Thanks for pointing that out. Not that I'm particularly keen to substitute the CL-60, but would I not need a higher energy rating given the higher cap bank voltage?
From post #22...
From Ametherm's website, I found a close equivalent of the CL-60 but with a higher energy rating of 75J - SL18 10005 (datasheet).
Other specs are: 10 ohm resistance, 5A max steady state current (no minimum listed), and 180C temp. at max current.
As (my usual) luck would have it, it is out of stock everywhere.
I only know that the SE Class A with 2-750 VA Transformers blows each and every time the Main House Fuse, without inrush limiter..(PTC) The Amp usually draws 750Watts / Channel @idle.
In the meantime I have made a LOW CLASS A MODE SELECTOR where the CLASS A runs on 167Watts instead of 750/Channel. But it could be Switched to high mode any time.
Config of the Circuit is parallel to the PTC, a 10 WATT 10 Ohm Resistor, both which will be coupled out when the amp has received it's full voltage, by a Relay. I use two PTC Resistors each of them when Cold =5 ohms & 10 Ampere. The Relay is fed by 12Volts which is regulated after the MAIN RAILS +- which are feeding the Amp itself.
These PTC are in the AC 240 Volts Line. I think to cut off the PTC with a Relay, when the Amp is switched on makes sense because otherwise the PTC getting really hot. In the same time that the PTC is set to off the LINE Of LINE VOLTAGE have =0R or not more than the RELAY contacts itself, so full Voltage (240 Volts) is applied to the Transformers.
And there is no need to have the PTC's hot when not in use.
In the meantime I have made a LOW CLASS A MODE SELECTOR where the CLASS A runs on 167Watts instead of 750/Channel. But it could be Switched to high mode any time.
Config of the Circuit is parallel to the PTC, a 10 WATT 10 Ohm Resistor, both which will be coupled out when the amp has received it's full voltage, by a Relay. I use two PTC Resistors each of them when Cold =5 ohms & 10 Ampere. The Relay is fed by 12Volts which is regulated after the MAIN RAILS +- which are feeding the Amp itself.
These PTC are in the AC 240 Volts Line. I think to cut off the PTC with a Relay, when the Amp is switched on makes sense because otherwise the PTC getting really hot. In the same time that the PTC is set to off the LINE Of LINE VOLTAGE have =0R or not more than the RELAY contacts itself, so full Voltage (240 Volts) is applied to the Transformers.
And there is no need to have the PTC's hot when not in use.
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Check www.distrelec.ch - but distrelec is acutally from Netherlands.. meaning the supplier, or check Security Check
The ones I use have a common spec to these you want to buy.. I got them very inexpensive from distrelec.ch
Sorry that's not correct, !
It's 750+ Watts for both channels..
As in my living room the 240 Volts Line voltage variate, it's not always 750Watts, but most times it's above that, and there are days when it's a little bit less.. but for BOTH Channels, sorry for the misinformation respectively wrong information.
@Ground_Point_9, I recommend using the MS22-50004 which is in stock at Newark, Mouser, and DigiKey: octopart link
It's a brutally large device with 230 Joules of inrush energy rating. And they provide lots of good information on the datasheet.
Final steady state "hot" resistance, when running at 2 amps, is 0.79 ohms. However your transformer's final steady state primary current is less than or equal to (300VA / 230V) = 1.5 amps. I'd guess the "hot" resistance would be around 1.6 ohms when operating at 1.5 amperes. So you'll drop 2.4 volts across the ICL. Since 2.4 volts is only 1% of the 230V mains, it seems like an acceptably small voltage loss, at least to me.
Initial "cold" resistance, at 25C before inrush occurs, is 50 ohms. So if your mains inlet + fuse + switch + internal wiring + transformer primary were made of ideal perfect superconductors, with zero point zero milliohms of resistance, peak inrush current will be (230V / (50 + 0.0)) = 4.6 amperes. In the real world, with copper wires, peak current will be less than 4.6 amps.
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It's a brutally large device with 230 Joules of inrush energy rating. And they provide lots of good information on the datasheet.
Final steady state "hot" resistance, when running at 2 amps, is 0.79 ohms. However your transformer's final steady state primary current is less than or equal to (300VA / 230V) = 1.5 amps. I'd guess the "hot" resistance would be around 1.6 ohms when operating at 1.5 amperes. So you'll drop 2.4 volts across the ICL. Since 2.4 volts is only 1% of the 230V mains, it seems like an acceptably small voltage loss, at least to me.
Initial "cold" resistance, at 25C before inrush occurs, is 50 ohms. So if your mains inlet + fuse + switch + internal wiring + transformer primary were made of ideal perfect superconductors, with zero point zero milliohms of resistance, peak inrush current will be (230V / (50 + 0.0)) = 4.6 amperes. In the real world, with copper wires, peak current will be less than 4.6 amps.
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For the recommendation, and the helpful explanation of how to calculate for, and pick the part, many thanks Mark.
Sorry Guys, today it's not my day..
this should read NTC - was in a hurry.. This would be proper
Also I have two versions, on with 24Volts on the regulator for the Relay and one for 12 Volts.
The 24 volt works better.. With the 12 Version I had to change the relay several times already. But I think it's a problem for the Quality of the relay..
In the pic you can see the actual one.. the Parallel resistor is mounted near the Fuse Socket, not on this board.
Attachments
I was always calculating the hot resistance & current based on actual amplifier current consumption:
My Aleph J draws a 4A constant current at 2 X 24.5V (my rails are sitting between 24 - 25V loaded). So we arrive at 196W.
Ignoring the transformer losses, this real-life consumption translates to 196W / 240V = "only" 0.82A. This is the "hot" current that runs through the thermistor, which is perfect for CL-60 rating, ensuring very low hot resistance.
I'd assume that F4 runs at lesser Iq, which means even lesser hot current than 0.8A...
Assuming that F4 draws 3A (1.5A per AMP PCB), 24V DC rails when loaded, and 220V AC, we get:
144W of total power draw.
144 / 220 = only 0.65A of hot current, which is less than half of what you calculated by taking the nominal VA transformer rating, and not the actual amp power draw.
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This thread got me to do some calculations & I thought I'd share.
FW clone of F1J, 18,000uF x4 PSU caps per channel, 24V rail = 20.7 joules of inrush energy per mains thermistor (two parallel 115V primaries, each with a CL-60).
FW clone of F3, 15,000uF x4 PSU caps per channel, 46V rail = 63.5 joules of inrush energy per mains thermistor.
Aleph 30 clone, 22000uF x4 PSU caps per channel, 41.4V rail (+/- 20.7V) = 75.5 joules of inrush energy per mains thermistor.
I've been using CL-60's in all of these applications (max 36 joules energy rated). Something to think about.
FW clone of F1J, 18,000uF x4 PSU caps per channel, 24V rail = 20.7 joules of inrush energy per mains thermistor (two parallel 115V primaries, each with a CL-60).
FW clone of F3, 15,000uF x4 PSU caps per channel, 46V rail = 63.5 joules of inrush energy per mains thermistor.
Aleph 30 clone, 22000uF x4 PSU caps per channel, 41.4V rail (+/- 20.7V) = 75.5 joules of inrush energy per mains thermistor.
I've been using CL-60's in all of these applications (max 36 joules energy rated). Something to think about.