Among the regular KSA 50 boards on the internet with 2 pairs power transistors, I've found this one.
Free Shipping! 1pcs KSA50 Class A amplifier PCB gold seal plastic speaker protection integrated generic version-in Other Electronic Components from Electronic Components & Supplies on Aliexpress.com
Having 6 pairs of power transistors it seems to me more likely to push out some 50 watts in Class A quality. I've seen this one assembled to @ $190 per channel.
Does anyone know more about this board?
Free Shipping! 1pcs KSA50 Class A amplifier PCB gold seal plastic speaker protection integrated generic version-in Other Electronic Components from Electronic Components & Supplies on Aliexpress.com
Having 6 pairs of power transistors it seems to me more likely to push out some 50 watts in Class A quality. I've seen this one assembled to @ $190 per channel.
Does anyone know more about this board?
Thank you for the correction!esgigt.
0.50A bias pr device and 3 pairs in push-pull configuration is 36Wrms at 8ohm.
peak class A current is twice the bias. so 3A peak. (.5a*6devices)
3*0.707=2.12A rms
2.12*2.12*8ohm = 35.95Wrms in class A at 8ohm speakers.
(Math courtesy of AudioSan. Hope I didn't butcher it)
esgigt.
0.50A bias pr device and 3 pairs in push-pull configuration is 36Wrms at 8ohm.
peak class A current is twice the bias. so 3A peak. (.5a*6devices)
3*0.707=2.12A rms
2.12*2.12*8ohm = 35.95Wrms in class A at 8ohm speakers.
(Math courtesy of AudioSan. Hope I didn't butcher it
Close - the total bias is only 1.5A because the same current flows through the the N and P devices in each pair. So peak class A current is 3 amps, then the rest of the math is fine.
I though I used those numbers. Do you mean that peak is 1.5 amps because the same current flows through the N and P device? I'm not sure I understand.
So 1.5A or 1.5 *.707 = 1.06Arms
1.06*1.06*8 = 9Wrms in class A.
Mine setup runs 2 pair at 600ma quiescent current. So I have 1.2 amps peak or 1.2 * .707 = .848Arms. .848 *.848 *8 = 5.76Wrms in class A?
I stand corrected esgigt.
In my setup each rail has an idle load of 2.4 amps for left and right.
1.06*1.06*8 = 9Wrms in class A.
Mine setup runs 2 pair at 600ma quiescent current. So I have 1.2 amps peak or 1.2 * .707 = .848Arms. .848 *.848 *8 = 5.76Wrms in class A?
I stand corrected esgigt.
In my setup each rail has an idle load of 2.4 amps for left and right.
Well, isn't that the way we learn?
I learned that the actual Class A output is distant from a real Krell...
"Having 6 pairs of power transistors it seems to me more likely to push out some 50 watts in Class A quality"
Also proved to be overly optimistic
I also stand corrected I think the idea behind more sets is that they can share the load. So if you had 6 pair per rail or 12 devices you could run 3.5 amps of quinscent current and produce just under 50wrms in class A. So each device carries the 1/6 of the load and will dissipate 1/6 of the watage in heat.
You would need a couple of 600va toroidals
Or go bqk in time and get the Mr. Fusion from the DeLorean.
You would need a couple of 600va toroidals
Or go bqk in time and get the Mr. Fusion from the DeLorean.
P = I*V = I*I * R = V*V / R
That is for non varying DC.
If you use a varying current then you must substitute Irms for I and Vrms for V giving:
P = Irms * Vrms = Irms*Irms * R = Vrms*Vrms / R
For a sinewave there is a defined relationship between Ipk and Irms and between Vpk and Vrms. The power formula becomes:
P = Ipk * Vpk / 2 = Ipk*Ipk * R / 2 = Vpk*Vpk / R / 2
Note that one does not need to use the square root of 2 for a sinewave. This is because sqrt(2) * sqrt(2) = 2, exactly !
For a push pull amplifier the maximum ClassA output is ~=2times the output bias current.
set the device bias to 500mA. The maximum ClassA output is 1Apk
Using multiple pairs each @ 500mA bias results in just multiplying the above values.
6pair output stage biased to 500mA per pair, results in maximum ClassA output of 6Apk
The maximum ClassA power is Ipk*Ipk * R / 2 = 6*6 *8 / 2 = 144W
That is for non varying DC.
If you use a varying current then you must substitute Irms for I and Vrms for V giving:
P = Irms * Vrms = Irms*Irms * R = Vrms*Vrms / R
For a sinewave there is a defined relationship between Ipk and Irms and between Vpk and Vrms. The power formula becomes:
P = Ipk * Vpk / 2 = Ipk*Ipk * R / 2 = Vpk*Vpk / R / 2
Note that one does not need to use the square root of 2 for a sinewave. This is because sqrt(2) * sqrt(2) = 2, exactly !
For a push pull amplifier the maximum ClassA output is ~=2times the output bias current.
set the device bias to 500mA. The maximum ClassA output is 1Apk
Using multiple pairs each @ 500mA bias results in just multiplying the above values.
6pair output stage biased to 500mA per pair, results in maximum ClassA output of 6Apk
The maximum ClassA power is Ipk*Ipk * R / 2 = 6*6 *8 / 2 = 144W
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Please excuse a little thread hijacking - but 12 devices per side and two 500va toroids is exactly what I have in a BA-3 build. I came to that through the back door from a heat management perspective, but subsequently am interested in optimizing for a little more energy to push lower efficiency speakers. The BA has a separate bias board, but I believe the science is the same. I understand the concept of "sharing the load" as opposed to an additive effect of multiple pairs. What I'm not clear about is the relationship between increasing the bias and lowering the value of the resistors on the output devices. I suspect there is a window or fairly narrow range where the combination of those two adjustments maintains/developes more class A power - and keeps the amp operating at a safe temperature.
If this is correct, can someone clarify the relationship and/or present a formula to determine usable values? It's probably pretty basic stuff but I still don't have a handle on it.
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Are you referring to heat dissipation - or current rating of the driver transistors - or both?
BTW: I'm more than willing to go read some articles for a better understanding if someone points me in the right direction,
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BcmBob, you may be confused with class AB "optimimum bias" where you want to keep Vre roughly 0.025V. In class A operation Vre doesn't matter (except in the Pass Alephs). Lower emitter resistors or Source resistors in your case just mean you'll want to pay a bit more attention to matching the output devices.
Henry is talking about output current capacity which leads to both. The driver stage must be capable of driving the bases with enough current so that you get the benefit of all your output devices. A common mistake is just adding output devices without considering the extra drive current required. For your burning amp MOSFET stage it's a little different. Low drive ability tends to limit bandwidth. See the end of the F5T article for an explanation.
Henry is talking about output current capacity which leads to both. The driver stage must be capable of driving the bases with enough current so that you get the benefit of all your output devices. A common mistake is just adding output devices without considering the extra drive current required. For your burning amp MOSFET stage it's a little different. Low drive ability tends to limit bandwidth. See the end of the F5T article for an explanation.
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Wow. What was the quiescent current for the single pair
of transistors in the KSA50? What was the rated wattage in class A?