The Stack Overflow podcast is back! Listen to an interview with our new CEO.

Hot answers tagged

18

I'd suggest your B340A diode is avalanching since you are far exceeding its reverse voltage rating. The diode must standoff the voltage you generate, so you need a voltage rating above your capacitor voltage. I'd use something in the 75-100 volt range, and perhaps an ES07B would suffice.


12

I believe you are exceeding the peak inverse voltage (PRV) rating of D4, the 40 V Schottky diode. During your switching cycle when the SW pin on the 2586 goes to 0 V, D4 becomes reverse biased due to the level at the output at the top of C34. With the output set to 57 V, this exceeds the 40 V reverse rating of D4. This can only be ...


10

There are rectifier stacks that can handle (say) 100kV. Forward voltage might be 120V meaning that they're 100+ individual diodes in series. Here is some data from a random Chinese supplier (no experience with those guys, but it should give you an idea):


9

1) Indeed a suitable transformer to make the 160 VAC will be hard to find. And because almost no one needs 160 VAC it will be expensive. If you were not a newbee in EE I could suggest a 120 V to 40 V transformer and then connecting the secondary in series with AC mains but that needs you to be aware of the dangers of doing this so I didn't suggest this. 2) ...


6

I can suggest two approaches: You may be able to purchase a 1:1 isolation transformer that is rated at the current level that you need and then remove some of the turns from one of the windings. This would have to be done very carefully to maintain the insulation integrity of the high voltage transformer and would only really be applicable for a one-off ...


5

Pin 2 is not grounded so it does not work.


4

Yes you can use a boost converter. This would be difficult design that required experience. 3.3V requires carefully choosing the MOSFET for threshold levels, most logic level MOSFETS really want 4.5V. 8A of current will produce a lot of heat in a diode so you will need a heatsink and probably a fan. The loop formed by the MOSFET, diode, and output ...


4

Convert 3v to 5v very high amperage using boost converter If the input voltage is 3 volts and the inductor is 3 mH and the switching cycle is 15.625 us (inverse of 64 kHz) then the maximum switched current into the inductor at near enough 100% duty cycle is 3 volts x 15.625 us / 3 mH = 15.625 mA. This clearly and obviously is the DCM (discontinuous ...


4

There is no circuit schematic on the web page you linked but it seems that there are no active electronics in the lamps and that the LEDs are wired in some sort of series combination to run directly from the mains. Figure 1. Phase-controlled triac dimmer waveform. By varying the turn-on point from 0° to 180° on each half-cycle the voltage can be reduced ...


4

The switching controller 'knows' when current should flow, so it is indeed possible to replace the diode with a MOSFET that is switched on at the correct time. This feature is called "synchronous rectification". For example, this is how it is implemented in the TPS61322 (the inductor is connected to the SW pin): However, most chips do not have integrated ...


4

Only for about half a mains cycle. ie NO. Even running a 60 Hz mains transformer on 50 Hz causes it to run hot (ask me how I know) due to increased magnetising current. Power transformers are designed to use the core iron well (except i very special cases) and magnetising current is arranged to flux the core to the point on the BH curve where the core is ...


3

A relay (solid-state relay or FET) is the best way to switch high voltage from low voltage. Arduino has low current output not meant to do what you're describing. It's not possible. Use a relay or a FET. It is even safer to use an optocoupler for isolation.


3

If you are having to ask then be clear that you are not qualified to carry out the work. If you are seeking understanding so that you can request a quotation for someone qualified to do the work then: \$ V_{L-N} = \frac {V_{L-L}}{\sqrt 3} = \frac {208}{1.73} = 120 \ \mathrm V\$ so, yes, that will work. Generally the supply should be able to take an ...


3

For the boost to be in steady-state the inductor current delta (ramp up) during the on time has to equal the inductor current delta (ramp down) in the off time. Let's assume Vd is the input voltage and Vo the output voltage, and the diode is ideal. simulate this circuit – Schematic created using CircuitLab Rearrange your equation $$Vd * tOn + (Vd - ...


3

No-one can know what will happen to your module in that case. That's because from "Update 2" in my answer on a previous question about the same "XY-016" module, there is evidence of multiple different MT3608 clone ICs being used on those modules. The behaviour could be different depending on the specific clone that you have. Either don't risk it (which is ...


2

This will work with some modifications. First, the inductor needs to be increased to 680uH according to the datasheet, or else system may be unstable. Output cap should be rated for 100V. If you get flicker, you may need to increase size of output cap. You need to select compensation resistor value of about 1K based on your load, min input voltage and max ...


2

You will struggle to find a 120 -> 160 VAC or 240 -> 160 VAC transformer as there aren't a huge number of use cases. You should be able to find a suitable variac, which you would be able to set to an appropriate output voltage. This would be able to function as a dimmer also.


2

D4 should have a voltage rating 3 times the expected DC output just for a 50% safety margin. The reason is that when C4 has 57 volts on it the diode sees both the 57 volts and the zero volts when the internal MOSFET is ON again. Now it has 57 volts on it in reverse. Then another surge of forward current. The inductor is overheating due to D4 shorting out, ...


2

You can buy a Hammond 290HX which has 120:356CT windings. You can use half the 356V winding to get 178V @420mA, and buck that down to 171V using the 6.4V CT winding. Cyan wires are the output.


2

if I could use a Step-up Power Converter connected directly to Arduino´s Digital Pin. Is it possible? That will not work, the Arduino's outputs cannot deliver enough current. An Arduino's output can deliver around 10 mA and that's enough to light up an LED but little more than that. In theory a 5 V to 12 V Step up converter can be designed and made but at ...


2

Am I right that I don't need a high current rating at the transformer? You need to pass enough energy to power the device, but other than that no. Induction heaters use a resonant circuit meaning the current (and voltage) in the work coil will be higher than in the supply. Are my hopes realistic here (financially)? Not enough info. What's the danger ...


2

I would have to assume that you are using some sort of XL6009-based converter module that has all the required passives. I have no information about the other module you are using. A quick look at the XL6009 data sheet shows that the minimum operating voltage is 5V. Although there would be a bit of design wiggle room, there is no expectation that it would ...


2

If you have a ground plane, then the guard ring will have little impact. Guard rings are usually to protect high impedance inputs like specialized op-amps, not so much for EMI radiation. National Semiconductor Application Note 241 ELECTRICAL GUARDING The effects of board leakage can be minimized using an old trick known as guarding. Here the ...


2

I provide this only because you have been given several other options and I think there's a chance this would be less lethal should you make a mistake. Please make every effort you can to learn about electrical safety before you attempt this project. It is very difficult to comprehend the basics of electrical safety without first understanding voltage, ...


2

It's adjustable by means of the resistor divider used to sense the output voltage and compare to the internal reference. There's no digital interface to adjust the output voltage. One way to adjust the output voltage via microcontroller is to use a resistor to sum an external voltage (generated by a DAC via the micro) into the FB pin node. This will allow ...


2

Read the specifications in the datasheet. If the booster came without a datasheet, then you're on your own, and you'll have to measure it. One of the limitations of a step-up booster is the maximum input current. If you want to achieve a high output voltage with a low input voltage (a high boost ratio), the input current would be many times the output ...


2

As mentioned in a comment, here is an interesting video giving details about the internals of a step up converter. I bought this item, and here are more details: Internal: Apart from some resistors and capacitors, there are: a Schottky diode SS24 a 3R3 power inductor an unknown AL877 6-pin chip (?) I also bought this item ("MT3608 DC-DC Step Up Boost ...


2

Such converters exist. That link goes to the datasheet of the Texas Instruments LM2621. It takes in from 1.2 V to 14V, and puts out a regulated voltage at up to 14V. It can deliver up to 1A. This is the datasheet of the LT1073. The datasheet includes an example of boosting 1.5V to 9V. It is only intended for low current, though. Like 16mA when ...


2

You will cause the transformer core to saturate and the windings will overheat and the transformer will be destroyed if you try this. No, it will not work. Not with a cheap transformer, not with a high quality transformer. You can go down in voltage (from rated), but not up, certainly not 2:1 (for given mains frequency).


1

Yes but not ideal as it could age the battery much faster. If the charger stays on because the MP3 player draws near or more than the 100mA level where the charger is supposed to shut off then the automatic charger stays on. If the charger had more "smarts" with a timer as a safeguard to limit the duration at 4.2V, that would be best. - details - Another ...


Only top voted, non community-wiki answers of a minimum length are eligible