According to the trusty spreadsheet, we get the following: Dissipation remains at 200 watts for 3.5 ohm load and 50 watts class-A, not 310 watts...(I am scratching my head 😕 )
Inputs
Number of output devices: 3 Pairs
Voltage rails (per rail): 37.5 volts
Emitter resistance (per device): 0.68 ohms
Bias voltage per Emmiter resistor: 606 mv
Idle bias per device 891.2 mA
Speaker ohms 3.5 ohms
Results
Idle bias per device: 0.891 Amps
Total Amplifier bias (per rail)** 2.67 Amps
Total Amplifier bias (both rails) 5.35 Amps
Total Dissipation (per channel) 200.5 Watts
Dissipation per device pair at idle 33.4 watts
Class-A output: Peak 100.1 Watts peak
Class-A output: RMS 50.0 Watts RMS
Efficiency 24.95 %
Max Class-AB RMS (1 kVA Toroid p/CH 5% reg) 180.78 Watts RMS
Inputs
Number of output devices: 3 Pairs
Voltage rails (per rail): 37.5 volts
Emitter resistance (per device): 0.68 ohms
Bias voltage per Emmiter resistor: 606 mv
Idle bias per device 891.2 mA
Speaker ohms 3.5 ohms
Results
Idle bias per device: 0.891 Amps
Total Amplifier bias (per rail)** 2.67 Amps
Total Amplifier bias (both rails) 5.35 Amps
Total Dissipation (per channel) 200.5 Watts
Dissipation per device pair at idle 33.4 watts
Class-A output: Peak 100.1 Watts peak
Class-A output: RMS 50.0 Watts RMS
Efficiency 24.95 %
Max Class-AB RMS (1 kVA Toroid p/CH 5% reg) 180.78 Watts RMS
Arif,
voltage of the rails will drop at such high output currents.
More like 140 in 3.5, 190 in 2 Ohms.(for a PS with very meaty hips)
voltage of the rails will drop at such high output currents.
More like 140 in 3.5, 190 in 2 Ohms.(for a PS with very meaty hips)
Hi Kamps,
you are putting the wrong values into the calculator.
It still shows 50W into 3r5.
What I am referring to is staying in ClassA until nearly at peak output voltage is delivered which I mentioned in my first post.
And the regulation I used was also 5%. But that is not our difference. It's whether it is valid to call an amplifier ClassA when it can only deliver 50W (-4.1db) rather than 120W into the specified load. This would be a high bias ClassA/B.
The original Klone is about 50W (-1.1db) cf 65W into 8r.
As an aside, the last line shows 5% reg and 180W.
Can you explain how this model reflects expectations from the real Klone?
you are putting the wrong values into the calculator.
It still shows 50W into 3r5.
What I am referring to is staying in ClassA until nearly at peak output voltage is delivered which I mentioned in my first post.
And the regulation I used was also 5%. But that is not our difference. It's whether it is valid to call an amplifier ClassA when it can only deliver 50W (-4.1db) rather than 120W into the specified load. This would be a high bias ClassA/B.
The original Klone is about 50W (-1.1db) cf 65W into 8r.
As an aside, the last line shows 5% reg and 180W.
Can you explain how this model reflects expectations from the real Klone?
I reversed engineered the input to get 50 watts class-A into 3.5 ohms. Which is more like 112 watts class-A into 8 ohms, however the amp cannot push 112 watts because it is limited by the Vrails. (and the spreadsheet does not cater for that).
However in both cases the dissipation remained at 200 watts or so with total bias at 5.35 amps (2.67 per rail).
The 180 watts at 3.5 ohm figure was just based on an input field I have that multiplies the result with a constant... it is a rough estimate. Looking back, i think at 3.5 ohms, a more realistic output would be about 150 watts with those parameters.
K-
However in both cases the dissipation remained at 200 watts or so with total bias at 5.35 amps (2.67 per rail).
The 180 watts at 3.5 ohm figure was just based on an input field I have that multiplies the result with a constant... it is a rough estimate. Looking back, i think at 3.5 ohms, a more realistic output would be about 150 watts with those parameters.
K-
Hi,
yes I know the spreadsheet shows impossible numbers at high bias currents. I reported it to the author last year. I even sent him a reworked version that looked identical to the user but eliminated the wattage error. He chose not to post the corrected worksheet.
yes I know the spreadsheet shows impossible numbers at high bias currents. I reported it to the author last year. I even sent him a reworked version that looked identical to the user but eliminated the wattage error. He chose not to post the corrected worksheet.
Hi Kamps,
is your speadsheet corrupted?
Iq=5.35A & +-Vrail=37V5 gives a dissipation of Pd~=401W
Try downloading a new version and use tools protection to prevent inadvertant errors being typed in. You will need to select the entry fields then format/cells/protection/unlocked before protecting. Don't use a password!
BTW, 5.35Aq is a bit high indicating 37.45Vpk at the output terminals. But this would give an allowance for speaker Re being below the 3.5ohm nominal impedance
is your speadsheet corrupted?
Iq=5.35A & +-Vrail=37V5 gives a dissipation of Pd~=401W
Try downloading a new version and use tools protection to prevent inadvertant errors being typed in. You will need to select the entry fields then format/cells/protection/unlocked before protecting. Don't use a password!
BTW, 5.35Aq is a bit high indicating 37.45Vpk at the output terminals. But this would give an allowance for speaker Re being below the 3.5ohm nominal impedance
AndrewT said:
Before I attempt to change the spreadsheet I need to clarify what you mean by Iq. and you arrival at the 400w Pd.
My calcs were either to multiply the bias current of (either ) the NPN or PNP bank by 2x rails = 200w i.e 2.67 x (37.5x2) = 206w
OR
Multiply the total bias current of both banks of PNP/NPN devices i.e. 5.35A by a single rail of 37.5 = 206w Pd.
Are these assumptions incorrect?
is Iq the total bias of one or both halves of the OP devices?
AndrewT said:
Actually, at 5.35 amps (2.675 amps bias per rail) the amplifier is in class A up to 100 watts in 3.5 Ohms. Think about the definition of class A.
jacco vermeulen said:
Since I am not an EE wanna be, I will say I am confused.
At 5.35 amps (2.675 per rail) I would have thought the amp is 100w class A at 8 ohms and half that (50w class-A) at 4 ohms.
The way I was told: (By chocoholic and NP) is that that the class-A operation is based on bias current not Vrail (though it is limited by Vrail).
So class-A Peak output is Iq squared X load. or 5.35X5.35X3.5= 100.2 watts and RMS output is Iq squared X 0.70707 X 3.5 = 50 watts RMS.
total dissipation is total Bias (5.35) X Vrail (37.5) =200.5
So what you are saying is that we differ in definition of class-A, you saying it is peak, I saying it is RMS. Yes/ No?
Hi,
this is wrong
ClassA output power = Iq*Iq/2Rload.=(Iq*0.7071)squared*Rload.
eg.
@ Iq=2A675 and Rload =3r5. PclassA=50W
Hi Kamps,
you almost quoted the formula correctly, but back to our previous discussion.
I can see where part of the confusion is. The spreadsheet reports two different quiescent current values. The first one is correct and not misleading. The current reported is a real entity that can be measured. The doubled value for both rails is of dubious value and does not exist in the quiescent circuit of the KSA. I cannot see any way of measuring this imaginary current.
Based on this, your method
The method allows you to design and select components for the build and on completion to check the conditions that you have built.
The other part of our discussion is whether the amp can be called ClassA if it only delivers 38.5% of it's rated power into the specified load.
The original KSA50 is specified as 50W into 8r for full ClassA and this is fully justified. It also happens to be able to drive slightly more than this into 8r as a ClassAB amplifier (I believe this is between 60W and 70W).
Does anyone want to jump into either camp? Or leave us to put on the gloves.
I wonder if the doubled value comes from a slight confusion on the part of the spreadsheet designer.
A push pull classA amp (as the KSA50 is) with no signal on the input or output and with no DC offset on the output has both the upper and lower halves balanced i.e. the same Iq in both halves.
As the input/output signal is increased the balance is lost and the current in one half progressively increases while the other half decreases. The difference in current between the two halves is delivered to the load.
This continues as the signal is increased until one half just cuts off (zero current) and at this point the other half is delivering double the initial (quiescent) value. This doubled value is delivered to the load because none of it is sunk(or sourced ) into the other half.
The maximum value of ClassA load current (=2*Iq) can be used to predict the maximum voltage and maximum power into the chosen load.
Max ClassA peak voltage = 2*Iq *Rload (2*2.675*3.5) = 18.725Vpk
Max ClassA peak power = 2*rms based power = (2*IQ)squared*Rload = 100W (there's the Jacco number).
Max ClassA rms based power =Iqsquared * 2 * Rload = (2*Iq)squared * Rload / 2 = 50W
You can see how the 2Iq is used in both the peak voltage and power formulae. In single ended the times 2 does not apply, the amp simply clips and distorts. However in a push pull amp the same hard limit due to clipping does not exist and if the power supply can supply the extra current demanded by driving a higher voltage into the load the amp then reverts to ClassAB where only one half is supplying (sinking or sourcing) current to the load. This overhead is what allows KSA50 to deliver that extra 10W or 20W in ClassAB mode.
Sorry to all for the length of this.
Appologies to all who already understand this ClassA vs ClassA/B stuff.
this is wrong
The correct formula is
ClassA output power = Iq*Iq/2Rload.=(Iq*0.7071)squared*Rload.
eg.
@ Iq=2A675 and Rload =3r5. PclassA=50W
Hi Kamps,
you almost quoted the formula correctly, but back to our previous discussion.
I can see where part of the confusion is. The spreadsheet reports two different quiescent current values. The first one is correct and not misleading. The current reported is a real entity that can be measured. The doubled value for both rails is of dubious value and does not exist in the quiescent circuit of the KSA. I cannot see any way of measuring this imaginary current.
Based on this, your method
the voltage you have used is also a real entity and you can easily measure it.
The method allows you to design and select components for the build and on completion to check the conditions that you have built.
The other part of our discussion is whether the amp can be called ClassA if it only delivers 38.5% of it's rated power into the specified load.
The original KSA50 is specified as 50W into 8r for full ClassA and this is fully justified. It also happens to be able to drive slightly more than this into 8r as a ClassAB amplifier (I believe this is between 60W and 70W).
Does anyone want to jump into either camp? Or leave us to put on the gloves.
I wonder if the doubled value comes from a slight confusion on the part of the spreadsheet designer.
A push pull classA amp (as the KSA50 is) with no signal on the input or output and with no DC offset on the output has both the upper and lower halves balanced i.e. the same Iq in both halves.
As the input/output signal is increased the balance is lost and the current in one half progressively increases while the other half decreases. The difference in current between the two halves is delivered to the load.
This continues as the signal is increased until one half just cuts off (zero current) and at this point the other half is delivering double the initial (quiescent) value. This doubled value is delivered to the load because none of it is sunk(or sourced ) into the other half.
The maximum value of ClassA load current (=2*Iq) can be used to predict the maximum voltage and maximum power into the chosen load.
Max ClassA peak voltage = 2*Iq *Rload (2*2.675*3.5) = 18.725Vpk
Max ClassA peak power = 2*rms based power = (2*IQ)squared*Rload = 100W (there's the Jacco number).
Max ClassA rms based power =Iqsquared * 2 * Rload = (2*Iq)squared * Rload / 2 = 50W
You can see how the 2Iq is used in both the peak voltage and power formulae. In single ended the times 2 does not apply, the amp simply clips and distorts. However in a push pull amp the same hard limit due to clipping does not exist and if the power supply can supply the extra current demanded by driving a higher voltage into the load the amp then reverts to ClassAB where only one half is supplying (sinking or sourcing) current to the load. This overhead is what allows KSA50 to deliver that extra 10W or 20W in ClassAB mode.
Sorry to all for the length of this.
Appologies to all who already understand this ClassA vs ClassA/B stuff.
I believe the Lang 20 watts mosfet amplifier was thermally designed to run full class A in 4 Ohms.
The ML2 even ran full class A in lower impedance loads.
Looking at peak output voltage, that would even make the original KSA a high bias Class AB amplifier in 4 Ohms.
If only peak class A output voltage counts the NUTTR and me are the only ones here with a full class A amplifier.
The ML2 even ran full class A in lower impedance loads.
Looking at peak output voltage, that would even make the original KSA a high bias Class AB amplifier in 4 Ohms.
If only peak class A output voltage counts the NUTTR and me are the only ones here with a full class A amplifier.
Hi Jacco,
your observation is correct.
I do not have access to their literature but I believe they claimed 50W ClassA into 8r.
My Krell Klone will be biased to 3A to allow ClassA into 6r (the minimum DCR of my speakers). The output stage and PSU components selected accordingly. I hope this will ensure it stays in ClassA for any music I care to play (too loudly).
Yes, that's 108W ClassA into 6r. I will continue to think of my Klone as an 80W amp (80W ClassA into 8r).
Are my assumptions correct? i.e. it will never leave ClassA when driving 8ohms speakers of up to 60 degree phase angle.
your observation is correct.
BUT, did Krell specify it as ClassA into 4r? I think not.
I do not have access to their literature but I believe they claimed 50W ClassA into 8r.
My Krell Klone will be biased to 3A to allow ClassA into 6r (the minimum DCR of my speakers). The output stage and PSU components selected accordingly. I hope this will ensure it stays in ClassA for any music I care to play (too loudly).
Yes, that's 108W ClassA into 6r. I will continue to think of my Klone as an 80W amp (80W ClassA into 8r).
Are my assumptions correct? i.e. it will never leave ClassA when driving 8ohms speakers of up to 60 degree phase angle.
AndrewT said:
Correct, no intent to dispute the numbers.
The way i perceive it is that an amplifier is called a full class A one if it fully does so into 8 Ohms.
If i were in Rick's shoes i'd consider raising the bias for 3.5 Ohms use and dropping the rail voltages as that much power is not needed for the 92 dB/W/M efficiency of the Phoenix LS's.
I didn't manage to get 7 amps bias at 35Vdc rails without turbine fan noise, the heat is just too much, 5 amps looks obtainable.
AndrewT said:
OK, I tweaked the spread sheet and came up with the same numbers as before: (but changed the formulae).
It is understood that the spreadsheet will not cater for the excess of power output (in excess of the Vrails) but will calculate the class-A output for a given bias current for a given rail.
Question is: How is the dissipation calculated? this is what I have:
Inputs
Number of output devices: 3 Pairs
Voltage rails (per rail): 37.5 volts
Emitter resistance (per device): 0.68 ohms
Bias voltage per Emmiter resistor: 606 mv
Idle bias per device 891.2 mA
Speaker ohms 3.5 ohms
Results
Idle bias per device: 0.891 Amps
Total Amplifier bias (per rail)** 2.67 Amps
Total Amplifier bias (both rails- Phantom figure) 5.35 Amps
Total Dissipation (per channel) 200.5 Watts
Dissipation per device pair at idle 33.4 watts
Class-A output: Peak 100.1 Watts peak
Class-A output: RMS 50.0 Watts RMS
Max Class-A peak voltage 18.7 volts
Efficiency 24.95 %
Max Class-AB RMS (1 kVA Toroid p/CH 5% reg) 150.65 Watts RMS
File attached for reference of formulae:
K-
Hi Kamps,
all your numbers look OK. The spreadsheet must still be right.
Now increase the bias voltage to 939mV.
This will give Iq=4A14
Vpk=29Vpk (This allows a margin for Vrail droop, cable losses, transistor loss and Re Vdrop. And probably a bit to spare just like the original KSA50).
PoutClassA=120W.
What dissipation is it reporting?
all your numbers look OK. The spreadsheet must still be right.
Now increase the bias voltage to 939mV.
This will give Iq=4A14
Vpk=29Vpk (This allows a margin for Vrail droop, cable losses, transistor loss and Re Vdrop. And probably a bit to spare just like the original KSA50).
PoutClassA=120W.
What dissipation is it reporting?
AndrewT said:
Dissipation = 310.7 watts.
Class-A output is 120.1 watts into 3.5 rms, rest is clas-AB...
Which is the same as your original dissipation number, however My assumption was that the guys wanted at least 50 watts class-A (into 3.5 ohms) and not 120 watts class-A.
If he wants 120watts class-A, then dissipation is 310 watts.
Also, Vrail droop must be factored in the input, i.e. 37.5v must be fully loaded not unloaded vRail, otherwise you will over state your required class-A number... I guess thats where we are differing.
are there still any PCBs left to be had? the KSA50 seems like a really cool amp. i have a few questions though:
I can solder but i have never DIY'd anything. how hard would a KSA50 be to build?
If i built a KSA50 it would be part of a 19-year-old university student's(mine) dual-purpose audio setup...i want it to sound great for personal listening between myself and my 4 housemates/friends, but we also like to throw loud drunken parties on weekends. My firmly held opinion is that great sound makes for a great party, and i've already proven my point to everyone i know by powering my friend's pair of B&W DM220s with my 35wpc Superscope receiver. i've heard the argument that the environment being a party, nobody cares about sound too much...i don't buy that. i.e., the amp will not be part of a "listening room" where the speakers are toed in for a sweet spot on a sofa and someone sits in the middle with a cup of brandy. is there any reason a big class A amp is a bad idea here? i am definitely fan cooling it.
I don't want to spend money on two transformers since i have to build a preamp too. has anyone done dual mono on a KSA50 using a single transformer with dual secondaries? or should i just do conventional stereo?
thanks.
I can solder but i have never DIY'd anything. how hard would a KSA50 be to build?
If i built a KSA50 it would be part of a 19-year-old university student's(mine) dual-purpose audio setup...i want it to sound great for personal listening between myself and my 4 housemates/friends, but we also like to throw loud drunken parties on weekends. My firmly held opinion is that great sound makes for a great party, and i've already proven my point to everyone i know by powering my friend's pair of B&W DM220s with my 35wpc Superscope receiver. i've heard the argument that the environment being a party, nobody cares about sound too much...i don't buy that. i.e., the amp will not be part of a "listening room" where the speakers are toed in for a sweet spot on a sofa and someone sits in the middle with a cup of brandy. is there any reason a big class A amp is a bad idea here? i am definitely fan cooling it.
I don't want to spend money on two transformers since i have to build a preamp too. has anyone done dual mono on a KSA50 using a single transformer with dual secondaries? or should i just do conventional stereo?
thanks.