Hello there,

I have made a DC-DC converter which (my aim was) steps 72V to 12V. I have used a 555 IC to drive a MOSFET (IRF540N). I have made the circuit in LTspice first and it works! (24khz and 17% duty cycle). I added an inductance (0.01mh real world value for a wire and a bulb?!?) and resistance (6ohm) to see how it works under load... I added a resistor to the gate (0-1kpot) to slow down the rate of change and limit spikes.

I the 'real world' however I have exploded 3 bulbs and showered myself in glass... exciting!!! (viva la safety goggles).

I have since reduced the duty cycle to a point where the 12V bulbs no longer explode and tested it with my multi meter. I have now got 6.9VDC and 14.3VAC... Why is there so much AC?

 

Hello,

Can you post a schematic ?
Do you have a capacitor on the output ?

Greetings,
Bertus

 

Umm... No Capacitor. I have not electronics training. Just a 555 book and a soldering iron.

I accidentally destroyed my IRF540N when I shorted the circuit out with my butter fingers... Oops. I am now using a 12N604a4 IGBT as the IRF540N was out of stock.

 

Hello,

Take a look at this tutorial about DC-DC converters.
http://www.maxim-ic.com/appnotes.cfm/appnote_number/2031

This page comes from this page from the EDUCYPEDIA.
http://www.educypedia.be/electronics/circuitssmps.htm

Greetings,
Bertus

 

Right... So having skimmed through some of that I now have a splitting head ache and no idea where to put the capacitor...

There is some fairly complex stuff there; feedback loops to maintain a steady voltage, noise reduction etc...

A totally steady voltage and noise doesn't really matter: I am only going to be running vehicle lights and the horn. 12VAC would not be a problem but I also need to run a 5V DC regulator (LM2940). I don't think they like AC... Do they?

 

Hello,

When you use this part of the schematic, you will have a capacitor on the DC.

The coil stores the energy and gives it to the capacitor.
The switch shown is the FET or IGBT you use.

Greetings,
Bertus

 

Thanks, I'll have to meditate on it... c'mon brain get me through this and I'll get back to killing you with beer.

 

When you use this part of the schematic, you will have a capacitor on the DC.

The coil stores the energy and gives it to the capacitor.
The switch shown is the FET or IGBT you use.

I hope you do realize that for the above circuit +Vout will be higher than +Vin because the circuit configuration is a boost converter.

I thought the OP is trying to reduce 72V to 12V.

 

Ok, Sorry.

You need a buck converter.
That has a transistor (switch) at the power supply side.
Also described on that page.
Or take a look at the LM5005.
http://www.national.com/pf/LM/LM5005.html

Greetings,
Bertus

 

After staring at it long enough the penny has dropped... Thanks brain

However... I am using N-Channel switch, which won't operate on the power supply side... (no positive Volts at gate to emitter/source).

Is there a way to do it with an N-channel switch?