F5 power amplifier

[Nelson Pass]

Yes, although it will not track quite as well. In fact you can use no
thermistor at all, as long as you don't mind the extra effort of
adjusting the circuit over a longer period of time and the additional
warm up time when you turn it on.

 

[pro]

Thanks mr. Pass.
At this point I go without (I'm very patient )
Regards

 

[npapp]

PS Question

I have a 600VA transformer with 20v secondaries. Will this work with the F5 as long as I use higher voltage capacitors in the PS? Is this too much voltage (27-28v DC) for the jfets?

Sorry to bump the question but I really want to try the bigger transformer.

 

[bobodioulasso]

Can't you unwind a little bit?

 

[EUVL]

> Is this too much voltage (27-28v DC) for the jfets?

Yes, but then you can now use a Zen-type regulator (actually MOSFET follower regulator) to lower it to 24V.

http://www.passdiy.com/pdf/zen-ver3.pdf

Just perfect.




Patrick

 

[labjr]

> Is this too much voltage (27-28v DC) for the jfets?

Yes, but then you can now use a Zen-type regulator (actually MOSFET follower regulator) to lower it to 24V.

http://www.passdiy.com/pdf/zen-ver3.pdf

Just perfect.




Patrick

Unless you need the transformer for something else, I would carefully and evenly unwind some of the secondaries.

 

[npapp]

Thanks for the confirmation that I shouldn't use the higher voltage and also for the suggestions. I perfer not to unwind the secondaries so I may give the regulator a try first.

Patrick have you built that reg and if so can post or email me a pic of your layout?

Thanks Nick

 

[bobodioulasso]

A regulation shoes a 7 to 8 volts drop out. In your case 3 or 4 volts would be nice.
You can use the Zen V3 PSU without the zeners.
Or a capacitance multiplier.

 

[EUVL]

> Patrick have you built that reg and if so can post or email me a pic of your layout?

I have built a version of this :

http://www.diyaudio.com/forums/showthread.php?s=&threadid=96737


Patrick

 

[bobodioulasso]

http://www.diyaudio.com/forums/showthread.php?s=&threadid=142114
http://sound.westhost.com/project15.htm

 

[tinitus]

Trying to learn
Probably looks awful, but its fun
Recommendable to try it out

 

[Renron]

I'm very new to electronics, and have no formal education with circuits. So, if I ask a question in the wrong format, I know you'll understand.

I'm trying to understand how the the wattage (normally) increases when the Ohm load (speakers) decreaces.

Example: From the F5 PDF from Papa;

When a positive voltage appears at the Gates of Q1 and Q2, it makes the current through
Q1 increase and the current through Q2 decrease. The resulting voltages across R3 and
R4 make the current through Q3 increase and the current through Q4 decrease. This
makes the output voltage go positive. As the positive input voltage increases, you
approach the point at which Q3 is conducting 2.6 amps and Q4 is conducting 0 amps –
and all of the 2.6 amps goes through the loudspeaker.


OK, so far so good, I understand how this works.


The power of 2.6 amps into 8 ohms is I^2 * R, or 2.6 * 2.6 * 8 = 54 watts. This is the peak
value, and the nature of an undistorted sine wave is that the peak wattage is twice the
average, so this circuit would operate 27 watts average Class A into 8 ohms. At currents
above 2.6 amps one of the transistors will shut off, leaving the other to continue to
increase beyond the 2.6 amps in what is known as Class AB.


OK got the 54 Watts output, so would the current (I) double if the Ohms were 4 rather than 8? Or is the current ^2 (squared)?

Trying not to get lost .......

Thanks for the help.
Ron

 

[bobodioulasso]

let V the available voltage.
Here about 20v

I=V/R (Ohm's law)

When V constant and R becoming R/2


I becomes 2*I

 

[jacco vermeulen]

this circuit would operate 27 watts average Class A into 8 ohms.

A bit less than 27W, actually.
The last little bit doesn't count because that's part of the crossover region. That is why Papa set the ruler at 2.6 and not 2.5A for 25W Class A in 8 Ohm.

The amp is in Class AB beyond an output current of 2.6A : one output device conducting and the other one shut.
The output current can increase with lower impedance loads up till the voltage level that the power supply can sustain.
The power supply is 24Vdc, it can deliver 20 Volts in a 4 Ohm load, so the continuous Class AB power level in 4 Ohm will be 50W.

Class A power is limited to the set bias current : 2.6A
Dropping the load impedance does nothing to alter that peak output current, beyond 2.6A out one of the output devices shuts off and the ampy no longer operates in Class A.
2.6 times 2.6 times 4 = 27W peak = 13.5W continuous Class A power in 4 Ohm.
(in reality more like 12.5W, because of the crossover zone story )

 

[Renron]

Bobo,

Thanks for confirming what I thought I knew, the current ( I ) is directly effected by the resistance ( Ohms )

I becomes 2*I = Sweet!


Thank you,

Ron

 

[Renron]

Above 2.6 A is class AB. Papa wrote that one could not tell the difference when listening that the amp had changed from Class A to Class AB. ( I could find it if you like)
Don't flame me here.....

Why do we build heat monsters that are primarily Class A?
Less Distortion than AB ?

Thanks for the above lesson Jacco.

Ron

 

[cviller]

Above 2.6 A is class AB. Papa wrote that one could not tell the difference when listening that the amp had changed from Class A to Class AB. ( I could find it if you like)
Don't flame me here.....

Why do we build heat monsters that are primarily Class A?
Less Distortion than AB ?

Thanks for the above lesson Jacco.

Ron

Try lowering the bias current of your F5 to around 10-50mA, then you'll hear how cross over distortion sounds. This amount of bias current is quite common in normal class AB amps, but then they use plenty of gain and feedback to repair the bad sound.

 

[jacco vermeulen]

Why do ?

The Mastah has all the answers, read his article on Class A biasing : HERE

The whole amp scene revolves around the concept of enhancing the amplitude of a signal without distorting it, there's a number of ways to reduce distortion levels.
One is making the gain stages more accurate by linearising them, examples of this would be to add current sources, current mirrors, use cascoding, local feedback.
Two is correcting the non-linear output through manipulation of the amplifier's gain, using global feedback from output to input.
Three is the easiest and oldest way, Class A biasing.

All discrete preamplifiers are mainly focused on Class A operation.
Virtually nobody complains about that because the heat disposal level of most preamplifiers is modest.

Amusing thing : there's an optimal bias level of the output devices of Class AB power amps for the lowest distortion level, mainly determined and set by Mr Boltzmann.
Also, the most accurate Class AB amps require the lowest possible value emitter resistors that will still keep current sharing of the power devices equal.
Both criteria result in 260mA bias current for each output device used.
Use enough output devices, and the heat/dissipation level of a Class AB power amp will become serious business as well.

Example is the Parasound Halo JC-1, originally it had 0.1R emitter resistors.
Optimally biased, the thing burned as much energy as it is rated for in 8 Ohm.
Still, only half of the heat level it would require for full Class A mode.
Duhh, did i say HALF ?

 

[Renron]

Chris,
I'll try the lowered bias.....after I build my F5.....from your boards I just ordered. LOL
Thanks, it help me understand Class A / Class AB differnence.

Jacco,
"Amusing thing : there's an optimal bias level of the output devices of Class AB power amps for the lowest distortion level, mainly determined and set by Mr Boltzmann."

Mr. Boltzmann (distribution function) > "commonly used in statistical mechanics in order to determine the speeds of molecules"

OK, now your just showing off!

"Still, only half of the heat level it would require for full Class A mode.
Duhh, did i say HALF ? "

In your subtle way of telling me that in Class AB one of the devices shuts off from Class A @2.6A, so ONE device is 1/2 the heat of operating 2.

Right?

Thanks for the baby steps guys. I appreciate it.
I also Googled "descete amplifier"-------seperate devices for each channel.

Thank you both,

Ron

 

[AndrewT]

half the heat requires half the current at the same supply voltage.

If the Halo JC1 were the same voltage as the F5 then 1.3A of bias would generate half the heat.

0r1 emitter resistors with 26mV across each of them would pass 260mA through each output device.
5pairs would therefore have 1.3A of total bias.
6pairs and Vre~22mV would require dissipation of the same heat.