I would aim for 2/3 to 3/4 charge to be reached as then the charging current would be a lot lower. This should only require a few seconds. As you say a class-a amp might present problems to trying to charge the caps, I had not considered this because I have no interest in class-a monsters 😉
Now I am confuse. Does somebody use large capacitpors wihtouth any protection from inrush current? My transformer is not toroidal if it makes some changes. If there is no solution for my problem I'm gona by the new transformer with 16VAC voltage to be sure that the voltage fall in the CRC will be compensated ;o(((
Neither do I have any interest in such monsters 😀
However, class AB/G/H monsters for PA are quite fun and I've examined a lot of these. Almost always there is only a primary side soft start. Some use resistors that are bypassed, while some others use thermistors with bypass. I've seen thermistors without bypass too, but that's uncommon. Also, I think it was mostly cheap brands that did this.
I have seen secondary side soft start once however. This was in an amplifier made by D.A.S. Audio which had an EI core transformer so there was no need for primary side soft start. I've never seen it used on both sides of the transformer. Have you?
Cutting out the resistor when at 3/4 charge on the capacitors sounds like it should work well even with a primary side soft start. There is a lot of difference between a little bit of saturation towards the end of a half cycle and being in full, deep, saturation during a whole half cycle which is essentially what happens when power is applied at the zero crossing of mains voltage and the transformer has full remanence in the same direction.
I've got some 1kVA transformers here and they have 1 ohm of primary resistance. That's a half-sine pulse of 300A peak, or 450A²s, if (when) this happens if the electric utility is stiff. For reference, from what I can find, 10A gG Diazed fuses as usually used in swedish fuse boxes blow at about 100-200A²s.
( A fuse datasheet: http://www.ifoelectric.com/Ifo_D-sak_TD_2006.pdf )
However, class AB/G/H monsters for PA are quite fun and I've examined a lot of these. Almost always there is only a primary side soft start. Some use resistors that are bypassed, while some others use thermistors with bypass. I've seen thermistors without bypass too, but that's uncommon. Also, I think it was mostly cheap brands that did this.
I have seen secondary side soft start once however. This was in an amplifier made by D.A.S. Audio which had an EI core transformer so there was no need for primary side soft start. I've never seen it used on both sides of the transformer. Have you?
Cutting out the resistor when at 3/4 charge on the capacitors sounds like it should work well even with a primary side soft start. There is a lot of difference between a little bit of saturation towards the end of a half cycle and being in full, deep, saturation during a whole half cycle which is essentially what happens when power is applied at the zero crossing of mains voltage and the transformer has full remanence in the same direction.
I've got some 1kVA transformers here and they have 1 ohm of primary resistance. That's a half-sine pulse of 300A peak, or 450A²s, if (when) this happens if the electric utility is stiff. For reference, from what I can find, 10A gG Diazed fuses as usually used in swedish fuse boxes blow at about 100-200A²s.
( A fuse datasheet: http://www.ifoelectric.com/Ifo_D-sak_TD_2006.pdf )
between 5 and 10seconds. Unless you require a slower (less current) charging rate.
One way is to use a low Rds-ON FET in series with caps...a RC time constant on the gate will bring it up slowly to soft-start the cap charging.... Nedd some protection diodes for turn-off...
The quick and dirty way is to use small toroidial inductor in series...You can calculate the inductance based on the block time needed to reduce the input current surge to a reasonable limit...then the toroid will saturate and be equivelent to just a wire when the amp is in use...Upon shuting the amp off, the core will re-set and be ready for the next block time....
Chris
The quick and dirty way is to use small toroidial inductor in series...You can calculate the inductance based on the block time needed to reduce the input current surge to a reasonable limit...then the toroid will saturate and be equivelent to just a wire when the amp is in use...Upon shuting the amp off, the core will re-set and be ready for the next block time....
Chris
Can someone who understands transformers and inductance confirm this?
A bank of 5 10r5W wirewound resistors is far cheaper than a small toroidal transformer.
you are right and I don't know a better solution. I know about a guy, that switch on three Krell KSA 250 at the same time about such TSRL.
Here the appropriate weblinks:
http://www.tauscher-transformatoren.de/assets/pdf/R_63_64.pdf
http://www.emeko.de/uploads/media/06-voltage-dips-description-short-e.pdf
http://www.emeko.de/index.php?id=11&L=1 (basic description)
http://www.fsm-elektronik.de/pdf/090807DatasheetTSRL.pdf (Datasheet)
www.tauscher-transformatoren.de/assets/pdf/TSR.PDF (Datasheet German)
www.tauscher-transformatoren.de/assets/pdf/R_43_44.pdf (Instruction Manual)
Addresses:
1) FSM Elektronik GmbH | Scheffelstraße 49 | D-79199 Kirchzarten
Telefon +49 (0) 7661. 98 55-0 | Telefax +49 (0) 7661. 98 55-11
info@fsm-elektronik.de | www.fsm-elektronik.de
2) Emeko Ing.Büro M. Konstanzer | Britzingerstraße 49
D-79199 Kirchzarten
Telefon +49 (0) 170 2410655 | Telefax +49 (0) 761 441888
emeko@t-online.de | www.emeko.de
Sorry for being unclear, but what I meant was to ask how high the capacitor voltage has become in percent of nominal voltage just before the resistors are cut out. If you let the capacitors charge until they are almost full, there will be no significant advantage as compared to a primary-side only soft start with the same delay and equivalent resistance.
If the capacitors are far from full, there will be an advantage. But in that case the capacitor charging when the resistor is cut out would create a large surge anyway.
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a primary side resistor designed to soft start a transformer for 100ms will get seriously hot if you ask it to remain in the 240Vac circuit for 5seconds to 10seconds. What if you ask for a 20second slow charge?
In the primary and in the secondary are not equivalent.
They are there for different purposes and designed for different duties.
Not really....
The small toroid inductor would be no bigger than 2" diameter....
You can put on the turns yourself by hand in 15 minutes....
The core you can get free if you request it as a sample....
It would not cost that much even if purchased...
I worked this same issue many times using this method...These are still working in critical applications where failure is no allowed...
For example there was a case where a surge current was being pulled from a BUS by a power supply's filter caps, that caused tripping.... I measured the inrush current with a current probe to find it was 28 Amps... The requirement was it needed to be below 18 Amps...
After modeling the circuit and step load I calculated the most inductance and block time for the given space I was given and the in-rush current was reduced to peak at 13 Amps in analysis... Then when it was measured with the current probe it also peaked at 13 Amps...
The little inductor was not ment to remove the surge completely, but instead reduces this peak in-rush to a reasonable level....
The best part of this is that once the toroid saturates it no longer has an impedance just the small DC resistance of the wire durring normal operation..
If you use a row of resistors you will have losses and poor dynamic regulation....
Chris
These resistors have a very strong negative temperature coefficient. The hotter they get the lower the resistance. So at first they are at room temp. and offer a high resistance then the in-rush current passes through them and they heat up and drop their resistance by a factor of 100X. You have to select a device with a thermal time constant that matches the size of the capacitor bank. The device has a chunk of ceramic inside that requires about as much energy to heat as can be stored in the capacitor bank. So you select the device based on "total energy". So it self-regulates and find an equalinrium. So even after hours the temperature is still the same.
Back in the vacuum tube era. They sold a thermal delay switch that looks like a tube. There was a bi-metallic spring inside that would be heated by current and after a some time flip the switch. Kind of like a thermostat. This was used to switch an in-rush resistor out of the circuit after it had done it's job. But now days we have these special in-rush limiting resistors that in effect turn into 1 ohm or 0.5 ohm resistors after a built-in time delay so the thermal delay tubes are only for antique equipment now. Not still it might be fun to design one of these into a modern amp.
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What are your component values? I suggested that the primary resistor be chosen for a good rate of capacitor charging, not that the "soft starts the transformer in 100ms" value were to be used. The worst case energy absorbed by a resistor soft starting an unloaded toroidal transformer depends on its value. The lower value you choose, the more energy will be deposited in the resistor.
A resistor soft starting an unloaded capacitor bank from a voltage source will absorb the same amount of energy regardless of the resistor value. (within reason) Worst case is that the same amount of energy as the caps store at full voltage goes into the resistor.
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For high power 2Kw+ we always used a primary on-rush circuit. A series resistor on the AC supply with a set of relay contacts to bypass it when the relay in parallel to the primary reached working voltage.
When power is engaged the voltage drop across the resistor would reduce the primary voltage below the relay trip point. As the power supply filter slowly charged the voltage the across the primary and relay would reach the trip point of the relay and close the contact giving full power to the transformer.
Now this was mainly in tube driven power amps that did not have a major load on the power supply at startup.
When power is engaged the voltage drop across the resistor would reduce the primary voltage below the relay trip point. As the power supply filter slowly charged the voltage the across the primary and relay would reach the trip point of the relay and close the contact giving full power to the transformer.
Now this was mainly in tube driven power amps that did not have a major load on the power supply at startup.
Why so long? Even a 4 A bridge rectifier can cope with a tau of 5 ms. The resistor value that leads to a tau of 1-2 s will prevent the capacitors from ever getting charged to a reasonable voltage level, when the amplifier is connected to the PSU.
The second peak that ocurrs, when the resistors are bridged, should not be bigger than the initial peak. It usually does not take more than a few mains cycles to achieve that.
The simplest solution is to use transformers, fuses and rectifiers that can cope with the capacitor inrush current and forget about soft-start circuits.
I think you are referring to the transformer start up delay system.
This is the delay for the secondary located Thermistor+ resistor bypass that slow charges the main smoothing bank.
Apply 1A to 0.1F and it will charge @ ~10V/s in 5 to 10 seconds it will charge to 50 to 100V if the 1A can be maintained.
As the capacitor voltage rises the charge rate becomes exponential.
Increasing the initial charge rate to 10A will ensure that almost all the charging is exponential since after just 100milliseconds the 10A constant rate current would have raised the smoothing cap voltage to ~10V or about 10% to 25% of final charge voltage.
A cold resistance of 5r will reduce the initial charging rate to ~10Apk if the transformer has a maximum output voltage of 50Vpk when delivering a current of 10Apk.
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When the capacitor is charged, there is still a voltage drop across the resistor. The moment the relay bridges the resistor, the capacitor will be charged further and a second current peak will ocurr. For a longer delay, a bigger resistor is needed. That means a bigger voltage drop, hence a bigger second current peak.
There is no gain in reducing the first peak to 10 A, when that leads to a much bigger second current peak. The correct way is to aim for two peaks, where the second is smaller than or equal to the first one, which must remain within the rectifier rating. That leads to a smaller resistor value and a shorter delay.
Of course a longer delay will reduce that second peak, but who wants to wait 10 s or more for an amplifier to start?
There is no gain in reducing the first peak to 10 A, when that leads to a much bigger second current peak. The correct way is to aim for two peaks, where the second is smaller than or equal to the first one, which must remain within the rectifier rating. That leads to a smaller resistor value and a shorter delay.
Of course a longer delay will reduce that second peak, but who wants to wait 10 s or more for an amplifier to start?
nobody.
I have a pre-amp that takes about 7seconds to start up. I find that a nuisance, but the manufacturer decided that is what they wanted it set to. I have not tried to over-ride their compromise.
The designer whether us or them make choices based on compromises.
I think you are wrong about needing a bigger resistor."a bigger resistor is needed. That means a bigger voltage drop"
I think you are wrong about the voltage drop. "When the capacitor is charged, there is still a voltage drop across the resistor"
I think you are wrong about no gain. "There is no gain in reducing the first peak to 10 A"
I think this conclusion does not follow from your logic. "The correct way is to aim for two peaks, ................... That leads to a smaller resistor value and a shorter delay.
What is wrong about it? Slower charge means lower current. Lower current means higher impedance.
The amplifier draws a quiescent current, which discharges the capacitors. The transformer will recharge them through that resistor. Any current through a resistor leads to a voltage drop across it. E. g. 100 mA quiescent current will give 0,5 V across the 5 Ohm resistor you proposed. When the relay contact closes, the capacitors have to increase their charge by those 0,5 V. A second current peak will ocurr.
There will be some gain. Let me reformulate it to: You may be led to believe in more gain, than you actually get, if you ignore the second peak. The second peak should not be bigger than the first, because the capacitor will already have heated up at that time. You'd rather aim for let's say two peaks of 15 A, instead of a 10 A followed by a 30 A peak.
There is also no need to reduce the inrush current to 10 A. Rectifier diodes usually withstand surge currents of several hundred A. Reduce the delay to a maximum of 1 s and exploit some more of the rectifier rating.
In most cases the transformer's inner resistance will limit the inrush current sufficiently without an additional capacitor soft start even for extreme capacitance values like in this case.
Hold the amplifiers in standby while it starts up. The inverter in an EV is essentially a very big 3 channel digital amplifier, but it only draws a tiny leakage current when the logic is not operating it. Therefore, the precharge current drops to almost zero when the precharge is finished.
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