35Vac will make the amplifier run much hotter.Adi2009 said:
3pairs with a total bias of 1.9A will be able to dissipate the heat, but the heatsink will require to be much larger than a KSA50.
You will end up with a high bias ClassAB amplifier capable of ~200W into 4r0 and ClassA output for loads >=16r0.
I could not edit my last post due to DIYaudio database problems.
The 16r0 should read:
A ClassA amplifier set to 1.9A bias (KSA50) can send upto 3.8A of ClassA current to the load.
50W into 8r0 ClassA requires that the amplifier should be able to reach 28.3Vpk into an 8r0 load and simultaneously pass 3.5Apk to the same load.
If you connect a 4ohm load to that 50W into 8r0 ClassA amplifier, then you will have a maximum ClassA output of 25W due to reaching the 3.5Apk ClassA current limit into that 4r0 load.
The push-pull amp is capable of delivering even more power if the power supply is able.
The KSA50 will easily deliver 100W into 4r0 as a ClassAB amplifier when the PSU is as standard.
You are proposing a 35Vac transformer and this will give ~+-50vdc from the supply. When biased to 1.9A, the PSU will probably drop to ~+-49Vdc. The amp will easily deliver 100W into 8r0 and very likely be able to deliver 200W into 4r0. All of these maximum powers are in ClassAB mode.
If you want to know the maximum ClassA power available you have to use the ClassA current limit to determine the ClassA power capability.
Check what Vpk and Ipk you will need to get 70W into 11r0.
Try to drive a lower impedance load and the amp will go into ClassAB when driven to maximum output.
The 16r0 should read:
A ClassA amplifier set to 1.9A bias (KSA50) can send upto 3.8A of ClassA current to the load.
50W into 8r0 ClassA requires that the amplifier should be able to reach 28.3Vpk into an 8r0 load and simultaneously pass 3.5Apk to the same load.
If you connect a 4ohm load to that 50W into 8r0 ClassA amplifier, then you will have a maximum ClassA output of 25W due to reaching the 3.5Apk ClassA current limit into that 4r0 load.
The push-pull amp is capable of delivering even more power if the power supply is able.
The KSA50 will easily deliver 100W into 4r0 as a ClassAB amplifier when the PSU is as standard.
You are proposing a 35Vac transformer and this will give ~+-50vdc from the supply. When biased to 1.9A, the PSU will probably drop to ~+-49Vdc. The amp will easily deliver 100W into 8r0 and very likely be able to deliver 200W into 4r0. All of these maximum powers are in ClassAB mode.
If you want to know the maximum ClassA power available you have to use the ClassA current limit to determine the ClassA power capability.
Check what Vpk and Ipk you will need to get 70W into 11r0.
Try to drive a lower impedance load and the amp will go into ClassAB when driven to maximum output.
Yes but I asked for a sound. The sound quality will be the same on MJ15003/04 and MJ15024/25 transistors?
I calculate and:
Number of output devices: 2 Pairs
Voltage rails (per rail): 50 volts
Emitter resistance (per device): 0,5 ohms
Bias voltage per Emmiter resistor: 475 mv
Idle bias per device 950,0 mA
Speaker ohms 4 ohms
Results
Idle bias per device: 0,950 Amps
Total Amplifier bias (per rail)** 1,90 Amps
Total Amplifier bias (both rails) 3,80 Amps
Total Dissipation (per channel) 190,0 Watts
Dissipation per device pair at idle 47,5 watts
Class-A output: Peak 57,8 Watts peak
Class-A output: RMS 28,9 Watts RMS
Efficiency 15,20 %
Max Class-AB RMS (1 kVA Toroid p/CH 5% reg) 249,97 Watts RMS
Max Class-AB factor (100%=ideal load, Krell/565's = 91%, 555=70% regulation of rails at max power) 80 %
If the total dissipation is 190Watts it will be ok for my transformer. I have 2x450VA for both channells.
I have 8r0 loads so I'll have 57 Watts in Class A.
I calculate and:
Number of output devices: 2 Pairs
Voltage rails (per rail): 50 volts
Emitter resistance (per device): 0,5 ohms
Bias voltage per Emmiter resistor: 475 mv
Idle bias per device 950,0 mA
Speaker ohms 4 ohms
Results
Idle bias per device: 0,950 Amps
Total Amplifier bias (per rail)** 1,90 Amps
Total Amplifier bias (both rails) 3,80 Amps
Total Dissipation (per channel) 190,0 Watts
Dissipation per device pair at idle 47,5 watts
Class-A output: Peak 57,8 Watts peak
Class-A output: RMS 28,9 Watts RMS
Efficiency 15,20 %
Max Class-AB RMS (1 kVA Toroid p/CH 5% reg) 249,97 Watts RMS
Max Class-AB factor (100%=ideal load, Krell/565's = 91%, 555=70% regulation of rails at max power) 80 %
If the total dissipation is 190Watts it will be ok for my transformer. I have 2x450VA for both channells.
I have 8r0 loads so I'll have 57 Watts in Class A.
Sorry for my questions but I don't speak English well and I don't understand some thinks.
I hope that now I understand it well, so:
I will use 3 pairs transistors MJ15024/25 per channel,
bias @1,8A
supply voltage ~2x50V
I'll have 70W ClassA with loads 11r0 and 50W ClassA with loads 8R0.
When the amp will driven to maximum output for 8r0 loads = 50W, he will go into ClassAB. Am I right?
And another question:
Is it necessary to use matched pairs transistors? The outputs ones and drivers?
I hope that now I understand it well, so:
I will use 3 pairs transistors MJ15024/25 per channel,
bias @1,8A
supply voltage ~2x50V
I'll have 70W ClassA with loads 11r0 and 50W ClassA with loads 8R0.
When the amp will driven to maximum output for 8r0 loads = 50W, he will go into ClassAB. Am I right?
And another question:
Is it necessary to use matched pairs transistors? The outputs ones and drivers?
the driver devices are singles, they don't need matching.
The output devices are 3pair.
They need matching.
At higher current the emitter resistor gives some measure of reducing current mismatch.
At low current (bias value = 600mA) the devices must be matched for Vbe.
It is easier to hold both devices at the same Vbe and measure the error in collector current. Keep the Vce low to minimise Power and thus temperature.
Match as pairs using one of the pair as a fixed reference. Then find devices that have similar Ic. You can now match the selected devices in pairs until you find devices that are close.
Do not try to match the NPN to the PNP.
The output devices are 3pair.
They need matching.
At higher current the emitter resistor gives some measure of reducing current mismatch.
At low current (bias value = 600mA) the devices must be matched for Vbe.
It is easier to hold both devices at the same Vbe and measure the error in collector current. Keep the Vce low to minimise Power and thus temperature.
Match as pairs using one of the pair as a fixed reference. Then find devices that have similar Ic. You can now match the selected devices in pairs until you find devices that are close.
Do not try to match the NPN to the PNP.
