Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Dc-Dc step up Converter

Status
Not open for further replies.

rednwhite

New Member
Hi guys

Built a step up converter but my output just seems to slowly increase and not create a constant DC output and the mosfet used for switching gets hot quickly.

Components I am using:
inductor = 12uH
Diode = Byt08p400
Load R = 10k
Cap = 100uF
Fet = IRF510
driver = mcp1404

Using a duty cycle of 0.5 at the minute. Any ideas of the problem?
 
What is the pulse frequency? If your pulse frequency is too low you will be saturating the inductor with resulting high currents.

If you look at the data sheet for your MOSFET you will see that it requires 10V, not 5V to fully turn on.
Edit: If you want to operate from 5V, then you will need a logic-level type MOSFET.
 
Last edited:
I was using 50kHz, would you suggest using a frequency more like 500kHz then? The ammeters I was using to measure input/output current were not showing anything
 
It will take about 9 µs to get to saturation current with those values.

It means that 50 kHz would be a bit too low to get the most out of the circuit. If you run with a 50 kHz and 50% duty cycle, that is 10 µs on and 10 µs off, so that is to long and it will saturate the inductor.
 
It will take about 9 µs to get to saturation current with those values.
How did you calculated that out? Wouldn't saturation time be dependent of the iron core?
 
12 µH * 3.8 A / 5 V = 9.12 µs

Getting the basic information out of Rednwhite is a slow and painful business.

12 µH Post #1
5 V Picture referenced in post #3
3.8 A Post #7

The rate of rise of current depends on the voltage and the inductance. The current limit depends on the construction of the coil.
 
perhaps some reading is required:

https://www.electro-tech-online.com/custompdfs/2010/11/dcdcconv.pdf
**broken link removed**
**broken link removed**
Flyback Converters for Dummies

Thats for high voltage, but the principles are the same, regardless of your output voltage. Your duty cycle is determined by the ratio of Vout to Vin. Frequency is chosen depending on your inductor, current load, and various other characteristics. Perhaps if you specified exactly what you wish to do, and what parts you have available to you, we could be more help. Example, you want to boost 5v to 12v, @ 0.5A using a 12uH inductor?
 
How did you calculated that out? Wouldn't saturation time be dependent of the iron core?
A normal assumption is that the core starts to saturate at its rated current, so you use that as the limit. Then substituting into V = L di/dt for V = 5V, L = 12µH, and di = 3.8A, you get a dt of about 9µs to reach 3.8A.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top