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Old 5th November 2009, 05:47 PM   #1
a.wayne is offline a.wayne  United States
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Default Soft start question

Hello ,

I'm about to add more caps to the power supply and my question is , how does one know if the current soft start will handle the extra capacitors . I'm adding 140K to the already 70 K there .

Regards,
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Old 5th November 2009, 09:03 PM   #2
a.wayne is offline a.wayne  United States
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Huh ! ........... anyone ?

Ok i will give it a try hopefully the smoke will stay inside .....
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Old 5th November 2009, 09:40 PM   #3
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Hi

If you are using a resistor for current limiting, I would upgrade it to 30-50W, metal clad types. That's what I use for 1/2 sec delay before the relay is activated.

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Old 5th November 2009, 10:21 PM   #4
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don't do it, not without doing some checking of specs first.

1) basic capacitance voltage/current relationship is I=C*dv/dt. Simplistically, you're going from 1x to 3x the capacitance, so the peak current through all the soft-start circuit and transformer will triple. Assuming that your soft start circuit is a negative temp coefficient thermistor bypassed by a relay, you need to check the spec on the thermistor to see if it can handle the additional current. You also need to check the transformer spec. Chances are you have maybe a 30 - 50% margin, it may work the first time and the first N times but at the very least it would be a long term reliability problem.

2) The second problem is that the voltage rails will take longer to come up to their steady state value, so the delay for engaging the relay would have to be lengthened. Ideally the relay doesn't switch until the caps are at full charge so the current is very low. Also any other delays like output relays would have to be changed.

3) Once you have a sufficient amount of capacitance to handle the expected load, adding more isn't going to do you any good. I have been exploring this area lately, in my simulation test bench with 50V rails and 20,000uF, full power sine waves will make the rail droop about 10V. Real music, with lots of heavy bass will make it droop 2-3V. More capacitance won't change the end result, only the slope of the droop. It is much more effective to raise the supply voltage than throw more capacitance at it. Cheaper too.
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Old 6th November 2009, 12:21 AM   #5
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Can your rectifiers handle the increased peak currents? Try modeling your power supply in PSU designer - you might be shocked at the peak current you get with that much capacitance.
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Old 6th November 2009, 02:32 AM   #6
a.wayne is offline a.wayne  United States
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Thanks for the response guys,

Here is more info on what it has .

1. The A/C input board has a power delay relay , there are jumpers to set the voltage supply range and an amp bias delay circuit.

2. The turn on delay relay is rated @ 125v/30A/24vdc

3. PS rectifier 400v/35amp

4. Trans is rated @`1,800VA max /ch. ( mono bloc)

5. Main resistor 4.7ohm/ 20W ( there is a mention not to use a variac to slowly bring up the voltage without first shunting this resistor , failure to do so will burn it out )


Based on what Michael suggested , it seems i should update the 20 W resistor to a 50 W.. I would also think the 30 amp rectifier /ch would be enuff .. if not suggestions ? as to lengthening the delay , how would i approach that ?


Bob where would i obtain that software ? PSU designer .............

Last edited by a.wayne; 6th November 2009 at 02:43 AM.
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Old 6th November 2009, 04:57 AM   #7
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what is the transformer output voltage?

The resistor and relay are on the AC line side of the transformer?
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Old 6th November 2009, 05:31 AM   #8
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i would use a seperate soft start for the extra caps
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Old 6th November 2009, 05:58 AM   #9
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Here's a really simple simulation result, showing a basic power supply. The transformer is 120VAC input, 35VAC output (50V p-p), a 4.7ohm resistor shunted by a relay on the line side, 4 diode bridge and 35000uF capacitor.

The relay is modeled with a spice switch primitive and turns on at t=2s. At t=7s a 30Hz sine wave, peak 8 amps into 4 ohms, loads the rail. Current through the 4.7 ohm resistor (R1) and the rail voltage is shown.

You see how the rail voltage quickly droops from about 48.7V to about 40V when the load is applied. More capacitance will make the decay take a little longer is all.
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Old 6th November 2009, 06:17 AM   #10
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Quote:
Originally Posted by a.wayne View Post
Hello ,

I'm about to add more caps to the power supply and my question is , how does one know if the current soft start will handle the extra capacitors . I'm adding 140K to the already 70 K there .

Regards,
Just curious why? Do you have a design which is sensitive against ripple or is it a powerful amp? Are you aware of that you will create lot's of unwanted harmonics on the mains?
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