How to use 4 pins inductor

Hi I placed it from my post above
 

Oops, I must have missed that somehow. I'm sorry about that.

I can't imagine using 479pF rather than 470pF would make too much of a difference, but where is the 417uF capacitor?

Also, did you have anything connected to the output when you turned on the power (when the chip fried)?
 
The 417uF Elec Cap is the electrolytic capacitor connected on the output of the diagram. BTW, I didn't placed the "Optional Filter" from the diagram.

The only load connected is 1kohms resistor and a 3mm red LED.
 
I've been playing and experimenting with inductors and SMPS for a while and I found that if you don't have a load, the output voltage can go extremely high and that will fry FETs and ICs, I lost several FETs before I figured that on out, I was looking for about 40 volts out and used 60 volt FETs figuring that was plenty, finally tried a 400 volt FET and I was getting an output of 190 volt and that was killing my 60 volt components. I've been using a 555 chip to make 40 watt, 40 volt circuits driven by 12 - 14 volts DC.
Kinarfi
 
lloydi12345 said:
Thanks for the input kinarfi, I'll take note of that. Anything else please?
Click to expand...
What's the goal of this project? self education or are you trying to make a 9 volt power supply.
Kinarfi
 
I'm actually building a project efficiently stepping down the voltage from 32V to 9V. It's just about the inductor thing. I already asked someone to buy me the IC again and I could do tests again this monday. I hope I could get everything right without seeing a smoke puff.
 
How do you mean pins 2 and 4 are connected? Do you mean you checked this with a multi-meter in continuity mode?
 
Sceadwian said:
How do you mean pins 2 and 4 are connected? Do you mean you checked this with a multi-meter in continuity mode?
Click to expand...

I believe that is what he did.

lloyd, an inductor is nothing but a coil of wire wrapped around a ferrous core. There will be very little resistance, so the continuity test will show it is connected. The inductor only opposes current if it is AC. DC will pass straight through it without a problem, and that is how the continuity test works--It sends a voltage through, checks the current, and calculates the resistance. If it's below a certain value (say, 5 ohms), it will beep. An inductor will have little to no resistance.
 
I was going to show you an alternate schematic and googled for the price of a MC34063A, $.079, and ran into **broken link removed** and plugged in your values. I also reread the spec sheet and came up with the following and looked in Spice for MC34063, then modified what they had to match your values. According to spice, you'll have a 100 mv ripple at .5A out.
 
Last edited:
I assume the same thing Derstrom, this doesn't help the original poster in figuring out why their circuit smoked though. I'm guessing that there was a wiring error in the construction of the example circuit, or otherwise some component that was out of spec so far that it caused the failure as opposed to the inductor itself.

I'm not sure the logic that caused the original poster to assume it was the inductor in the first place that caused the failure.

No schematic has been posted as to the original posters TRUE design (not intended) nor any image posted of the actual hookups, once again we're back at the point of not enough information.
 
Sceadwian said:
No schematic has been posted as to the original posters TRUE design (not intended) nor any image posted of the actual hookups, once again we're back at the point of not enough information.
Click to expand...

He said he wired it exactly as it was shown in the schematic, but a photo of the setup would be a huge help. I would guess that there might be something mis-wired or mis-read, but we can't know for sure without a picture.
 
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