I have a question about ripple current of reservoir capacitors.
When using LTSpice, if I make the output current to be 7A, then there are 75A (ptp) bursts going through the first 10000uF! If I reduce the output current to 2A, this is reduced to 30A.
There is no 10000uF capacitor existed on earth with a ripple current rating of 30A at 120Hz.
It is a class AB amp so how am going to calculate it? using the average?
The first capacitor has ESR of about 0.05R at 120Hz so theoretically, there can be as much as rail / ESR current going through, and this can be as high as 1,750A for a 85VDC rail!
When using LTSpice, if I make the output current to be 7A, then there are 75A (ptp) bursts going through the first 10000uF! If I reduce the output current to 2A, this is reduced to 30A.
There is no 10000uF capacitor existed on earth with a ripple current rating of 30A at 120Hz.
It is a class AB amp so how am going to calculate it? using the average?
The first capacitor has ESR of about 0.05R at 120Hz so theoretically, there can be as much as rail / ESR current going through, and this can be as high as 1,750A for a 85VDC rail!
What is the source impedance of your AC supply? it should be at least 100 mOhms (100' of #12 I think) and that is presuming that everything upstream from the box is connected with superconductors. The peak currents are from the really short conduction angle, only possible with a zero Ohm source. The real world transformer would limit that some. Years ago I measured 400A peak on turn on into a stack of home theater power amps (8 channels worth). In operation it never got to nearly that much current. It did make for an impressive demo and durability test for the relays switching the load.
Thanks, Demian.
I have no idea what the AC source impedance is at the IEC socket. The transformer primary winding resistance is 2R. Secondary winding resistance is about 0.12R. The transformer is a 625VA 240V 50V-0-50V. In my LTSpice modelling, I only included the 0.12R, which is obviously wrong, but I don't know the total AC source impedance. Do you have any suggestion / approximation?
I have no idea what the AC source impedance is at the IEC socket. The transformer primary winding resistance is 2R. Secondary winding resistance is about 0.12R. The transformer is a 625VA 240V 50V-0-50V. In my LTSpice modelling, I only included the 0.12R, which is obviously wrong, but I don't know the total AC source impedance. Do you have any suggestion / approximation?
You can include the primary resistance reflected in the (turns ratio) ^2 as a parasitic element on the secondary side. In your case this looks to be about 350mohms. This seems rather high though because power transformers are normally designed to give equal copper loss in primary and secondary windings so I'd expect the transformed primary resistance to equal the secondary....
To reduce the reading error, check the resistance between the two 50V ends. Then include double this value in your sim because its more likely than not the trafo designer did a fairly decent job of optimizing the pri/sec resistances
As Demian suggests you can go on to include leakage inductance (leakage goes in series with the primary) though the leakage inductance is very unlikely to significantly reduce your peak currents, particularly for a toroid which usually have very low values.
As Demian suggests you can go on to include leakage inductance (leakage goes in series with the primary) though the leakage inductance is very unlikely to significantly reduce your peak currents, particularly for a toroid which usually have very low values.
No - I was just estimating what the primary resistance would normally be from your secondary measurements and the turns ratio. To include any impedance on the primary side as a secondary parasitic you must transform it by the square of the turns ratio, not use it natively
To reduce the reading error, check the resistance between the two 50V ends. Then include double this value in your sim because its more likely than not the trafo designer did a fairly decent job of optimizing the pri/sec resistances
Good idea. I just got back to the bench and measured again and confirmed that each of the two primary windings has 0.12R and the primary has 2R.
Thanks a million. That has really been a great help!
There is one thing I am not so clear. I understand now how the 0.59R is derived. But since this impedance is seen by both the two opposite +V and -V rails, when modelling one rail only should the impedance be 0.35R / 2?
I have several pieces of information on that, and a question too.
On my LM1875 projects thread, it is popular for constructors to ask "why did my amp blow up?" along with a photo of an amplifier the size of a thumbnail powered by a transformer the size of a teapot. Mystery solved! This needs fuses added.
If the rails fuse blows, one rail is output to the speaker or if the amplifier blows, one rail is output to the speaker. So, there's no difference to the speaker, but fuses might save the amplifier.
Although crude, fuses could be informative and educational, and might succeed where transformer sizing directions have failed. Approximately, about 2% of constructors have used appropriate current on that particular project. May I assume that the success rate of fuses could do better than ~2%?
Question:
What size and type fuses should I put on the rails for one chip LM1875 amplifier board?
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Dan,
put a T fuse in the mains supply to the primary of the transformer.
Ensure the rating is for close rating or lower.
i.e. two 60W channels running from a 160VA on 115Vac needs a fuse of T1.4A, or smaller. The nearest lower value is T1.25A.
The transformer will probably not start reliably with this size of fuse. Use a soft start with 60r to 100r as the current limiting resistor.
put a T fuse in the mains supply to the primary of the transformer.
Ensure the rating is for close rating or lower.
i.e. two 60W channels running from a 160VA on 115Vac needs a fuse of T1.4A, or smaller. The nearest lower value is T1.25A.
The transformer will probably not start reliably with this size of fuse. Use a soft start with 60r to 100r as the current limiting resistor.
Now back to the topic, i.e. Power Supply Reservoir Size.
The way I do is that I check the PSSR of my amplifier and try to come up with a power supply PSSR such that the two PSSRs combine together to produce at least -110dB from 20Hz to 10MHz. From there I can determine how much ripples the power supply can tolerate when the maximum current is drawn then determine the Reservoir Size.
Would -110dB be good enough for you? What is your target?
The way I do is that I check the PSSR of my amplifier and try to come up with a power supply PSSR such that the two PSSRs combine together to produce at least -110dB from 20Hz to 10MHz. From there I can determine how much ripples the power supply can tolerate when the maximum current is drawn then determine the Reservoir Size.
Would -110dB be good enough for you? What is your target?
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