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Old 3rd February 2011, 04:27 PM   #1
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Default SOFT START FOR POWER AMPS

CAN ANYBODY RECOGNIZE THIS SOFT START. IS IT NEED EXTRA POWER FOR WORK?
THERRE ARE 18VAC AND 24V DC .IS IT NEED EXTRA POWER FOR WORK.
CAN ANYBODY EXPLAIN THIS SOFT START.
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Old 3rd February 2011, 04:41 PM   #2
AndrewT is offline AndrewT  Scotland
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What's with the capitals? You do this in many of your posts !
I prefer to use low value resistors in series rather then high value resistors in parallel.
1.) it ensures that each resistor sees a maximum voltage that is a fraction of the mains voltage, whereas the parallel see the full mains voltage ~ 360Vpk + spiking transients.
2.) the wire inside a low value resistor will be considerably shorter and thicker than the wire in an equal power higher value resistor. That thicker wire will be more tolerant of overload current.

Good separation of Mains from LV side.
Don't like the RC timer. Time varies with mains voltage.

Don't like the lack of current going to the transistor switch. The relay may not pull in and the transistor may overheat.

Whoever designed this forgot about the practicalities of transistor switching and reliability.
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Old 3rd February 2011, 04:46 PM   #3
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Yes, you need to supply 24V DC or 18V AC power for this circuit to work. Either one is OK but you need these to power the 24V relay. Also, the GND connections at the bottom right corner must be connected to a ground point for the relay to trigger.

The incoming AC power is connected to the Active and Neutral connections on the end of the board. Your power amp is connected to the "Slow Active" and Neutral connections on the side of the board. Notice that there are several resistors between the Active and "Slow Active" connections. When the AC power is started to the board, the resistors restrict the current flow. After a very short time (for instance 100milli seconds or less) the relay should cause the connection between the Active and "Slow Active" areas to be connected with a short, which bypasses the resistors.

Andrew has already brought up a few potential problems with the circuit so I will defer to him. The following are some general comments on the board:

Since the connection for the relay is made only via the traces on the circuit board, they must carry all the current for the amplifier. You should make sure that the trace width and thickness is sufficiently large to not overheat, etc. Also, make sure that the relay is rated for your AC voltage and current demand of your amplifier. You might also want to connect the same wire to all three terminals of the Active, Neutral, and "Slow Active" terminal blocks to share the current flowing through each one.

-Charlie
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Old 3rd February 2011, 05:14 PM   #4
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Quote:
Originally Posted by CharlieLaub View Post
Also, the GND connections at the bottom right corner must be connected to a ground point for the relay to trigger.
That is what the transistor does - closes the circuit to ground after a short delay.

Could be better. I don't like resistors - thermistors are much better and safer. I don't agree with Andrew on the RC timer - it is not greatly affected by supply fluctuation.
If a thermistor is used, a much longer time period can be used before the relay closes.
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Old 3rd February 2011, 05:26 PM   #5
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Quote:
Originally Posted by MJL21193 View Post
That is what the transistor does - closes the circuit to ground after a short delay.

Could be better. I don't like resistors - thermistors are much better and safer. I don't agree with Andrew on the RC timer - it is not greatly affected by supply fluctuation.
If a thermistor is used, a much longer time period can be used before the relay closes.
There's nothing wrong at all with using resistors in this way. The magnetic field in the transformer will begin to stabilize after a few tens of milliseconds and by 200mS the regulator caps have partially charged and current demand will drop to normal levels. So you only need to limit current inflow (via the power resistors) for a fraction of a second.

Rod Elliot has some good advice regarding/against thermistors at these links:
Soft-Start Circuit For Power Amps
Inrush Current

-Charlie
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Old 3rd February 2011, 05:31 PM   #6
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The relay that would fit that footprint would be a typical OMRON type G2R-1 24V with an operating current of 21 mA, This is a Darlington pair with gain >750, base current of 1 mA would be more than sufficient to drive the relay which has a coil resistance of 1 K Ohm. The time constant would be the time required to charger the 220uF capacitor to about 1.5V. The 1.5 K Ohm resistor is used to discharge the capacitor quickly should the circuit be turned off then on to ensure that the switch on delay is ready to be reactivated.

This circuit will work fine for your application.
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Old 3rd February 2011, 05:33 PM   #7
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Quote:
Originally Posted by CharlieLaub View Post
There's nothing wrong at all with using resistors in this way.

Rod Elliot has some good advice regarding/against thermistors at these links:
Soft-Start Circuit For Power Amps
Inrush Current

-Charlie

Rod Elliott is talking about using thermistors on their own - that is not what I'm saying here. Replace the resistors in the circuit with the correct size thermistor.
Resistors can burst into flames if the relay doesn't close, a thermistor will not.
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Old 3rd February 2011, 05:36 PM   #8
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John. I would to agree with you in using thermistors, we use them extensively in soft starting battery chargers and DC to AC inverters. For the exact reason you mention
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Old 3rd February 2011, 05:37 PM   #9
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Originally Posted by Nico Ras View Post
John. I would to agree with you in using thermistors, we use them extensively in soft starting battery chargers and DC to AC inverters.
They are the ticket and really not much more expensive than the resistors. I learned my lesson on that
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Old 3rd February 2011, 05:38 PM   #10
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Furthermore, the thermistor preserves the relay contact almost indefinitely
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