New answers tagged

0

Does anyone know how to build a synchronous boost converter with IR2104 mosfet driver? and... Any help would be much appreciated. It's going to look something like this: - Q2 charges the inductor from the main power rail by shorting it to ground and, when Q2 releases, Q1's diode will push the stored inductor energy into the load capacitor rapidly ...


6

A 5 ohm resistor will try to draw a current of 4.8 amperes from a 24 volt power supply. You say you measure currents of 280 mA and 400 mA with 2 different supplies. Clearly neither supply is providing 24 volts. If you want to test your supplies at the same current level as your motor, you need a much larger resistor (24/0.18 = 133 ohms). Also, have your ...


0

I believe that your batteries may be too small. I use modules that look very much like (probably identical) for my LED yard lights. However, my lights must be smaller than yours - mine are 38V @ 1.7 Amps. I removed the original power supply and use these modules instead. The smallest battery I use is Dewalt Flex-Volt batteries - I have both 6Ah & 9Ah ...


0

If you want to have a reversible convertor, imagine a buck or boost topology (no matter which one of them, one is a reverse to the another). Generally, it consist of two switches and if you don't want the reverse capability, one of the switches is a diode so it drives itself and you have only to drive the other one. If you drive both switches in anti-phase, ...


0

Can I simply connect a Zener diode in reverse in parallel with the boost converter, which breaks down at some voltage above 36V, e.g. 37V? You can do that, but it won't have any effect unless the motor is being mechanically driven at higher than normal speed. To brake the motor you can switch a resistor across it. The lower the resistance the more ...


0

Simple low side current sense with a non-inverting amp. Power the amp from the same power source as your mcu's ADC, scale the amplification as you see fit (max failure current should make a signal near max of your ADC input). As scaled here, you get ~2.3V for your max 12mA signal. I've simplified your circuit as the current source in this diagram as you'...


3

If the Arduino and load share a common ground then you may be able to put a current sense resistor in the load's ground wire and measure voltage across it directly, or if that is not possible measure current at the ground side of the voltage converter output. However in the general case where grounds are not shared a fully isolated measurement will be ...


0

There is also this schematic that I don't really understand too... First, take a look at the internal diagram given in the datasheet: The IC is is a simple boost (step-up) converter. Here's a simple diagram: Image Source (There's a detailed explanation about how a boost converter works) Pin 1 : I can leave it on fly ; Nope! Here is what the datasheet ...


0

The circuit is a boost switched-mode power supply. Basically it stores energy in the inductor and the capacitor, switching back and forth with a transistor at high speed according to a clever signal which depends on the output voltage. (Ask a question about "boost switching regulator" if you want more on this). The basic circuit is like this (which you can ...


5

Current only flows when there is a potential difference (voltage) between two nodes. Assuming this is an "ideal circuit", the switch in your schematic has zero resistance, and therefore the anode of the diode is at 0V (assuming the bottom line is the circuit "ground" reference). The cathode of the diode will either be at 0V or higher, and in either case the ...


1

It is possible. You need a huge battery though. Below you may find a simple boost design. Pay attention to the input current. There are MOSFETs/IGBTs and diodes that can handle this current available on the market. However, heat dissipation may be an issue.


0

Looking at the motor you want to use I would not advise on doing what you have planned for a few reasons : Conversion efficiency can be as high as 90-95%, but 5% loss at the power you talking that is a lot of heat. Motor nominal voltage is 24V, running it at 48V to get more power/rpm out of it will most likely result in over-heating of its winding. You are ...


1

The days of through hole are slowly coming to an end. The advantages of SMT are just too compelling - cost, size, performance, and compatibility in assembly with other SMT. For experimenters, there are circuit board adapters to convert surface mount to through-hole. They’re large enough that you can also mount some passives on them too. For a switch mode ...


1

You will be out of luck. Modern SMPS systems use a high switching speed, which simply won't work reliably with the high stray capacitances and inductances of breadboard. Many things you can prototype on breadboard. Others you can't. Many chips make sense in DIP packages. In this case, neither works out.


0

You said, "I wanted to build an excruciatingly simple boost converter circuit". I wanted to do the same thing, and built many a Joule Thief in LTSpice, and I put it into the same category -- The Joule Thief is really a self-optimizing boost converter disguised as a hobbyist circuit, but I've learned a lot about boost converters from stepping the Joule Thief ...


0

Will something like MT3608 be able to provide 5v when the input is also 5v from a wall charger? Unfortunately the MT3608 doesn't do a very good job when the input voltage gets close to the output voltage. I tested one set to produce 5 V, with a 10 Ω load (~0.5 A current draw). Below 4.85 V in it produced a nice stable 5 V out. From 4.85 V to 5 V the ...


1

No, it will not work if you connect the 5V charger to it. The module in your post is a boost converter, which means the input voltage has to be lower than the output voltage. If you connect the lithium battery to it fully charged (4.2V) you can use the module, since 4.2V is less than 5V. You don't mention if you want your system to be automated or ...


0

L = 22 μH \$t_{on}\$ = 26.11 μs I have tried to take the solve the equation for the inductor currents at each state But it does not seem right The problem might be your inductor value of 22 μH It should be much higher for a \$t_{on}\$ of 26.11 μs (D = 0.7833 at 30 kHz). For instance, with an applied DC voltage of 325 volts, after 26.11 μs, the ...


0

You wrote that no ripple is needed to take into the account. Then you simply calculate how long it takes until the loaded current of your inductor is diminished to zero when the voltage over the inductor is Vout-Vin. If your loaded current in the inductor is Ix then the needed time is Ix/((Vout-Vin)/L). You have actually tried this but got stucked with ...


Top 50 recent answers are included