21 votes

Why should I route the feedback trace under the output capacitor?

The reason for this is that there's a very large ripple current flowing from the output inductor down into the output filter capacitor. The trace between these two isn't perfect (it has resistance and ...
Jonathan S.'s user avatar
  • 15.5k
11 votes
Accepted

Why should I route the feedback trace under the output capacitor?

I doubt it matters. Appnotes love to go overboard with layout instructions, but without explaining what and why; let alone giving quantitative values, or comparing alternative layouts with ...
Tim Williams's user avatar
5 votes
Accepted

Tantalum capacitors in synchronous DC/DC converter

It is not necessary to use a tantalum. Generally, for the input side ceramic capacitors are sufficient, combined with bulk electrolytic or Al-polymer if needed to handle load transients. The DC-DC ...
hacktastical's user avatar
  • 51.1k
2 votes
Accepted

Running a TV Using Only Solar Panel, Without a Battery

Quick observations: looks to be pretty well designed, though you haven’t put in didoes to protect the PV panels (not sure if they’re in the PV panels, they may be already fitted). You haven’t looked ...
Puffafish's user avatar
  • 4,086
2 votes

I want to control the voltage and current of a LTC3780 module using an Arduino

There are a ton of Chinese modules that say they use the LTC3780. Without more information (specifically, a good schematic) it's hard to say exactly what could go wrong. The pricing suggests the ...
Spehro Pefhany's user avatar
2 votes

Do I need to add a filter for an isolated DC-DC converter?

The test configuration is a circuit that is used to test the performance of the DC-DC converter under controlled conditions. It is not necessary to use the test configuration for normal operation of ...
Mustafa Türk's user avatar
2 votes

Clock Driver - What's wrong with my H-bridge design?

Q1 dies because its gate-source voltage goes outside its limits. As you can see from the datasheet, its maximum allowed gate-source voltage is ±12V. But once you enable it, the gate-source sees -24V (...
Rohat Kılıç's user avatar
2 votes

How to tell if buck converter outputs are isolated?

A buck converter is per definition not isolated. For the broader term of converter being able to down-regulate but with transformer (and transformer ratio), it would be a forward type converter.
winny's user avatar
  • 13.7k
2 votes
Accepted

Sampling gain effect in buck converter

Let me try to give a simple answer to this question. Current-mode-controlled converters are made of two loops: an inner current loop and an outer loop which ensures voltage regulation. The error or ...
Verbal Kint's user avatar
  • 20.9k
1 vote

Sampling gain effect in buck converter

Pulse width modulation used in the control of power supplies essentially causes a phase delay in the loop, similar to the sampling delay in a pure digital system. When the inernal ramp voltage reaches ...
MrPWM's user avatar
  • 132
1 vote

Building a constant current circuit

The best circuit for a fine-tunable constant current source would be a Howland Current Pump. There are various versions of it, and the version I prefer using in particular is the following: Image ...
Pxl's user avatar
  • 326
1 vote

Boost converter output instability

Yes, this will happen on your design because without input current limiting (or an inductor bypass circuit) the initial transient of supply voltage will form an RLC low-pass filter of fairly decent Q ...
Andy aka's user avatar
  • 443k
1 vote

Circuit to convert 14.8v battery with max 4a discharge current into 12v, 9v and 5v output?

To reduce unwanted interactions (e.g., motor noise vs. control circuits), it's an advantage to have three separate power supplies. One might derive a 5 V supply from the 12 V rail with resistors and ...
DrMoishe Pippik's user avatar
1 vote

Can you power 2 voltage rails simultaneously using a buck-boost converter?

You can use a linear regulator along with a boost converter to get 3.3V and 5V respectively.
Shamurov's user avatar
  • 153
1 vote
Accepted

Boost converter duty cycle

This is my thought process, run Pavg equation for the diode. I’m getting 714 A. Don’t think my setup is correct. It won't be anywhere 714 amps. If the forward volt drop is 0.7 volts then, if ...
Andy aka's user avatar
  • 443k
1 vote
Accepted

5V USB-Power without Battery makes MCP73832T toggle the LED

Most likely C2 is the problem. When the battery is not connected, the MCP wrongly thinking there's a battery due to C2, so it drives Stat low (turning the LED on) then it very quickly charges the ...
LordTeddy's user avatar
  • 1,995
1 vote
Accepted

Can I use one 12 V 8 A power supply to provide 5 V 10.5 A power via buck converters?

That will work just fine. The buck converters will need to be at least 52.5/96 = 54.7% efficient, which is such a low requirement that practically any buck converter on the planet can do it. The Pi 4B ...
Jonathan S.'s user avatar
  • 15.5k
1 vote

Buck converter with external reference voltage?

Probably more complex than you'd like but the LTC3886 can be digitally programmed (I2C). It has 12-bit digital programmability of output voltage of 2 to 13.8V in 4mV steps and 16-bit digital readback ...
Spehro Pefhany's user avatar
1 vote

Why have 3 AAA batteries not one AA and raise voltage (how expensive is it to raise voltage from 1.5 to 5v)?

I offer an alternative solution here, which worked for me. I have a head torch that I use on the bike, designed to run off 3xAA cells. It works just as well with 1.2V NiMH cells as it does with 1.5V ...
Chris H's user avatar
  • 2,541
1 vote

Why have 3 AAA batteries not one AA and raise voltage (how expensive is it to raise voltage from 1.5 to 5v)?

White LEDs run at around 3V. the current will depend on the size of the LED. For example, 0.33A for a 1W LED, or 1A for a 3W LED. It's certainly possible to boost the voltage from a 1.5V cell to 3V ...
Simon B's user avatar
  • 18.9k
1 vote

How can I amplify an analog DC voltage signal?

Subtract 0.6V and scale by 9 and add 1V. Or in math terms: VOut = (VIn - 0.6V)x9 + 1V-> Vout [1;10]V with Vin [0,6;1,6]V Or in circuit terms: simulate this circuit – Schematic created using ...
ElectronicsStudent's user avatar

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