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0

The source did, in fact, need to be connected to ground although the 10k resistor was drawing too much current. Switched to a 180k resistor and everything started working as expected.


1

A circuit like this usually works well. The problem is you need current carrying capability, BJT's have a voltage drop and heat up with higher current. A mosfet could be used, like a Pmosfet which is voltage controlled, but the voltage needs to be able to reach the voltage of the mosfet. So you need a switch that works well below 3.3V (BJT) and a mosfet ...


-1

The problem PV has enough power to charge a smartphone, but every day when power reduces two (2) phones failed ( in succession?) with damage due to PV charger. Why When PV source dims and voltage drops, smart phone's charger cuts out. THen PV voltage rises to an adequate threshold and a smartphone charger turns on again with a surge demand power and PV ...


0

I would recommend you research UVLO a bit more (Under Voltage Lock Out). As that is the thing you're describing. Are you sure your current chips don't already have some feature like this built in? If not, it's fairly easily done with a comparator and a couple of resistors. But there are a couple of challenges. Firstly when you power on, there could be a ...


1

The device you're looking for is called a "voltage supervisor" or sometimes "reset controller". It monitors the supply voltage and asserts the reset input on the microcontroller until the voltage is within a particular acceptable range. Some also add a short delay after the voltage becomes acceptable before they release the reset signal Many ...


1

While I don't think for this product we need to technically be concerned about EMI, I'm curious if a little bitty switching regulator like this generates all that much noise? Yes, it does, or it can. And if this commercial product is going to be sold, it will most likely need to conform to IEC 61010 and pass the FCC unintentional radiators testing at an ...


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If you already have some sort of programmable microcontroller on board, something like an OR-ing PMIC might be another possible solution. If you used an LM5050-1 with an external N-channel MOSFET, you could get away with much lower losses than a diode would have. Programming would be as simple as writing the shutdown pin high on your primary battery until ...


-1

simulate this circuit – Schematic created using CircuitLab A circuit similar to this I take it. Where each of the CLK signals go to an arduino to generate the PWM? The drive resistors need to be (5v-(Vf of LED)-(Von of transistor))/10mA (V=IR or V/I=R) and the LED usually need around 10mA for good full drive. Note: Each input will turn on the LED when ...


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The only mistake is using the NFET as a high side switch when it should be a low side switch with Vs=0V then with Vgs>=10V you pull down the load cathode and series R from the supply with the drain. So transistors used as switches (FETs and BJT’s) are always inverting. Vgs is chosen from the specs or as a rule of thumb Vgs>2.5 x Vt(max) the threshold of ...


7

TL;DR: I was driving my MOSFET gate at the supply voltage I was switching/ so I didn't maintain a high enough VGS. I'd failed to consider that Vs is not 0V once the MOSFET is on. Per the accepted answer the simplest solution to that is putting the MOSFET on the low (ground) side so that VS ~= 0, making VGS easy to keep high. See corrected (but warning, ...


1

The graph gives the energy dissipated in the device for each pulse of current where the shape of the pulse is a half sine wave. The current marked on each curve is probably the peak on-state current per pulse. Perhaps the Westcode literature will confirm.


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The energy per pulse is the energy loss in the thyristor during a pulse with the given shape (a half sine wave) and duration (in milliseconds) and peak current (in amperes). For example, a 1000A 1ms pulse will cause 1J of dissipation. If you repeat that at 1kHz you'll get 1000J/s or 1kW dissipation in the thyristor if Tj = 125°C.


1

Note that I'm not an expert in thyristors specifically, which is why this answer sounds general. The general rule for stacking things like this is that it's your responsibility to make sure that the voltage is balanced (or current, if it's a bunch of devices in parallel). This means that you either need to look for the manufacturer's recommendations, that ...


0

To answer your question, with a straight up ideal inductor, it can possibly result in zero current only during switching because the current will be delayed. To achieve this you need a saturable element, when the switch turns on, this component will hold DC voltage until it saturates, this time until saturation will be the delay between current and voltage. ...


1

Does this mean there will be relatively lossless turn on? From the point of view of the power supply feeding the converter, energy will be delivered to the load and also stored in the inductor. The energy delivered to the load is somewhat delayed by the inductor charging up so, for a short period of time less energy is delivered to the load and this may or ...


2

Sounds like impedance mismatch and standing waves between your SPTS circuit and the antenna. So, technically, not mainly your circuit's fault: You didn't match your monopole antenna. A monopole by itself does not have a 50Ω feedpoint impedance (50Ω is a very common wave impedance); your patch antenna is most probably already matched to your system; you can ...


0

simulate this circuit – Schematic created using CircuitLab ths one should turn oun at around 1.5v in


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I am wondering if there is a negative to negative (and/or positive to positive) switch. simulate this circuit – Schematic created using CircuitLab Figure 1. A double-pole, single-throw (DPST) switch can be used to connect negative to negative and positive to positive. I know mechanical relays work, but I need an electrical way of doing it. Is ...


5

You're mixing up some things. The flyback diode is needed at the relay's coil because of the coil's inductuctive behavior and the fact that the current through the coil is switched of instantly (not gradually) and the fact that a transistor is used to do that on/off switching. Transistors are sensitive to overvoltage. Such an overvoltage can occur ...


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