I would like to use a transformer like the one in the drawing to build a large class A amp(KSA50 clone) with two secondaries per channel, quasi dual mono I guess. The voltages are too high for use with an unregulated supply and I wanted to use a capacitance multiplier anyway. Can I use one side of each winding(and tape off the other side) and the center tap for 39v(instead of 78v, which is unmanageable) without losing power capacity, i.e. will I still be able to use the full 1250 VA that this transformer is rated for even though I am only using half of each secondary winding?
Thanks.
Edit: I looked over the diagram again, and did some math...I think I am wrong.
Would it work to parallel pairs of secondaries, to end up with two center-tapped secondaries, each with 39-0-39 @ 8A?
i.e. Connect 1 to 3 and 2 to 4 and also connect their center taps, and do the same thing with the other side.
- Tom Gootee
http://www.fullnet.com/~tomg/index.html
i.e. Connect 1 to 3 and 2 to 4 and also connect their center taps, and do the same thing with the other side.
- Tom Gootee
http://www.fullnet.com/~tomg/index.html
Yes, actually I hadn't thought about that...the only issue I see is the KSA50 is a big class A beast which dissipates about 150 watts per channel idle. If I go with this strategy I will only be able to use a single bridge rectifier per power supply instead of two like I would be able to use if I had two 39v secondaries. Original KSA50s also had two rectifiers in the power supply. I'm worried about stressing the rectifier diodes over time. Do you think this might be a potential source of failure even if I use a highly overspecified bridge?
Well, I guess that if the bridge is "highly overspecified", then, by definition, it should not be a potential source of failure. 🙂
I ran a quick simulation of one channel, with LTspice, mainly to see what the diodes would be doing:
I don't know anything else about the KSA50 amplifier. But if it dissipates 150W per channel at idle, and you have the power supply topology shown in Figure 5 at http://sound.westhost.com/power-supplies.htm#rectifiers ,
and the secondaries are both modeled as ideal sine voltage sources with +/- 55.146-volt peaks (i.e. 39 x sqrt(2)) with a series 0.3 Ohm resistor (a guess for the winding resistance), and the load is replaced with a resistor of 68.7 Ohms (and using, say, 2 x 3x 10000 uF smoothing caps [i.e. 30000 uF per side], each with .025 Ohms ESR @ 120 Hz, and 15 nH ESL), then, if silicon rectifiers are used, you get about +/- 50.8 VDC outputs, and about 1.48A through the load, with the load dissipating around 150 Watts.
In that case, the four rectifier diodes, which I modeled with MUR460 600v/4A types (highest-rated silicon diode spice model I had on hand), each diode has current peaks of about 9.7 Amps. But each diode's average current (after startup, in steady state) is only about 740 mA, and their RMS currents are about 2.4 Amps each. Each diode's average power dissipation is about 900 mW.
If I change the diodes to MBR20100CT 100V/10A Schottky types (highest-rated Schottky spice model I had on hand), then the output voltages go up to about +/- 51.3 VDC. Changing the load resistor to 70 Ohms puts its power dissipation back to about 150W, with about 1.47A avg current.
In that case, each Schottky diode's RMS and average currents are about 2.4 Amps and 730 mA, respectively, with peaks of about 9.9 Amps. Each diode's average power dissipation is about 510 mW.
I don't know if any of that information is useful for you, at all. (And note that the transformer was not modeled well.) But it was quick and easy to do, with LTspice (available free from http://www.linear.com , or use the direct download link on my spice-modeling webpage at http://www.fullnet.com/~tomg/gooteesp.htm .)
- Tom Gootee
I ran a quick simulation of one channel, with LTspice, mainly to see what the diodes would be doing:
I don't know anything else about the KSA50 amplifier. But if it dissipates 150W per channel at idle, and you have the power supply topology shown in Figure 5 at http://sound.westhost.com/power-supplies.htm#rectifiers ,
and the secondaries are both modeled as ideal sine voltage sources with +/- 55.146-volt peaks (i.e. 39 x sqrt(2)) with a series 0.3 Ohm resistor (a guess for the winding resistance), and the load is replaced with a resistor of 68.7 Ohms (and using, say, 2 x 3x 10000 uF smoothing caps [i.e. 30000 uF per side], each with .025 Ohms ESR @ 120 Hz, and 15 nH ESL), then, if silicon rectifiers are used, you get about +/- 50.8 VDC outputs, and about 1.48A through the load, with the load dissipating around 150 Watts.
In that case, the four rectifier diodes, which I modeled with MUR460 600v/4A types (highest-rated silicon diode spice model I had on hand), each diode has current peaks of about 9.7 Amps. But each diode's average current (after startup, in steady state) is only about 740 mA, and their RMS currents are about 2.4 Amps each. Each diode's average power dissipation is about 900 mW.
If I change the diodes to MBR20100CT 100V/10A Schottky types (highest-rated Schottky spice model I had on hand), then the output voltages go up to about +/- 51.3 VDC. Changing the load resistor to 70 Ohms puts its power dissipation back to about 150W, with about 1.47A avg current.
In that case, each Schottky diode's RMS and average currents are about 2.4 Amps and 730 mA, respectively, with peaks of about 9.9 Amps. Each diode's average power dissipation is about 510 mW.
I don't know if any of that information is useful for you, at all. (And note that the transformer was not modeled well.) But it was quick and easy to do, with LTspice (available free from http://www.linear.com , or use the direct download link on my spice-modeling webpage at http://www.fullnet.com/~tomg/gooteesp.htm .)
- Tom Gootee
Thanks for the help Tom, it was helpful. Also, I had a brain freeze - original Krells only had one bridge per PS and used a center tapped transformer. Looks like I'm in the clear.
Hi,
if you use half of the secondary copper you end up with half the VA rating (roughly).
I tried paralleling a four secondary, 37+..... 37 to give a high current supply for the KSA100. 37Vac is a bit low for the KSA100, 39Vac would be ideal for +-52Vdc when biased.
The 35A rectifier ran hot.
I rewired it so that one rectifier operated on one pair of secondaries and the second rectifier operated on the other pair of secondaries and tied all the free ends to form the zero volt centre tap. The outputs from the rectifiers are paralleled into the smoothing bank. The rectifiers run cold with just a small heatsink when biased to 2.5A.
A beefy output stage could run a KSA50 at KSA100 voltage and bias, if the sinks are big enough.
if you use half of the secondary copper you end up with half the VA rating (roughly).
I tried paralleling a four secondary, 37+..... 37 to give a high current supply for the KSA100. 37Vac is a bit low for the KSA100, 39Vac would be ideal for +-52Vdc when biased.
The 35A rectifier ran hot.
I rewired it so that one rectifier operated on one pair of secondaries and the second rectifier operated on the other pair of secondaries and tied all the free ends to form the zero volt centre tap. The outputs from the rectifiers are paralleled into the smoothing bank. The rectifiers run cold with just a small heatsink when biased to 2.5A.
A beefy output stage could run a KSA50 at KSA100 voltage and bias, if the sinks are big enough.
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