports 5.5V Max. This would allow 0V centered signals to pass through the switch. You will need to arrange the switch control signals to swing from the negative rail to the positive rail.

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There should be no problem passing the CAN signal through a switch, you don't need the complexity and cost of the optos if you are not driving it from a logic source. The switch should be a signal (not power rated) switch, with gold contacts for low resistance at low currents and voltages. I don't see why you need DPDT, DPST is sufficient, and more commonly ...

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If you did this to some SxS switches, the wiper would fall off the top or right side of the bank of contacts depending on if you do it for the final number or somewhere before. The selector was then out of service until someone at the switching office put it back. A friend of mine asked himself the same question many years ago as a child in a small town. ...

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Hitting a reed switch that's only rated for 500mA with 442mA frequently will pit the contacts on your switch, eventually causing it to fail. There's a number of cheap MOSFETs that will switch the low side. I picked up a handful of IRLZ44 parts for 29 cents each at a local surplus store (\$2 each from DigiKey). Even cheaper is the AO Semicon AO3400 or ...

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