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Old 11th October 2005, 08:39 AM   #1
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Default Setting the relay time on softstart

Hi,

The softstart allows 5A current until the relay clicks in.

I am running +-56VDC rails with 180,000uF capacitor bank on one amplifier and +-80VDC rails with 120,000uF capacitor bank on another amplifier. The transformers are probably rated at 625VAC.

How can I calculate how much time is necessary to fill the capacitors up?

Should I set the relay to 2 seconds? 3 seconds? 5 seconds?

Regards,
Bill
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Old 11th October 2005, 08:44 AM   #2
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I'll guess your diodes are >> 10 A which means they can cope with 100-200 A short term. Therefore you'll need to limit the transformers inrush current only and this will take a rather short time. 1 second is enough in order not to blow your fuses in the wall.
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Old 11th October 2005, 11:32 PM   #3
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Thanks Peranders.

My worry is not about the fuses or the diodes. It is the power switch. Without the softstart, the amp has been running for over a year, using the wall socket switch. The house lights are dimmed when it is powered on, and sometimes sparks came out from the wall socket. A number of high power rated switches installed on the amp all have been destroyed. I hope the softstart will protect the switch and provide stability for many years to come.

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Bill
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Old 12th October 2005, 08:09 PM   #4
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Default soft start

If you get sparks from the wall socket you have a fire hazard that should be corrected ASAP ! !
This can cause a fire ! !
Get a better wall socket or have an electrician install one, your house will thank you.
Ed
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Old 13th October 2005, 12:29 AM   #5
tlf9999 is offline tlf9999  United States
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if sparks are all you need to get rid off, why cannot you just parallel a small capacitor (like 0.01u 600v) to the switch? it will take care of your sparks quite well.
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Old 13th October 2005, 03:00 AM   #6
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The inrush current is due to caps not fully charged so can anyone tell me how long it will take to fully charge the caps?


tlf9999,

My amps will have the X2 cap on the main so I guess that serves the function of your small cap.

Regards,
Bill
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Old 13th October 2005, 05:06 AM   #7
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The inrush current consists of two components, magnetizing the transformer and charging the caps.

The volatge increasing speed is:

I/C = V/s

So the answer is one or two mains periods => 17-34 ms if you exclude the softstart.

May I ask why you have so huge smoothing battery? Do you have a low power supply rejection ratio of the amp? "Much" isn't better always. The drawback is more harmonics and more losses in the transformer. Most likely your transformer will be able to deliver half of the VA rating due to the extra losses.
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Old 17th October 2005, 02:56 AM   #8
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Peranders,

That is because

(1). The caps are pulled from an old amp that sounded very good so in a way it costs me nothing to put the extra uF in; If I don't use them they will be wastes.

(2) If it sounded very good with 180,000uF in the old amp I presume it is OK with the new amp with the same transformer; and

(3) The amp is to drive an averaged 3 ohm load (minimal 2.5 ohm) and I play organ music.

Question:

I am aware of possible higher harmonics but not transformer losses due to higher filter capacitance. I will be very interested in your explanations or pointing me to the relevant articles on the web.

Regards,
Bill
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Old 17th October 2005, 07:33 AM   #9
AndrewT is offline AndrewT  Scotland
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Hi,
second time today I have said use http://www.duncanamps.com/psud2/index.html
You can set the time period to allow you to examine in detail how the currents and voltages vary during the charging and subsequently.
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Old 17th October 2005, 07:46 AM   #10
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Linky no worky Andrew.
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