No hum , your DC is rectified output of 50Khz AC. Could not hear that hum !
At the top tier of tech is that Connex SMPS.
- It's controller modulates 70 - 111khz to keep 96% efficiency.
- Its sine wave "switcher"
has feedback from the trafo to develop a magneticequivalent in the trafo's core itself. No 50/60hz trafo does that , you use the dirty line
(sine) lots of garbage.
-just about burn up proof.
So , spend 200$ on 15 amps of 50/60hz raw dirty rectified AC , or
60-80$ for lab grade 70Khz rectified DC at 1/20 the heat and weight.
The audio purist would gravitate towards the bulky analog ... what for ,,
nostalgia ?
For the badger or any other DIYA AB amp ... even class A , 70Khz rectified
(and monitored by both voltage and magnetic feedback) DC is "lab grade".
No PSRR issues (extra 40-50 DB) to add to the Badger PSRR.
BTW , most audio grade PSRR's can peak out 1.6 - 2X rated current.
OS
The "off the shelf" ones will usually barely exceed rated output.
Link LLC Current-Resonant Controller IC with Built-in Full Protection Functions - PowerPulse.net
is to that Connex class of smps controller. The rectified output of an SMPS
built around that IC would increase a Badger's PSRR to nearly un-measurable
levels.
With careful design and sourcing 2X peak currents and even integration into
21st century
protection systems could be possible.Fun , Fun ....
OS
On a real cheap one , single mosfet smps. They all have some crude form
of regulation - that is the purpose of the opto in the typical.
But , still - the rectified HF DC is still isolated from the mains. Like the old-school Isolation
transformer.
Better IC's have that cool magnetic resonance negative FB loop that rejects
many more DB of input noise (and better opto/voltage reg behavior.
Like having a pre-conditioned 50/60hz lab grade line source , for a analog
analogy. Under load , 50/60 hz ... ripple increases greatly. SMPS , just a little
more HF "noise" (which can be filtered out much more easily).
OS
I'm repeating my question to Johno: How did you calculate this? Is this just a rule of thumb of your own (output power is half the DC input power)?
I think there has to be a more serious algorithm, a formula that describes the relationship between output power into a certain impedance and the DC current consumption, doesn't it? It would help me that much, 'cause as I've said yet, I've found out a way to tweak my bricks' DC output voltages
.@ Ostripper: These bricks feature a true active PFC. After opening one of them with a hand saw, I've found a pair of controllers by TI, one of them being a PFC controller that also monitors output current and features overvoltage protection. Later this day, when at home again, I'll have a look again and tell you the designations.
Best regards!
I'm repeating my question to Johno: How did you calculate this? Is this just a rule of thumb of your own (output power is half the DC input power)?
I think there has to be a more serious algorithm, a formula that describes the relationship between output power into a certain impedance and the DC current consumption, doesn't it? It would help me that much, 'cause as I've said yet, I've found out a way to tweak my bricks' DC output voltages
@ Ostripper: These bricks feature a true active PFC. After opening one of them with a hand saw, I've found a pair of controllers by TI, one of them being a PFC controller that also monitors output current and features overvoltage protection. Later this day, when at home again, I'll have a look again and tell you the designations.
Best regards!
You will get a max PSU output of 480W pr ch. But i doubt this makes room for any peak currents above this numbers. With 55Vdc rails and 4 ohm load you will have a peak power draw of 756W pr ch. That is way above the limits.
Kay wrote "I'm repeating my question to Johno: How did you calculate this?"
Each switch mode power block is current limited to 5A, each power block in series must carry the total current. That is, Kirchhoff Law applies. Therefore the max power out is [(2 x 24) x 5] = 240W. If you want to do the math you will see that it is the energy but not the current which doubles.
In series add the voltages and in parallel add the currents together. To double both the voltage and current you need two series pairs in parallel (4 power blocks).
Hope this helps
Each switch mode power block is current limited to 5A, each power block in series must carry the total current. That is, Kirchhoff Law applies. Therefore the max power out is [(2 x 24) x 5] = 240W. If you want to do the math you will see that it is the energy but not the current which doubles.
In series add the voltages and in parallel add the currents together. To double both the voltage and current you need two series pairs in parallel (4 power blocks).
Hope this helps
that is just talk, i am familiar with smps psu's having repaired a lot of those ATX psu's in the past, the ripple frequency of the mains filter is 120 hz and is just at the primary side of the choppers.....the ripple frequencies at the secondary side can be in the 50khz or more....
Honestly, I doubt this. This holds true for an ideal class A amplifier, where the maximum output power is half the DC input, but not for class AB and B. I believe to recall that the efficiency of an ideal class B amplifier that is driven by a sinusoidal signal is something like π/4, which equals to about 78.5%. Does this seem reasonable?
So, with a DC input power of 480 watts an output power of at least 300 watts could be expected. For 300 W @ 8Ω the rail voltage needs to be +/-54 V at least. Is this correct?
Best regards!