I am experiancing a strange problem with the soft start circuit in this classe power amp i am working on when used with a variac.
The soft start is a 10 ohm 25 watt resistor in series with a 800+va toroid that gets bypassed 1 second after the power is switched on.
if i leave the soft start resistor inline and bring up the power very slowley with the variac, the resistor gets very hot. hot enough start smoking the enamel. and the variac shows about 3 amps of current draw at 30-40 volts ac.
Of course i am looking for shorts and there arent any. if i bypass the resistor and bring the amp up it draws little current and works just fine. if i just flip the switch on with the variac up and the soft start in it works just fine.
Now the only thing i can think of is that i drain the PSU caps after every run just to make sure. i use a 100 watt light bulb and drain all caps.
So the caps are at or near zero at first power. so i brought the variac up slowley watching the meter thinking that it would slowly start to drop as the caps charged. thats when the resistor let me know it was a little too hot.
Could it be that the caps just arent charging fast enough with that resistor inline and are appearing as shorts? and that is what causes that large current draw?
Let me assure you there is nothing shorted. i have checked it a dozen times and as i said, bypassing the resistor and bringing the amp up on the variac does not cause this condition. the meter is not calibrated in small steps and the first mark is 1 amp, and it stays below 1 amp the whole way up as such.
I wouldnt think a 10 ohm resistor would cause enough of a current drop to cause the capacitors to not keep up with the rising voltage of the transformer. strange i need to figure out the math on that.
The problem with this design is that if the power line sags enough, it will kick the soft start relay out and start the timer over, during that time if the amp is under heavy enough load to cause the power line sag, it is going to smoke that resistor again.
and now that i have increased the size of the power supply capacitors. i wonder if i should change that resistor to something larger and maybe change it from a brown devil ceramic type to an aluminum house dale and mount it on the chassis to help dissapate the heat.
strange problem.
Zc
The soft start is a 10 ohm 25 watt resistor in series with a 800+va toroid that gets bypassed 1 second after the power is switched on.
if i leave the soft start resistor inline and bring up the power very slowley with the variac, the resistor gets very hot. hot enough start smoking the enamel. and the variac shows about 3 amps of current draw at 30-40 volts ac.
Of course i am looking for shorts and there arent any. if i bypass the resistor and bring the amp up it draws little current and works just fine. if i just flip the switch on with the variac up and the soft start in it works just fine.
Now the only thing i can think of is that i drain the PSU caps after every run just to make sure. i use a 100 watt light bulb and drain all caps.
So the caps are at or near zero at first power. so i brought the variac up slowley watching the meter thinking that it would slowly start to drop as the caps charged. thats when the resistor let me know it was a little too hot.
Could it be that the caps just arent charging fast enough with that resistor inline and are appearing as shorts? and that is what causes that large current draw?
Let me assure you there is nothing shorted. i have checked it a dozen times and as i said, bypassing the resistor and bringing the amp up on the variac does not cause this condition. the meter is not calibrated in small steps and the first mark is 1 amp, and it stays below 1 amp the whole way up as such.
I wouldnt think a 10 ohm resistor would cause enough of a current drop to cause the capacitors to not keep up with the rising voltage of the transformer. strange i need to figure out the math on that.
The problem with this design is that if the power line sags enough, it will kick the soft start relay out and start the timer over, during that time if the amp is under heavy enough load to cause the power line sag, it is going to smoke that resistor again.
and now that i have increased the size of the power supply capacitors. i wonder if i should change that resistor to something larger and maybe change it from a brown devil ceramic type to an aluminum house dale and mount it on the chassis to help dissapate the heat.
strange problem.
Zc
Zero Cool,
Be very careful with this. I have an Adcom power amp and the service manual warns specifically against using a variac to apply power, because this will damage a power resistor in the soft start circuit.
This sounds verys similar to your issue! - you may not have a problem at all, it is just the way the soft start circuit is configured - the power resistor is not switched out until voltages reach a certain level as the power supply is energised. Using a variac leaves the resistor in circuit far longer than expected in nomal operation.
I hope this can be of use.
Cheers,
Ed
Be very careful with this. I have an Adcom power amp and the service manual warns specifically against using a variac to apply power, because this will damage a power resistor in the soft start circuit.
This sounds verys similar to your issue! - you may not have a problem at all, it is just the way the soft start circuit is configured - the power resistor is not switched out until voltages reach a certain level as the power supply is energised. Using a variac leaves the resistor in circuit far longer than expected in nomal operation.
I hope this can be of use.
Cheers,
Ed
Hi,
if the relay was timed to bypass the soft start resistor then the problem would disappear.
I suspect that Adcom have used voltage detection to pull the relay in and slow build up keeps the relay out and pulls continuous current through the resistor.
Is there a pair of extra contacts in the relay that discharge the caps quickly (through a resistor) on power off? These would draw extra current.
if the relay was timed to bypass the soft start resistor then the problem would disappear.
I suspect that Adcom have used voltage detection to pull the relay in and slow build up keeps the relay out and pulls continuous current through the resistor.
Is there a pair of extra contacts in the relay that discharge the caps quickly (through a resistor) on power off? These would draw extra current.
Just a thought. Reading your original post, you say that with the variac at 30-40v the current is about 3A. Well that is about 10 ohms, (dissipating about 90 to 100 W). The problem resistor is a 10 Ohm value, rated at 25 W
It would appear that somehow the resistor is drawing this current, but how is it configured in the circuit?
The fact that everything operates OK when power is applied normally suggestes that there is no faulty part at work here, but I understand your concern about the possiblity of the soft start circuit operating on a power line "dip" One way is (as you suggest) to calculate the max dispersion you expect (find the voltage at which this fault would appear) and fit an appropriately rated resistor. There are also solid state "soft start" circuits nowadays - used in power tools and the like. I wonder if these might provide an alternative solution?
Ed
It would appear that somehow the resistor is drawing this current, but how is it configured in the circuit?
The fact that everything operates OK when power is applied normally suggestes that there is no faulty part at work here, but I understand your concern about the possiblity of the soft start circuit operating on a power line "dip" One way is (as you suggest) to calculate the max dispersion you expect (find the voltage at which this fault would appear) and fit an appropriately rated resistor. There are also solid state "soft start" circuits nowadays - used in power tools and the like. I wonder if these might provide an alternative solution?
Ed
Hi Zero Cool,
That is a very common thing with soft start resistors. Every manufacturer that uses one indicates in the service manual to short the resistor out when using a variac (at least they used to do that). Marantz, Nakamichi, Denon, Yamaha, Adcom. All the same.
So. Don't worry. Short the resistor out and bring it up on the variac when you need to do that. Don't forget to remove the short when you are done. I use bright yellow clip leads (hard to miss).
-Chris
That is a very common thing with soft start resistors. Every manufacturer that uses one indicates in the service manual to short the resistor out when using a variac (at least they used to do that). Marantz, Nakamichi, Denon, Yamaha, Adcom. All the same.
So. Don't worry. Short the resistor out and bring it up on the variac when you need to do that. Don't forget to remove the short when you are done. I use bright yellow clip leads (hard to miss).
-Chris
Since I built a soft-start circuit from scratch and I recently posted the operation here:
http://www.diyaudio.com/forums/showthread.php?postid=1096285#post1096285
It's not a simple delay circuit. My time constant is determined by when the rails reach a given voltage, not by time.
The caps are shorts when they turn on. The voltage across a capacitor can't vary instantaneously.
Larger caps will require:
Adjusting the resistor - to reduce the current spike
Adjusting the time - so that the caps can nearly charge.
The rule is full charge is reached in 5 time time constants. R*C. C would be the addition of the values of the filter caps. R is the AC resistor. The time should probably be about 5 * RC. You can work backwards and compute n in the n * RC = 1 sec for the original circuit and then figure out what R can satisfy this equation. You may have to adjust the resistor power rating or the timer duration. Use the temperature of the resistor as a gauge.
If you let the resistor charge too long it will blow.
The circuit needs to:
Keep the input and output off while this occurs.
Reset the timer when the amp is fully up.
http://www.diyaudio.com/forums/showthread.php?postid=1096285#post1096285
It's not a simple delay circuit. My time constant is determined by when the rails reach a given voltage, not by time.
The caps are shorts when they turn on. The voltage across a capacitor can't vary instantaneously.
Larger caps will require:
Adjusting the resistor - to reduce the current spike
Adjusting the time - so that the caps can nearly charge.
The rule is full charge is reached in 5 time time constants. R*C. C would be the addition of the values of the filter caps. R is the AC resistor. The time should probably be about 5 * RC. You can work backwards and compute n in the n * RC = 1 sec for the original circuit and then figure out what R can satisfy this equation. You may have to adjust the resistor power rating or the timer duration. Use the temperature of the resistor as a gauge.
If you let the resistor charge too long it will blow.
The circuit needs to:
Keep the input and output off while this occurs.
Reset the timer when the amp is fully up.
If I remember right, I used something <50 ohms and <3W metal oxide resistor (40,000 uf capacitance, 50V rails).
If the inputs are open and the speakers are open the resistor should be able to hang on longer. Once there is a signal and a load, the resistor pops easily. I = 120/10 limits the current to 12 AMPS for 1 second so it seems that it's just preventing a spike so the lights don't dim. If you increased C, then you probably need to increase the delay time.
If the inputs are open and the speakers are open the resistor should be able to hang on longer. Once there is a signal and a load, the resistor pops easily. I = 120/10 limits the current to 12 AMPS for 1 second so it seems that it's just preventing a spike so the lights don't dim. If you increased C, then you probably need to increase the delay time.
anatech said:
Thanks, unfortunatly i don't have a service manual only service notes that the factory assembly people used during final testing.
In those notes they do talk about bring up the voltage with a variac during various test's but no mention of bypassing the softstart resistor. But now i know!
But now i have a fun project. I have some nice 250 watt resistors mounted on a very large heatsink. I am going to set them up to total 10 ohms and do some test's to see how much current the caps draw as such. and i can do some power drop test's to see what line voltage the softstart circuit will kick out and try and re time itself with out fear of smoking the 25 watter. if nothing else my shop could use the extra heat
This is an interesting problem actually. for another project i have quite a few caps in the power supply that will need to be charged up on power start. how to charge them without blowing a circuit breaker after flipping the switch should be interesting. i need to review my plans.
Zc
Hi Zero Cool,
Well, if you have a large capacitor bank you can stage the controlled current charge. I would normally have the relay across the secondary DC and have it pull in as the voltage comes up. A 48 VDC coil plus resistor will accomplish this in most amps. If your DC drops the relay will drop out, frying the resistor. You want this as it may reduce the damage. Resistors are cheap.
Keep in mind that the longer the resistor stays in circuit, the hotter it will get. So figure on a 1/2 sec charge time an size the resistor for that. I commonly see values from 2R2 to 10 R in 120 VAC circuits.
-Chris
Well, if you have a large capacitor bank you can stage the controlled current charge. I would normally have the relay across the secondary DC and have it pull in as the voltage comes up. A 48 VDC coil plus resistor will accomplish this in most amps. If your DC drops the relay will drop out, frying the resistor. You want this as it may reduce the damage. Resistors are cheap.
Keep in mind that the longer the resistor stays in circuit, the hotter it will get. So figure on a 1/2 sec charge time an size the resistor for that. I commonly see values from 2R2 to 10 R in 120 VAC circuits.
-Chris
Here is the schematic for my soft start.
Power from the ac line is routed through the 10 ohm 25 watt resistor, and then (not show) to the power switch and to the transformer.
An Auxilliary winding on the Transformer is rectified and regulated for various ciruits in the amp, this supply also powers a 555 timer. The timer is set for 1 second, this then turns on a relay in the S.S. circuit and bypasses the resistor.
Power from the ac line is routed through the 10 ohm 25 watt resistor, and then (not show) to the power switch and to the transformer.
An Auxilliary winding on the Transformer is rectified and regulated for various ciruits in the amp, this supply also powers a 555 timer. The timer is set for 1 second, this then turns on a relay in the S.S. circuit and bypasses the resistor.
An externally hosted image should be here but it was not working when we last tested it.
An externally hosted image should be here but it was not working when we last tested it.
Two things to note. One i do not know if the relay will drop out under a power line sag. this i need to do some testing for, but if it does. Then i want to make some changes to prevent that from burning up that resistor.
Two, if there should be an output stage failure the only fuse in the amp is the AC mains fuse. which is an 8 amp fast blow. Lets say this does in fact blow and someone replaces the fuse and turns the amp on again, not knowing there is a major problem in the amp, then the S.S. resistor will burn up instantly due to the shorted output stage and before the mains fuse blows again.
This is exactly what happend to this amp. the Previous owner shorted out the speaker cables and blew up the amp. then tried a new fuse and burned up the soft start circuit.
I wonder if i could put a fuse in line with the resistor in the soft start circuit to prevent that resistor from burning up under these circumstances? I wonder if maybe a 3-4 amp slow blow, or maybe a fast blow fuse would work. it would have to be large enough to absorb the turn on spike yet fast enough to blow before the resistor overheats.
Hmm, or maybe....I have some airpax thermal breakers, maybe i could mount a thermal breaker in contact with the resistor and should it get too hot, the thermal breaker will open and shut off power to the amp. hmmm i like that idea better.
Zc
Two, if there should be an output stage failure the only fuse in the amp is the AC mains fuse. which is an 8 amp fast blow. Lets say this does in fact blow and someone replaces the fuse and turns the amp on again, not knowing there is a major problem in the amp, then the S.S. resistor will burn up instantly due to the shorted output stage and before the mains fuse blows again.
This is exactly what happend to this amp. the Previous owner shorted out the speaker cables and blew up the amp. then tried a new fuse and burned up the soft start circuit.
I wonder if i could put a fuse in line with the resistor in the soft start circuit to prevent that resistor from burning up under these circumstances? I wonder if maybe a 3-4 amp slow blow, or maybe a fast blow fuse would work. it would have to be large enough to absorb the turn on spike yet fast enough to blow before the resistor overheats.
Hmm, or maybe....I have some airpax thermal breakers, maybe i could mount a thermal breaker in contact with the resistor and should it get too hot, the thermal breaker will open and shut off power to the amp. hmmm i like that idea better.
Zc
Upupa Epops said:
The auxiliary winding is on the mains transformer. with the power switch off, there is no power to the mains transformer so there is no power to the 555. when the power switch is turned on, the auxilary winding powers the 555. 1 second later the 555 turns on the transistor/relay bypassing the resistor.
What i dont know, but am afraid of, is that if a large enough power sag, OR a power glitch happens, long enough to reset the 555 that the resistor will burn up again.
Zc
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