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THIS CIRCUIT WILL KILL YOU Or your family Or your friends The capacitor MUST be X or Y rated for AC mains use. If it is getting hot you are using the wrong cap. Regardless of what formulae and calculations say, if the resistor is getting too hot you must use a larger wattage one or cool it better. The same applies to the zener diode. This circuit ...


5

The polarity protection works correctly as explained in Mosfet in reverse polarity protection. The rest is the Typical Applications given by Microchip in the MCP16301/H datasheet. So, I don't see any issues there. I don't know if you have considered the inrush current when applying 30V while C2 initially forms a short: it should not exceed max Pulsed Body-...


4

It seems to me like you have both channels of your power supply connected in parallel. I notice that one channel shows a voltage of 59.40 volts and the other shows a value of 59.38 volts and, given that there is a slight discrepancy, being in parallel, there could be a real negative current flowing into the right hand channel fed from the left hand channel. ...


4

Those two diodes perform a useful function when the external load is connected between the power supply outputs (this configuration doubles the voltage potential of the positive rail). In this case, if the load becomes short circuit then the two outputs will fight each other, one will become dominant and in this situation the diodes prevent one of the ...


3

You have come across the reason why high voltages are used in the home (230V, 110V), even higher are used to transmit power over longer distances (10kV and higher). This is also the reason many cars are moving away from 12V power to 48V, and lorries/trucks use 24V rather than 12V. As you mention, Power is the current times the voltage, P=IV. If you remember ...


3

Make sure that your cap is a safety cap ,like say X rated 275VAC .Not 4oo VDC.Your Zener will get very hot when the supply is not loaded .If your hot zener goes open circuit the cap will get too much volts and get hot and fail.These non isolated supplies can present a shock hazard .Seek local advice and do not get electrocuted.


3

Yes that’s what it means. Planes often use 400 Hz generators and the plane’s frame becomes “Earth ground”. As always, Gnd just means 0V anywhere in the universe and Earth bonded implies 0V to earth ground with some impedance <100 Ohms each to another grid location with an earth ground, unless in dry remote areas with poor earth bonding. Earth is used ...


3

It'll be the braided shield and the thickest red cable, but: You'll be a bit disappointed in the current you'll be able to draw. The USB-C specification allows for chargers to deliver high currents, but only to devices that speak the USB-PD profile. That's a complex protocol, and just cutting off the wires will not do it. See this answer that discusses ...


3

The proper way would be to use... a buck converter. 10A at 12V is only 120W. Assuming 80% efficiency, you're looking at around 3A out of the 48V, which is pretty manageable. You will need to design it carefully, a bit of simulation and prototyping, as you're looking at dumping around 25W of heat. A good design could take that lower, but plan for worse and ...


3

'Chirping' comes from the power supply restarting over and over again. The typical cause is an external overload, but defective internal parts can cause the same effect. The power supply is trying to run but is encountering an over current condition that prevents the output voltage from rising to a preset level by a preset timeout of one second or less. You ...


3

The "specs" are likely for no heat load, the graph below is likely the information for the same module but there is no way to be sure. Either way, the graph shows the relation between heat flow and temperature across a module (if not the module, if you want to be sure, go find the manufacturer and see if the data is supplied). As the heat load increases, the ...


2

The simplest way to implement reverse polarity protection is to put a diode in series with the PTC fuse. If the voltage on pin 1 is negative with respect to ground, the diode will no conduct, and the circuitry downstream won't see the negative voltage. Make sure the reverse voltage rating of the diode is large enough to handle any negative voltages you ...


2

You made the battery, shouldn't you be telling us what we need in order to charge it? Typically I would say you need a 400mA constant current with a voltage of 1.41V * number of cells. You would then charge it for 16 hours with some sort of timer. At that rate it would be fairly safe regardless of discharge level, but you still don't want to leave it ...


2

A line and neutral of a 208-volt distribution system will provide 120 volts. That is only a bit more than half of the needed 220 volts. Any 220-volt load will operate very poorly or not at all. If you can get two 120-volt hot lines from a 208-volt system, you will have 208 volts. That would probably provide reasonably acceptable operation for many 220-volt ...


2

With 12Vdc drawn from 230Vac, you have a current-limit by 100Ω and 2.2uF @ 100Hz = -j800 Ohms for an apparent current of ~170 mA . with 230mA rms thru 100 Ohms the power dissipation is I^2R-Pd=5.3W exceeding its rating and operating a finger burning max temps of 150'C+ Conclusion You must use an e-Cap rated for > .4A ripple current with low ESR for ...


2

The problem is probably interference between the power supply and the telephone line. Telephone lines and power lines should be isolated from one another, but cheap devices (the DTMF decoder, or the power supply, or both) may skimp on isolation - if it were a circuit with a speaker, you would hear a loud humming noise. This goes along with the USB ...


2

Use 12V and wire the fans in parallel. If you use a 36V power supply and wire the fans in series, the voltage might not be shared equally. This could even cause the fans to burn out. In parallel, they will all get 12V.


2

As mentioned above, mosfets have leakage currents that might be in the same range as your minimum current requirements. The blocking mosfets have a Drain-to-Source Leakage Current and the conducting mosfets will have a Gate-to-Source Leakage Current. If these issues can be tackled, then your suggested circuit will work nicely. Therefore: search internet ...


2

From a physical connector rating standpoint, it is possible to use a USB-A port for delivering more than the standard 5V - the quick charger for my phone (HTC brand) does just that. It can provide 5V @ 3A, 9V @1 .7A or 12V @ 1.25A when attached to the phone using a USB-A to USB-C cable. However, that particular charger is not using USB power-delivery to do ...


2

This looks like the device modulates the power supply by drawing current in a data pattern. To measure this you can place a small series impedance (resistor) in series and use a differential opamp to measure it. To remove the DC offset of the normal device current you can add a high pass filter C1 and R6. Something like this: simulate this circuit – ...


1

Downwards triangle is ground. All grounds are connected together unless specified otherwise. It does not appear that this circuit has separate analog and digital grounds. In fact, it connects the ground directly to the live mains! Hence all the dire warnings about not connecting it to your PC at the same time. The circuit is intended to be in a sealed box ...


1

If you care about matching brightness (which I assume you do based on the nature of your question) then what you want to shoot for is equal current through each bulb. So there’s two ways of solving that problem. One is to wire the bulbs in series and use a boost LED DC-DC driver, and let Kirchhoff’s Law do the rest. This is arguably the cheapest and most ...


1

Get a single 6V wallwart and assume each letter is drawing 140-180 mA of current, so you find one that can source enough current for 9 letters. Put a 5W resistor that can drop 6V down to ~3 V and give you ~1.6A (a real big 2 ohm should work for that, or even better a couple of 5W 5ohms in parallel), noting that you're gonna dissipate a full 5W in that ...


1

I am quite certain it's from the Hirose HR-30 series. You should measure the dimensions to confirm.


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1- Wouldn't it be better to put the TVS diode in the output of the rectifier? No. These TVS diodes are to protect M1 which is rated for a max Vds of 1000V. That's why the design has three 300V TVS diodes, so they trigger first. 2- Could the number of diodes be reduced? Maybe using more modern diodes? The input line to line voltage can reach 480 VAC. ...


1

Further Datasheet research lead me to find that the popular 741 OpAmp was the replacement IC.


1

There are few caveats in your project. First, if the Anker charger is true Type-C charger, it won't output VBUS until it will see 5.1k pull-down from your Type-C receptacle. So you must provide two pull downs, on CC1 and CC2 pins, before you will see any power. Second, as soon as you signal to the charger that some consumer is connected, the charger will ...


1

You want electricity. What you want it for is kind of irrelevant. You want it, and for your own reasons, you're looking at solar as a starting point. Solar is lovely when space is plentiful and light isn't a problem. But storage of electricity isn't reliable at the temperatures you're looking at. So other options are: temperature controlled environment for ...


1

Yes, a 7805 would be acceptable for powering the HC-SR04. It only requires 15mA which is not much. The trick with voltage regulators is to not run too much power through them. Find the max power dissipation of the 7805, if you can't find it then 250mW or 500mW might be a good place to start. Lets say you have 12V going into the regulator, that would be a ...


1

Typically, you want the grounds (what you are referring to as the negative terminals) connected together anyway. In fact, it can often be dangerous not to. The connection between the two may be inside the PLC.


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