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Old 19th April 2013, 08:26 AM   #1951
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Originally Posted by 1audio View Post
There is also some inductance in the transformer. You can short the primary and measure the secondary's inductance at 120 Hz and 1 KHz to get a number to add to the model. I would definitely add something like 100 mOhms for the AC source at a minimum.
I have measured and included 29uH leakage inductance.
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Old 19th April 2013, 08:26 AM   #1952
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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.
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Old 19th April 2013, 08:30 AM   #1953
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Quote:
Originally Posted by HiFiNutNut View Post
So the combined AC source impedance is 0.12R + 1.4R = 1.52R?
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
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Old 19th April 2013, 08:35 AM   #1954
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Originally Posted by abraxalito View Post
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.
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Old 19th April 2013, 08:37 AM   #1955
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I assume then you meant 0.12R per secondary so your total secondary resistance is 0.24R. So that's the secondary part of the impedance - the primary part becomes 2/(2.4)^2 = 0.35R for a total simulation value of 0.59R.
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Old 19th April 2013, 08:59 AM   #1956
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Originally Posted by abraxalito View Post
I assume then you meant 0.12R per secondary so your total secondary resistance is 0.24R. So that's the secondary part of the impedance - the primary part becomes 2/(2.4)^2 = 0.35R for a total simulation value of 0.59R.
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?
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Old 19th April 2013, 09:03 AM   #1957
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Makes sense to split the total secondary resistance into two halves yes - 0.3R per side ought to be close enough.
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Old 20th April 2013, 06:46 AM   #1958
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Originally Posted by AndrewT View Post
I have not made up my mind which is better. I have amplifiers of both types. Fused supply rails and unfused supply rails.
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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Last edited by danielwritesbac; 20th April 2013 at 06:53 AM.
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Old 20th April 2013, 08:15 AM   #1959
AndrewT is offline AndrewT  Scotland
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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.
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Old 20th April 2013, 12:44 PM   #1960
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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?
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