75V power supply using a 555

RichardO

Joined May 4, 2013
2,270
We should probably start over. Rich, your circuit has the wrong duty cycle. That is why the inductor has to be small in inductance and high in current.
So if we start with the calculator with 12v in and 85v out with a constant load of .1a, we need about 1Mh. It will now run in continuous mode at much lower current in both the inductor, transistor and diode.
The value of the inductor does not set the requirement for the current. This is determined by the input voltage and the power into the load. In other words, the average inductor current stays the same if we juggle the convertor operating frequency, duty cycle and inductor value but keep input voltage and output power the same.

My assumption that the peak current for the inductor is 4 times the average current is based on the assumption that the drive to the FET is 50% duty cycle at the typical load voltage and current. I should have modified the circuit to reflect this.
 

RichardO

Joined May 4, 2013
2,270
Mmmhh... I'm getting a reading of 653 mA RMS on that... although the waveform jumps between 760 and 550 mA.. compared to the 100 mA RMS @ 75V, that would be 7.836 watts at the input and 7.5 watts at the output... which (if I'm not mistaken) would be an efficiency of 96%.... it's not that bad, is it?
96% efficiency sounds a bit too good. My gut feel is that the inductor and FET are dissipating more than 300 mW (4% of 7.5 watts).

... on second thought, if the inductor and FET both have resistances less than 50 millohms then losses might be that.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
Wait a minute here... I gave you the current at M1 0.609 A ... but when I measured the current at the Voltage source, I get an RMS of 1.86 Amps!
That means that the source is doing 12V x 1.86A = 22.32 Watts, while the output is doing 75V x 0.1A = 7.5 Watts...
Which gives us a far more credible (and troublesome) efficiency of less than 34% !!!!
I wonder... which part's gonna heat up the most?
 

wayneh

Joined Sep 9, 2010
18,104
Now you're seeing what I'm seeing. The inverter I worked on is intended to be battery powered, so that sort of loss is not acceptable.
 

ronv

Joined Nov 12, 2008
3,770
You must be counting the turn on transient. That only happens on power on, so doesn't really count.
If you press ALT and mouse over the part (you should see a little thermometer) you can measure the watts.
Then hold down the right clicker and draw a box around a large portion of the waveform (excluding the start) to get the part you want.
Once you have that you can control, right click the label at the top of the waveform and get average power. I think your 96% is pretty close. A better FET might pick up another 1/2 percent.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
If you press ALT and mouse over the part (you should see a little thermometer) you can measure the watts.
Thanks for the tip! What I did is I clicked on the time scale at the bottom of the graphic and then changed the axis' left limit to 5ms, and kept the right to 10 ms, so that got rid of the startup slope... can't seem to be able to adjust the right limit to anything above 10ms, though...

Yes, the RMS at the power source is back to 656 mA. Also, I'm reading 7.8341W at the source, and 7.3352W average at the output (why is it reporting it as average, and not as RMS?). That gives about 94% efficiency... pretty darn good...

I also replaced the 750 ohm load at the output for a 100K one (as a test for protection, to make sure that the circuit never runs unloaded), and the output voltage didn't change at all, plus the current of course went down to around 1 mA @ 110 mW.
Funny, if I place a 1MΩ resistor instead of the 750Ω one then the output voltage jumps to 105V !!! Though, as expected, current goes all the way down to 106 µA ...

I think it's time for me to actually build this thing (version 2.0 of your "555 boost" file) and see what happens. Though I'm gonna have to use the (sucky) IRF740, 'cause (as I've already said) that's what I have available...
Now... where can I find me a decent 1 mh inductor without going to Newark or Digikey? Guess I'm going to have to build one myself...
 

ronv

Joined Nov 12, 2008
3,770
Your right Rich. I like it. I have always stayed away from discontinuous mode.
Trading the higher inductor current for the lower inductance is probably a good trade and more stable.
 

ronv

Joined Nov 12, 2008
3,770
You could try Rich's circuit. It would be easier to wind an air core 30uH inductor. Your FET will get a little toasty though.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
You could try Rich's circuit. It would be easier to wind an air core 30uH inductor. Your FET will get a little toasty though.
I just measured Rich's circuit. Although power at M1 has peaks of 250W, it averages at around only 200 mW (I still ask... why is LTspice reporting average power and not RMS?). Current is at 1.2 Amps RMS

Current through the inductor is 1.28 Amps RMS, while power is at 205 mW average...
I'm reading 695 mA at the 12V power source... that means 90% efficiency, if I'm not mistaken..

It doesn't look that bad at all... maybe I can use 3 of the 100 µh inductors that I have available and connect them in parallel to get the 33 µh, and I'm all set! Besides, three inductors in parallel will be rated to carry up to three times more current, and hence run cooler... am I right?
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
I just checked Rich's circuit with an IRF740 (actually an equivalent with a 55 mΩ impedance)... efficiency goes down to 85%... still acceptable for my application...
I think I'll go ahead and use this circuit, and then let you all know how it went, pictures and all... Thank you all for your time and your support!
 

ronv

Joined Nov 12, 2008
3,770
I think that may be a problem. You just found a problem with spice. It doesn't show inductor saturation. With Rich's circuit the peak currents are pretty high because the inductor current goes to zero. So it takes an inductor capable of higher saturation current (but lower inductance) than the other one. There is a way to model this. I have to check where I have the formula.
So basically what happens is when you go over 1 amp in your inductor (3a for 3) the inductance goes down. This will make the current go up. You can see the problem. But let me look.
 

wayneh

Joined Sep 9, 2010
18,104
You must be counting the turn on transient.
Thanks for clarifying that issue. To avoid including startup, I use the same method as cmartinez; change the x-axis range to a steady state area and then repeat the control-left click. The upper end of the x-axis is limited to the range set for the simulation, to answer the question in #46.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
The upper end of the x-axis is limited to the range set for the simulation, to answer the question in #46.
Thanks for the info, Wayneh... I just checked and yes, in the pull-down "Simulate" -> "Edit Simulation Cmd" I found all configurable parameters for this purpose
 

ronv

Joined Nov 12, 2008
3,770
I can't seem to get it to work. I guess you could try it. If it gets thru the start up it will probably be ok.
One other note. The cap on the output (4.7 Ufd). should be ceramic.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
I can't seem to get it to work. I guess you could try it. If it gets thru the start up it will probably be ok.
One other note. The cap on the output (4.7 Ufd). should be ceramic.
I understand the advantages of ceramic caps (more thermal stability, among others), but I have an 4.7 µf electrolytic cap rated at 350 V .... wouldn't that be ok?
 

ronv

Joined Nov 12, 2008
3,770
It's the inductance and equivalent series resistance that hurts them. The ripple current is pretty high so with series resistance they heat up as well as not filter to well.
If you check and make the cap electrolytic (I don't think they have a 350 volt one, but you can get the idea) you will see voltage spikes on the output. Then measure the current in the cap and check it against the data sheet. Again, you can try it and see how hot it gets, but long term you want ceramic.
 

Thread Starter

cmartinez

Joined Jan 17, 2007
8,759
It's the inductance and equivalent series resistance that hurts them. The ripple current is pretty high so with series resistance they heat up as well as not filter to well.
If you check and make the cap electrolytic (I don't think they have a 350 volt one, but you can get the idea) you will see voltage spikes on the output. Then measure the current in the cap and check it against the data sheet. Again, you can try it and see how hot it gets, but long term you want ceramic.
Oh my God... you're right! the spikes are enormous with an electrolytic cap... 3 Amp spikes at the load, with 15V spikes at the output... Things go back to normal when I use an X7R or a tantalum type... I guess I'll just have to spend a couple of bucks at Digikey to get an X7R ... cause the Tantalum 4.7µf@100V are listed at almost $20 dlls!!!
One last question, why doesn't the inductor I chose have a current saturation value listed?
 

RichardO

Joined May 4, 2013
2,270
One last question, why doesn't the inductor I chose have a current saturation value listed?
I think that the spec sheet sort of does spec the saturation current -- the IDC (ma) max. spec.

Normally, the saturation current is specified as the current where the inductance is reduced by 10 or 20 %. If the maximum current in the spec _is_ the saturation current then they are sloppy not saying what the reduction in inductance is at that current.

Sometimes the maximum current rating for the inductor is at some maximum temperature. That could be what is specified for this inductor since they say:
"Temperature Rise ...............................................................................................................................................20 ˚C max. at rated current"

My guess is that they are specifying the maximum temperature with the IDC (ma) max. since the inductor seems a bit too small for a 1 amp saturation current.

Fortunately, the max temperature current and the saturation current are similar so you may be OK using your 3 100 uH coils in parallel to get 33 uH. The danger is that if the inductor saturates, the inductance drops dramatically and therefor the current goes up dramatically. This increase in current can easily destroy the MOS-FET. I would gradually increase the load current while monitoring the temperatures of the coil and MOS-FET. If either gets to hot to touch then the coil is saturating.


Edit: I forgot to mention that due to the high on resistance of the IRF740 you will need to parallel them. If you have them, you should parallel 4 or 5 of them. to get the on resistance down near 0.1 ohms or so.
 
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