7

Many smaller inverters (and my Honda EU2000 generator) have "floating neutrals". I don't think "floating" is quite the right term here, as the Neutral will alternately be connected to +170 V or Ground. What happens is that the inverter has a single high voltage supply (+170 V or so), and connects the Line to that supply on one half-cycle, and Neutral to ...


6

To add another specific to what The Photon said, error amplifiers in switching regulators are often implemented as gm amps. So they take the difference in voltage on the input and convert it to a current with a gain of gm (A/V). That's a really easy way for IC designers to save a pin as the compensation network goes between the output of the gm amp and ...


6

Pin 1 is on the right side of the connector which is a bit unusual, and from the pictures I guess you thought pin 1 was on the other side, so that explains the wrong polarity. The power supply has a minimum load on two channels, notice it says "0.4-5A" on channel 1, not "0-5A". This is common on this type of switching supplies. It only has feedback (and ...


6

My question is where should the V_sense resistors connect, before or after the secondary LC filter? It should almost certainly NOT go after the 2nd LC filter because you'll turn the circuit into a power oscillator. With feedback systems you have to take care of not introducing too much phase shift or delay or negative feedback (desirable) becomes ...


5

Whine or high-pitched whistling usually comes from coils and ceramic caps. Coils whine due to magnetostriction: core material expands and contracts according to the magnetic field, so they convert current ripple into sound. High-K ceramic caps (ie, not C0G) are piezoelectric: they expand, contract and flex depending on the voltage applied to them, which ...


5

This is a classic in switching power supplies. The auxiliary \$V_{cc}\$ is an image of \$V_{out}\$ and is dependent upon the transformer turns ratio linking the auxiliary and power windings. That is to say, if you have a 12-V output and a turns ratio of 1:1 on the auxiliary winding, then the auxiliary will be 12 V also. This is a theoretical approach because ...


5

Most (all that I used) of the impulse power converters have VOUT connected to the coil. This one has VIN connected to the coil. Apparently you are used to buck converterns, then. It might be worth a web search. A buck converter works by switching power on and off to the coil, so the coil input sees an average voltage less than the supply. Coils try to ...


4

It means that the output 115VAC is not connected to the input power in any way (not common with the minus, which is connected to the car chassis, nor to the +12V input). Similar to adding an isolation transformer to your mains that comes out of the wall (well, without the pesky design, safety testing and certification ... ). That's a fairly subtle ...


4

In a word: don't The best way to drop 10 volts at 0.25A is already answered by other people, so I will not cover that. Your question, however, aims at powering an ESP32 module specifically, and you don't disclose more details about the intended usage. However, if you are targeting an IoT scenario with WiFi communication, you need to consider that the ESP32 ...


4

Best? no , Simple Yes Get a junkyard Festoon lamp holder but instead of a 12V bulb, buy a 24V 10W bulb designed for 250mA constant current with just enough voltage drop to run any 3.3V LDO and only dump < 0.5W in the LDO and 2.2W out of 10W in the Lamp, so it will be dim. If the bulb is 10W rated at 24V it will be 3 Ohms cold and if rated at 28V, it ...


4

you're planning to burn up 2.5 watts a TO220 part is going to need a heatsink at that power level, but you can put a resistor in series with the input and make the heat there where it's not damanging anything simulate this circuit – Schematic created using CircuitLab at low current the reguilator sees near full voltage, but the current is low so ...


3

What it means is that the neutral output of the inverter is not connected to ground. On a car, ground would be the chassis. The ground pin might be connected to the car chassis, or it might be totally disconnected. If you are just plugging one appliance in, it isn't necessarily unsafe. If the appliance is double insulated, with a 2-pin plug, it's largely ...


3

The leading-edge blanking circuitry or LEB is a common circuit found in modern switching power supplies. It is usually built as shown in the below circuit: When the drive goes high, a short pulse of \$t_{LEB}\$ duration blinds the IC for a small period of time. We are talking about 250-350 ns for ac-dc controllers operating below 100 kHz and around 100-150 ...


3

The feedback loop around the TL431 is a classic. It works by pulling the opto LED cathode to ground which, in turns, modulates the peak current setpoint in the primary side via the optocoupler emitter or collector current. This works well for a constant-voltage output. Then, when the output current \$I_{out}\$ increases, the voltage starts going down and, ...


3

What you are hearing is most likely the piezo-electric effect from the MLCC capacitors - as various other comments and answers stated. Your oscilloscope shot shows a 3.81 kHz oscillation with 300 mV on the 5 V line. This is very much audible. As your switching frequencies are all outside the audible range, the source of the oscillation is very likely from ...


3

According to Onsemi conductors PDF manual for this IC, page 5, this IC has a ripple at the output of 400 mV. Their designs often suggest use of a low value inductor (LC network) to remove most of such ripple. However inductors and capacitors before any extra filter are subject to whine and chatter. This should only affect a few parts, and you CANNOT make ...


3

Here are the innards: This is a classic boost regulator. Figure 3. A classic boost regulator. Source.* The transistor shorts the inductor to ground causing it to "charge" up magnetically. when the transistor is turned off the inductor voltage rises to feed the output.


3

The LM431/TL431 is an adjustable zener diode, with somewhat more stability with temperature (TL431 is better). The suffix identifies the temperature stability and static accuracy. It works very well in SMPS loops because it is often driving an opto-coupler which throttles the PWM to regulate the output voltage, often with better than 1% load regulation. In ...


2

The TL431 is (probably) the lowest cost and best value for money IC in existence. A TL431 in even 500 quantity costs under $US0.01 in China (and probably not vastly less in large quantities). For that you get 0.5% accuracy voltage reference, 50 ppm/C variation including all parameters across whole temperature range, an internal comparator, useful ...


2

A TL431 acts pretty much like an op-amp plus a reference. You need a reference regardless of whether you use a discrete op-amp or not. The amplifier portion of the TL431 contains compensation much as a discrete op-amp. The TL431 is handy in a typical galvanically isolated supply because it can directly drive a feedback optoisolator. Image from the ...


2

I finally found the root cause: the HV5530 chips really need 12 V logic in order to work. Initially I gave them 5 V logic and that made the communication really susceptible to noise. I made a circuit to raise the logic level and the display looks perfect even with the snubber from the boost converter removed.


2

Not every triangle you see in a schematic is an op-amp. All kinds of amplifiers are drawn on schematics as triangles. In this case, the amplifier symbol likely represents the complete difference amplifier subcircuit, including both the op-amp and feedback resistors.


2

My question is why the turn on speed is linked to the diode reverse recovery time ? It isn't. What they are saying is: usually the FET can turn on faster than the diode can turn off, however that results in both the FET and the diode being on simultaneously, ie cross-conduction, and increased losses. So it is more efficient to switch the FET a bit slower to ...


2

"How to determine parasitic capacitance of a diode according to the datasheet?" There are two simple answers. 1) Read the data sheet and take the value you see . In the case of the data sheet you link, it is stated as being 12 pF. 2) Unless the data sheet provides a graph, you can't. In general, reverse-biased diodes show a capacitance which decreases ...


2

Functionally you can get a similar result in both your examples using a differential amplifier with some limited gain. simulate this circuit – Schematic created using CircuitLab


1

Often it's a good idea to read the connector drawings from the connector manufacturer -- I used to keep the Pin-1 markers of all kinds of connectors in my notebook. However, very surprisingly, Molex doesn't mark the Pin 1 on the KK-5.08 connectors, though they do for the KK-3.96! (Source: Molex drawing for KK-3.96 connector) (Source Molex drawing for KK-5....


1

If I understand this correctly, you intend to use this "PeakPSU", apparently a cheap PicoPSU knockoff: ...which is a board with a bunch of DC-DC converters intended to generate the usual voltages to power a PC motherboard from a 19V laptop power supply. The "specification": Note it doesn't spec all the stuff you'd really like to know about a switching ...


1

1) All your noise problems come from two issues: Timing cap* inductor pair are too big. Reduce L 20uH to 33uH low DCR. Move up to 30 kHz. I suspect you have resonance with your 4.7uF load and 100uH. The e-caps are STD rated for low ripple current and must be low ESR. Thus the current is limited and the piezo ceramic caps are audible with too much current ...


1

Everyone has their own preferred simulation tool depending on what their background is and what they are simulating most often. You can get Almost all simulation tools to simulate any circuit(within reason) some just are a little easier in some scenarios. I would suggest you look at what simulation software the people around you or the people that you are ...


1

The Miller effect is due to having a capacitance across two points with a negative voltage gain between them. [Intuitively you can think about one side of the cap getting pulled down by the gate-drain gain while the other side is pulled up by the gate driver, so you need to add the extra charge that the drain is trying to pull out.] It multiplies the ...


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