KSA50 AMP（New production Live in...）

[gxcnabin]

Married, that is my personal space.
1amp,Radiator is not big enough, during the test.

 

[Pocoyo]

Dude ,are you married? or is that your play room? My wife would kill me dead!

What final bias current did you settle on? 1amp ?

My wife will bury me alive

 

[Pocoyo]

Hi Gxcnabin & Tangmonster

What is function of VR1 & VR2 ?
For input (VR1 or VR2), how big ohm in there ?

How to bias this KSA ? Is that right we just
replace the fuse with multimeter, then turn vr1 or vr2 ???

Thank you guys

 

[AndrewT]

How to bias this KSA ?

read the KSA threads.
Krell specify the voltage across Re for correct bias.

 

[Pocoyo]

Those threads, another 8000++ post in 800++ pages

 

[Pocoyo]

read the KSA threads.
Krell specify the voltage across Re for correct bias.

output voltage depends on your +-Vrail and gain.

Assuming you have +-38Vdc and the losses from rail to max output voltage are 4v then you have 34Vpk to drive your load.

Divide by sqrt(2)=1.4142 to give rms output=24.04Vac (72w into 8R)

The gain of amp=r130/r129+1=22/1.5+1=15.67

Input required to produce 24.04V is 24.04/15.67=1535mV

From this you can insert any gain resistor values and max output voltage to determine the maximum input voltage.

 

[AndrewT]

measure the voltage across Re (output emitter resistor).
Krell tell what value of resistor to use and what voltage to set it to.

Post 386 seems to have nothing to do with your question, i.e. output bias

 

[Pocoyo]

measure the voltage across Re (output emitter resistor).
Krell tell what value of resistor to use and what voltage to set it to.

Post 386 seems to have nothing to do with your question, i.e. output bias

Hi AndrewT

Setting the bias is easy...turn the pot so the DC you measure across both the driver emitter resistors is at a minimum. Connect the driver board to the outputs, then if you can, slowly apply power to the whole shooting match, monitoring current draw, anything more than a few 10's of milliamps is bad, you'd want to turn everything off and check your wiring......If you can't do it slowly, then just turn it on...from a distance?

Anyway once you have all the parts connected, power applied and no smoke issuing from anything, attach a meter to measure the voltage across one of the 0r68 output emitter resistors (say Re1), and another to measure the output voltage relative to ground, and if you have 3 meters (and who doesn't?) attach another one to measure the total voltage across the driver emitter resistors r127/128...

Slowly turn the bias pot to increase the output idle current, you can also measure this as the voltage across the 25ohm driver emitter resistors, nothing happens at first, basically until the board output voltage reaches about 1.2v across r127 & 128, then the outputs start to turn on and you will read a voltage across Re1...keep increasing until you have about 0.2 volts across Re1...with 3 output pairs this corresponds to ~1A total idle current, you'd probably want to allow this to sit for a little while check temps, make sure the voltages across all the output emitter resistors are nearly the same, any big discrepancy should be investigated. Once you are satisfied that everything is OK, start increasing the bias until you see ~0.4v on Re1...as the temperatures on the sinks change this is going to move around, once it seems to have settled, adjust the other pot to reduce the DC offset to the lowest possible value (<50mv was easy to achive on all the boards I've made). This will wander as the temps change, so repeat the process a few times over the course of an hour or two...

(NOTE- .4v across Re1 is only valid with .68 emitter resistors, do not attempt this with lower value resistors-- compute the class A power you will get separately. With 4 output device pairs and .499 ohm emitter resistors 50 W class A into 8 ohms is about .22 V across the emitter resistors, into 4 its about .32V).

hmmm ... as i know pink mouse use 4pair output and now abin use 2pair output. It is different ???

 

[Pocoyo]

can anybody help me to set bias this amp
i dont have any electric background, thank you

 

[neychi]

can anybody help me to set bias this amp
i dont have any electric background, thank you

If your emitter resistors are 0.47 ohm and bias current, for example, is 1.8 A, then for 2 pairs of output transistors will be: 1.8 / 2 = 0.9 A, a voltage drop on each emittier resistor will be: 0,47 x 0.9 = 0.423 V, which is the value that you measured on the instrument 😉

 

[AndrewT]

those blue paragraphs seem to describe the process perfectly.
Now find out what voltage you should set up for the Krell Klone.

If you change from 2pair of MJ15003/4, then you must not assume that your modification will be as reliable as a Krell Klone.

 

[Pocoyo]

If your emitter resistors are 0.47 ohm and bias current, for example, is 1.8 A, then for 2 pairs of output transistors will be: 1.8 / 2 = 0.9 A, a voltage drop on each emittier resistor will be: 0,47 x 0.9 = 0.423 V, which is the value that you measured on the instrument 😉

Thank you AndrewT & Neychi



if Emitter Resistor 0,47 ohm
bias current 1.5A so a voltage drop in each RE will be 0.47 x 0.75 = 0.3196 V
bias current 1.9A so a voltage drop in each RE will be 0.47 x 0.95 = 0.4465 V

if Emitter Resistor 0,68 ohm & bias current 1.5A
bias current 1.5A so a voltage drop in each RE will be 0.68 x 0.75 = 0.51 V
bias current 1.9A so a voltage drop in each RE will be 0.68 x 0.95 = 0.646 V

maybe i would try with 2 Pair MJW 3801 & 1302 for each channel

The next question, how big power i get if RE = 0.646V with Bias 1,9A ?
or what should i do to get 50 W power amplifier ?

Thank you before

 

[neychi]

@Pocoyo, for 50 W in class A you will need at least 1,76 A.

 

[Pocoyo]

i realy hope now, those conrad can manage the temperature

btw Neychi, what happen if i give more or less voltage input ?
For example : RE 0.47ohm X Bias 1.8A must be 0.423 V
what happen if i give 0.5 V or 0.35 V ? clipping, smoke or explode maybe

 

[neychi]

i realy hope now, those conrad can manage the temperature

btw Neychi, what happen if i give more or less voltage input ?
For example : RE 0.47ohm X Bias 1.8A must be 0.423 V
what happen if i give 0.5 V or 0.35 V ? clipping, smoke or explode maybe

More or less voltage on emitter resistors will result with more or less bias current. With voltage drop of 0,5 V you will have 2,13 A bias and with 0,35 V it will be 1,49 amps. How much dissipation output transistor can handle depends on its power (SOA) and supply voltage. For a supply voltage of +/-38 V and the bias current of 2.13 A you will have dissipation per channel of 162 W which is a pretty high amount, and for that you will need a big heatsink. If you have 2 pair of output transistors, dissipation will be approximately 40 W on each transistor. The temperature of 60-70 ° C at which the crystal transistor is heated, power of output transistor will be considerably less than the declared 250W (25 °) for MJ21193 / 4 or 200W for MJL plastic versions. The technical characteristics data for each transistor contain a graph of power dissipation vs. temperature dependence. In your case, smaller bias seems like better solution.

Best regards

Dean

 

[neychi]

@Pocoyo, for 50 W in class A you will need at least 1,76 A.

50W / 8 ohm of course 😉

 

[tangmonster]

Hi Pocoyo

was on a long weekend trip with my lancia so did no get your private message until today.

It seems though that you have enough information to set the bias current?

It seems that with my 0.33c/W heatsinks that i might not be able to get more than 1Amp bias (2amp class A). for 50W (into 80hms) you need about 1.9A per rail (3.8amp class a)

i am currently experimenting with a fan system to see how loud it will be else i will settle for the lower bias.

With both my variable resistors the bias started coming up around lower than 1k. The pot is a 5k item. The other pot is just to set dc output to 0 ,as far as I understand.

Don't be fooled by a 40C heatsink after 10 or so minutes you need to run it for about 30 minutes to really see the temp.

 

[tangmonster]

FYI

http://www.diyaudio.com/forums/solid-state/171837-got-huge-heatsinks-amp-build-3.html

AndrewT gave a very nice explanation on how this push pull class A amplifier works in post 24. This should clear up a few questions on how this "overbiased" class AB amplifier works as class A up till a certain current until it leaves class A.

There is one mistake though I think:

"The +ve supply rail at this instant will pass 1.5A to the output stage.
The -ve supply rail at this instant will pass 1.5A from the output stage. The difference is 1A and it all goes out to the 8r0 load."

Should be:

"The +ve supply rail at this instant will pass 1.5A to the output stage.
The -ve supply rail at this instant will pass 0.5A from the output stage. The difference is 1A and it all goes out to the 8r0 load."

This is also some good reading:

http://www.passlabs.com/pdfs/articles/leaving_class_a.pdf

 

[tangmonster]

I am currently busy with the design of my enclosure. I cannot see a way using this design to the fullest without a fan.

To keep noise down I will use 1 120 x 120mm fan inside the box. I used a cheap pc fan I had for testing my idea. without any fan the heat sink go to 60C in about 20 or so minutes and settles there. I want this lower.

I want some airflow to flow over the heat sink fins to get the best result.

in the attached pictures you can see the 12cm fan with the front and rear panel bent around the corners with a 10mm(roughly) gap at the edges so air HAS to pass through fins around the corner and then out the back through the fan.

The air will enter in the front two gaps and the rear two gaps.

I tested my idea with LONG flaps nearly closing the heatsink and also short flaps.

The result.

long flaps: temp drops from 60C and settles around 42C

short flaps: temp drops from 60C and settles around 40C

I do realize that i have many air leaks around all the edges with this cardboard test. But if i am correct in a aluminum enclosure with less leaks the temp will only get better.

I think if settle for about 30mm flaps things should be fine.
The gap between the heat sink side and the front and back panel will be about 4mm. I will calculate area to make sure the intake and exhaust area is about the same size.

I can possible make the fan blow in rather than suck out?

 

[firestorm]

Tangmoster, I believe you will get better cooling, if you turn the heatsink around, so that it the fins point inwards. So it looks something like this
http://www.diyaudio.com/forums/solid-state/171837-got-huge-heatsinks-amp-build-4.html

scroll down to bottom.