900W H-class PA Amp with Limiter

[gantex]

Waylene doesnt listen to ADVICE given

maybe u alright andrew, and i see he is new one on electronic. Just learn and practise dont have anything basic of electronic.

 

[waylene]

Guys!! I fixed problem as you told me to do!!! Thanks for all your help! I check PCB over and over again. I saw one of my track, does not connected to my 1934 transistor, I solder track and start up amp with 220v ac bulb in series, there my one channel play 100% rite, with out any problems. But I have to thank all of you who help me. @ Gantex, yes you rite I'm new here at DIY. Guys I must tell you, with tears in my eyes, my biggest wish and dreams, was to be like your guys. But according my Parents, broke up, I'm the eldest one who have to leave school, to take care of my Mum and younger brothers and sister. That's the main reason why I can't go further to study for what you guys are now. But I appreciate much of every one of you who help me solve my problems with Electronics. Sometime I feel not to disturb you guys, but I'm excited to see what I do with my hands, must work. But still thanks you guys.

From a new learner.

Warren.

 

[kaylan]

friends, how do I increase the power amplifier H900? not change supply source 120-60-0-60-120dc.
I can do this by adjusting the bias to make H900 near a class B?
How do I adjust biasing? doing this will have more power?

Thanks!

 

[AndrewT]

adjusting the bias does not change the maximum power available to the load.

The supply rails and the losses from supply rails to the load, determine the maximum voltage at the load.

 

[waylene]

Hello Andrew!

 

[black]

can do tricks such power thermistor high by tempeatur heat sink Bias final stage to go to class b thus increasing efficiency Sorry for my English using google translate

 

[sandeep kayarkar]

My 3rd H900

Almost done.

 

[waylene]

nice!

 

[apexaudio]

Almost done.

Nice work,
Regards

 

[waylene]

Hi can some one explain me more, what is the function of the ldr and what does it mean. Because I saw that on H900 and B500.

 

[jaagut]

Hi can some one explain me more, what is the function of the ldr and what does it mean. Because I saw that on H900 and B500.

in NSL-32, there is LDR & LED inside the component.

LDR is light dependent resistor;
LED is used to produce light for LDR

LDR reacts to light:
more light, less resistance
less light, more resistance

in B500 or H900, when output signal is high, LED will glow more, LDR will decrease resistance, and because LDR forms a voltage divider it will decrease input signal.

imagine that this NSL-32 is similar to an auto-brake in car, when car runs faster, the brake will activate, and will slow down the car.

 

[wattaniya]

hi apex sir,
can u tell me what is the output ac volt for 3200w?

 

[AndrewT]

work it out from P = IV = V²/R = I²R
V=sqrt(P*R)

 

[wattaniya]

i dont know how to calculate, can u?
powersupply 95vdc dual.

 

[AndrewT]

V = sqrt(P*R) = sqrt (3200 * 4ohms) = sqrt(12800) = 113Vac.
Peak = sqrt(2) * 113Vac = 1.414*113 = 160Vpk

You need supply rails of ~+-180Vdc.

Buy a ready made PA amplifier.

 

[wattaniya]

thanks andrew for the reply,but this is 3200w * 1ohm.

 

[AndrewT]

Do the arithmetic.

 

[wattaniya]

V = sqrt(P*R) = sqrt (3200 * 4ohms) = sqrt(12800) = 113Vac.how come 113vac?

 

[AndrewT]

work it out from P = IV = V²/R = I²R
V=sqrt(P*R)

V = sqrt(P*R) = sqrt (3200 * 4ohms) = sqrt(12800) = 113Vac.
Peak = sqrt(2) * 113Vac = 1.414*113 = 160Vpk

You need supply rails of ~+-180Vdc.

Buy a ready made PA amplifier.

V = sqrt(P*R) = sqrt (3200 * 4ohms) = sqrt(12800) = 113Vac.how come 113vac?

Go back and insert 113Vac into the Power equation.

P = V²/R = 113²*4 = 3192W

 

[wattaniya]

ok ,i got it andrew,

so for 3200w *1ohm load -
V = sqrt(P*R) = sqrt (3200 * 1ohms) = sqrt(3200) = 56.5vac on output. is this correct andrew?
i think 95vdc power supply is ok.