Switching Power Supply
I got this power supply in 1996 with an used Grid 1750 P286 laptop computer which I bought for $300 and which came with Windows 3.1 but I hated it and I stuck with MS-DOS. This step-up (boost) switch mode power converter is intended to allow the use of the computer in the car, with a 12 - 14 V input.
Output: 17.25 Vdc, 1.25 A
My current Compaq laptop draws more than this adapter can give so I am studying the possibility of modifying this adapter so it will be able to supply 2.5 A which is twice what it says now.
It seems this power adaptor was originally designed with more capacity in mind and maybe the reason it is listed as 1.25 A is only due to the heat dissipation capacity. The course of action I would suggest would be:
1- Increase load slowly monitoring temperatures until current limiter kicks in.
2- Modify current limiter circuit to allow current up to 2.6 A
3- Check that the switching transistor is driven so that it switches as it should.
4- Modify heat sinks if needed.
I think the main modification required would be to lower the value of the main inductance and this could be done by removing some turns or, maybe much better, by putting another similar inductance in parallel.
We can clearly differentiate the three different conduction stages:
Scope traces: Vert: 5V/div, Horiz. 2 uS/div.
I got it to work giving 2.5 Amps! These are the changes I made:
1- Added a second Inductance in parallel with L1. I used one from the junk box which looked similar to the original:
2- Replaced R16 with lower value. As I did not have resistors of such low value I used a regular resistor as a mechanichal base and I wound some copper wire around the body to make the value I wanted. You can see it in the upper right of the photo.
3- I had to replace C1 (at the input) with a much higher value.
It seems the FET is not transitioning fast enough and that is the cause of the heat. Here you can see the waveform at pin 8 of the IC. The rising edge seems better but the falling edge takes too long. The rising slope takes about 1/2 uSec but the fall time is about 3 uSec which is 6 times as long.
I removed capacitor C4 and you can see the higher slope of the falling edge. Here you can see the old and new waveforms superimposed.
I assume the "step" on the falling edge is the moment when the FET stops conducting so it would be the duration of that step that is important. the duration of the step is about 550 nS and that would be the transition between ON and OFF. I would expect a similar thing in the rising edge but it is only a tiny spec so it seems the transistor can overpower it on the way up but on the way down there is only a resistor.
Next I tried this but still without C4:
And this is what the waveform at the gate looked like:
I still need to test the heat dissipation as I will probably need to enlarge the heat radiators.
This is what I have now:
I added a new exterior heatsink and placed the FET and the diode on it.
It is so overdimensioned that it barely warms up.
The following two circuits are from the MC34060 specsheet.
It seems the power supply I have is pretty much a variation of this first one.