If you decide to rely on the interconnecting cable to provide the 0.1Ω resistance, you might be surprised how long this needs to be. For example, 0.5 sq. mm CSA cable (22 AWG) has a resistance of 39Ω per kilometre. That would require 2.56 metres!Thanks Currentflow! This helps to make a more complete picture
A compact crc power supply with the bypass diodes
(For one supply per channel, such as monobloc, dual mono.)
-Or-
Without the bypass diodes here is an orthodox crc power supply with similar filter effectiveness
P.S.
To make it fit in the amp enclosure: Cable can be substituted for the 0.1R, such as using 2 boards (br board with first caps, then some cable, then tank/reservoir board, in which case the 0.1R-5W are not used), and that can help for easier layout but is also useful to heatsink the bridge rectifier for capacitor longevity.
That reminds me of some NiChrome wire I have at 1R/meter...
How can I tell what amperage it could handle?
How can I tell what amperage it could handle?
The limiting factor is the level of heat which may be tolerated and is determined by the way in which the wire is used. In free air, a straight length of wire would withstand more current (run cooler) than that of a tightly coiled wire. Heating is proportional to the power dissipated by the wire.That reminds me of some NiChrome wire I have at 1R/meter...
How can I tell what amperage it could handle?
Power = I^2 x R,
Or, rearranging to determine current for a given power and resistance:
I = square root of [P / R]
If you decide to rely on the interconnecting cable to provide the 0.1Ω resistance, you might be surprised how long this needs to be. For example, 0.5 sq. mm CSA cable (22 AWG) has a resistance of 39Ω per kilometre. That would require 2.56 metres!
Well, I think that resistance at DC was not important in this case, when actually the signal we want to trap in the series element is the higher pitched harmonics off the bridge rectifier, and up at that higher pitch, the pins of the resistor do nearly as much as the resistor itself.
I couldn't measure this too smartly with the ESR meter and a ordinary ruler; however, it is about 0.01R per each quarter inch of ~22ga pins on that particular resistor. Given the purpose of this filter, it does seem than an inductive series element may be more efficient than either a diode or a resistor.
P.S.
I tried this with 8" 20ga solid copper thermostat wire. Practically, it sounds just like an orthodox CRC. That doesn't match with my personal preferences; but, at least it is an easy way to fit a biggie size supply in a little enclosure, and also, it was very easy to heatsink the bridge rectifiers (they with the first caps simply bolted on where convenient, and then cable out to the tank/reservoir board).
Anyway, the technique works well, but is in need of some guidance for optimizing the cable-as-series-element gauge and length specifications.
Well, I think that resistance at DC was not important in this case, when actually the signal we want to trap in the series element is the higher pitched harmonics off the bridge rectifier, and up at that higher pitch, the pins of the resistor do nearly as much as the resistor itself.
Can you give some guidance as to how much attenuation are you trying to achieve here and at what frequency? After all, DC resistance is what makes the series element of a C-R filter, so its importance cannot be ignored.
For a low-pass filter the -3dB frequency is calculated: 1/(2 x PI x CR):
Using your original component values of 48,000uF and 0.1Ω gives 33.2Hz.
I couldn't measure this too smartly with the ESR meter and a ordinary ruler; however, it is about 0.01R per each quarter inch of ~22ga pins on that particular resistor. Given the purpose of this filter, it does seem than an inductive series element may be more efficient than either a diode or a resistor.
Inductors also have resistance. It may well be that the amount of inductance required by this application offers further attenuation by the very nature of its inherent DC resistance. This may be used to your advantage.
ρ = rho, the electrical resistivity (also known as specific electrical resistance or volume resistivity) of copper = 0.01724 ohm×mm^2/m.P.S.
I tried this with 8" 20ga solid copper thermostat wire. Practically, it sounds just like an orthodox CRC. That doesn't match with my personal preferences; but, at least it is an easy way to fit a biggie size supply in a little enclosure, and also, it was very easy to heatsink the bridge rectifiers (they with the first caps simply bolted on where convenient, and then cable out to the tank/reservoir board).
Anyway, the technique works well, but is in need of some guidance for optimizing the cable-as-series-element gauge and length specifications.
To calculate the resistance of a conductor:
R = ρ x L / A where L is the length and A the cross-sectional area.
e.g. 20AWG has a CSA of 0.52mm^2. L = 8” = 0.2032m
R = 0.01724 x 0.2032 / 0.52 = 0.0067Ω (Still very low)
or use thinner wire to couple the capacitors.
I have said this few times in the past.
A twisted pair of CAT5 (~24awg) between a pair of caps. Adds a little r between C & C to give CrC
The twisted pair wires from the transformer to the rectifier and and another twisted pair from the rectifier to the first capacitor add a little bit more r to the resistance of the transformer. This creates an rCrC from the assembly.
It costs nothing but a bit of time. Scrap the PCBs for Capacitor PSUs. Use thin twisted pairs.
I have said this few times in the past.
A twisted pair of CAT5 (~24awg) between a pair of caps. Adds a little r between C & C to give CrC
The twisted pair wires from the transformer to the rectifier and and another twisted pair from the rectifier to the first capacitor add a little bit more r to the resistance of the transformer. This creates an rCrC from the assembly.
It costs nothing but a bit of time. Scrap the PCBs for Capacitor PSUs. Use thin twisted pairs.
or use thinner wire to couple the capacitors.
I have said this few times in the past.
A twisted pair of CAT5 (~24awg) between a pair of caps. Adds a little r between C & C to give CrC
The twisted pair wires from the transformer to the rectifier and and another twisted pair from the rectifier to the first capacitor add a little bit more r to the resistance of the transformer. This creates an rCrC from the assembly.
It costs nothing but a bit of time. Scrap the PCBs for Capacitor PSUs. Use thin twisted pairs.
That would be a very simple method for adding a nominal level of resistance. There will also be internal series supply source impedances, i.e. the bridge rectifier and the mains transformer itself. This will form additional filtering stages in conjunction with the first set of 3300uF capacitors. The question is, how much of a problem is there that all this filtering is attempting to resolve?
The HF garbage that comes out of a rectifier needs attenuating.
The better the PSU does that job the more chance that the amplifier will perform as designed.
The better the PSU does that job the more chance that the amplifier will perform as designed.
This TI application note describes a PSU design that utilises snubber networks and additional filtering for noise suppression, but has no added series element:
AN-1849 An Audio Amplifier Power Supply Design
AN-1849 An Audio Amplifier Power Supply Design
Ti is showing an rCCCCC PSU.
A filter MUST have a series element.
If there is no series element then it CANNOT be a Filter.
A filter MUST have a series element.
If there is no series element then it CANNOT be a Filter.
I was referring to no "added" series resistor element (as in Daniel's design) and yes, the word suppressor is perhaps more appropriate in this sense.Ti is showing an rCCCCC PSU.
A filter MUST have a series element.
If there is no series element then it CANNOT be a Filter.
interference attenuation is a well understood concept.
Suppression applies to what in respect of an omitted series element?
Suppression applies to what in respect of an omitted series element?
TI's application circuit does not include a series resistor between the bridge and the output connections. Suppression applies to noise in this context.interference attenuation is a well understood concept.
Suppression applies to what in respect of an omitted series element?
The note states: "The design uses toroidal transformers, a fully integrated bridge, and various rail capacitors for ripple voltage reduction, noise suppression, and to act as high current reservoirs".
You stated in #1372: "Ti is showing an rCCCCC PSU". Where is the r you refer to?
r = the actual resistance of the source feeding the capacitors.
That resistance is made up of wiring between the components and the rectifier bridge terminals and the transformer leadouts and the transformer secondary and the reflected resistance of the transformer primary and the reflected resistance of the source feeding the transformer.
That r is NOT zero ohms.
In addition there is reactive impedance mostly due to inductance of the wiring and cabling.
That gives a two pole filter, LC
In BOTH cases, the C can ONLY work as a filter if it is fed from a source that has some impedance/resistance.
That resistance is made up of wiring between the components and the rectifier bridge terminals and the transformer leadouts and the transformer secondary and the reflected resistance of the transformer primary and the reflected resistance of the source feeding the transformer.
That r is NOT zero ohms.
In addition there is reactive impedance mostly due to inductance of the wiring and cabling.
That gives a two pole filter, LC
In BOTH cases, the C can ONLY work as a filter if it is fed from a source that has some impedance/resistance.
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where does this end? all the way back to the power station?
reflected primary impedance is good enough for me, any resistance of the source feeding the traffo
will just result in lower primary voltage, and i will not concern myself with that anymore..
reflected primary impedance is good enough for me, any resistance of the source feeding the traffo
will just result in lower primary voltage, and i will not concern myself with that anymore..
In summary, TI's design offers an overall relatively low impedance configuration by relying on internal component impedances, both resistive and inductive, the values of which may not be simple to determine.r = the actual resistance of the source feeding the capacitors.
That resistance is made up of wiring between the components and the rectifier bridge terminals and the transformer leadouts and the transformer secondary and the reflected resistance of the transformer primary and the reflected resistance of the source feeding the transformer.
That r is NOT zero ohms.
In addition there is reactive impedance mostly due to inductance of the wiring and cabling.
That gives a two pole filter, LC
In BOTH cases, the C can ONLY work as a filter if it is fed from a source that has some impedance/resistance.
At the expense of increasing the overall supply impedance, there would be a further noise reduction advantage by introducing additional series resistance.
Theoretically speaking, the further from the power station the audiophile purist lives, the better the PSU performance for a given set of capacitors! OTOH, longer transmission lines will collect more interference...where does this end? all the way back to the power station?
well, one thing is for sure, no amplifier power traffo can ever come close to the impedance of the pole mounted distribution transformer..
i lived beside one for several years, in my house then, line voltage was a steady 240 volts any time of day and night....
i lived beside one for several years, in my house then, line voltage was a steady 240 volts any time of day and night....
http://www05.abb.com/global/scot/scot252.nsf/veritydisplay/7fb68603fb9f47f8852577f4006474e9/$file/1lca000003-lte_singleph_overhead_10kva_167kva_rev01.pdf
Most of our rural ones are 50Kva 220-240V (secondary). 14kv is the primary
here.
If we have 100 huge amps , we may "tickle" it during onrush current !! 😀
OS
Most of our rural ones are 50Kva 220-240V (secondary). 14kv is the primary
here.
If we have 100 huge amps , we may "tickle" it during onrush current !! 😀
OS
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