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600 watt regulated DC power supply

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paulw271

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Here is a design for a 600 watt regulated DC power supply. The supply has 2 modes: 60volt 10 amp max, or 30 volt 20 amp max. Both current and voltage can be regulated, and the outputs are floating.

This page has more information about the design:

**broken link removed**

Here is the schematic:
**broken link removed**

-Paul
 

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Hi Paul,

You have claimed the supply can output 60V 10A but I have some doubts regarding your design.

If I have to design such a supply, I would have to use a single 66V or higher voltage winding or in your case two 33V windings. I don't see how you can obtain 600W output using two 24V secondary windings. The maths don't add up.

If the regulator could do its job properly, then at that output of 60V and 10A, there should be at least a few volts across the regulating elements for the control to "regulate". So if you can also let us know what is the voltage input to the high wattage pass elements during such loading.

Added: I have also found that your current measuring shunt is outside the voltage measurement loop. This would cause the output voltage to drop 10mV when 20A is flow and your voltage regulator cannot correct such error.
 
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That must get hot on the lowest voltage setting and highest current draw. Somehow I don't see how this can dissipate the power without meltdown. I can't see a fan or anything and would strongly advise adding one, plus some sort of thermal protection.

I don't like the look of that transformer, I wouldn't think it's suitable to be used without the output earthbonded because I doubt it meets the safety standards as far as double insulation is concerned.
**broken link removed**

Voltage ripple will also be a big problem, for 1V at 10A of ripple you'll need 0.1F smoothing capacitors. Those 40,000:mu:F capacitors will develop a 2.5V ripple (assuming their value is bang on, it'll be much worst if it's -20%), this will mean your regulator will need to have a dropout voltage of <1.4V, which it doesn't, >3V is probably more likely.
 
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No spec's what-so-ever. 50VDC at 10A with lots of ripple but remove the load and see how high the voltage goes.
Short the output with it set to 60V/10A and see how much smoke 600W makes.
 
Hi everyone, thanks for the feedback.

The transformer output spec is 24v RMS which should make almost 34v peak. I measured the output of the 40000uF caps, and they were 30.4v (with a very small load, and thus very little ripple). So yes, you might be correct that regulation will not work properly at 60v (or 30v depending on the mode). Maybe I can change the spec to 25v or 50v max (and add a series resistor between the voltage regulation pot and the 5v reference to limit the max voltage).

Concerning total power dissipation, each output transistor has its own heat sink measuring 5"x4"x1.5", and the chassis has a fan in the back. That said, I just completed the construction and have not done much testing to see if the airflow is sufficient. I have run it in the 30v mode with the outputs shorted and the current limit set to 20amps for 1 minute, and could still hold my fingers to the heatsinks, but they were still getting hotter (not a good sign). The design should work so long as the output transistors stay below 80 deg c. (Each MJ15024 is rated for 250wats, which derate at 1.43w/deg.c above 25deg.c, which is still 171 wats each)

I don't understand the comment about needing 0.1F caps for smoothing. The circuit has a feedback loop to control the pass transistor, which will regulate the output voltage. The voltage at the output of the 40000uF caps can ripple, that's ok.

I also don't understand why the transformer is unsafe. What does earthbounded mean, and what needs to be double insulated?

Thanks for pointing out the current measuring shunt being outside the voltage measurement loop. I can move the bottom of R6 to the -output to correct this.

Again, thanks for all the comments.

-Paul
 
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The transistor amplifier doesn't have much gain. It is important for a regulated power supply to actually regulate, therefore needs a lot of gain.
How much does the output voltage drop from no load to full load and what is the ripple voltage at its output at full load?
 
Paul,

You wrote "60volt 10 amp max, or 30 volt 20 amp max", means the supply is variable and will be used to deliver voltages anywhere between 0-30 or 0-60 V.
The worst case will be when you set the output volatge lower than 30 V (say 10 V) at 20 Amps. There will a lot of power dissipation in the regulating transistors. The effieciency of the power supply will be very low. You should use diffrent winding tapes (either with manual or automatic switching) for low voltage settings.

Why dont you go for some Switch Mode Power Supply?

Aily
 
I just ran a few more measurements:

In the 30v mode with a 2.5 ohm load, the maximum voltage the supply could properly regulate was 25.2 volts. Higher than that it ran out of headroom. Output had 100mV of noise, but no ripple. The mean voltage across the 40000uF caps was 26.8v, with 1.4 volt peak-peak ripple.

I tested the 60v mode with and without the 2.5 ohm load (0 amps and 8.25 amps at 20 volts) without adjusting the voltage pot in between, and saw a 0.2 volt change.

I did see 1 problem, however: with the 2.5ohm load (I didn't try with other loads) and the output voltage set to below 0.3 volts, the output is unstable. It develops a 2 volt peak spike that quickly decays down to about 0 before repeating. With the voltage setting above 0.3 volts, or with no load the output is stable. Any ideas?
EDITED: I found the problem. The ground connection of the voltage control pot was miswired. With this corrected, the output is stable.

As far as building this supply vs. a switching supply, I just wanted to learn about linear supplies. Maybe my next design will be a switcher.

-Paul
 
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paulw271 said:
I don't understand the comment about needing 0.1F caps for smoothing. The circuit has a feedback loop to control the pass transistor, which will regulate the output voltage. The voltage at the output of the 40000uF caps can ripple, that's ok.
The problem is the ripple needs to be above the regulator's minimum dropout voltage.

[latex]C = \frac{I_{LOAD}}{2FV_{RIPPLE}}[/latex]

So what's the dropout voltage of your regulator?

Say the transistor has a saturation voltage of 1V, and Vbe is 1V plus 0.4V for the ballancing resistors, this will give 2.4V. The diodes will drop 1.2V each at full load so that's 2.4V as theres two diodes in series. This will give a total dropout voltage of 4.8V and this is if you're lucky!

Now let's work out the minimum capacitor value.

I was wrong earlier on because no capacitor will be big enough, the peak voltage is 34V so at full load the peak output from the regulator will be 29.2V.
 
illitirate.is.me said:
is there anyway to mod this to output 12-13.5v??
Of course, its output voltage is adjustable. Adjust it to 12.0V or 13.5V or anywhere in between. Then calculate the voltage for a suitable transformer so its transistors don't get too hot.
 
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