LME49810 on ebay
 

Hi guys, I noticed this for sale on ebay and was wondering if anybody may be able to fill me in on some specs for it

New - 300W LME49810 Mono channel Audio Amplifier Board - eBay Amplifiers, Audio, Video, Car Parts, Accessories, Cars, Bikes, Boats. (end time 15-Apr-10 05:13:34 AEST)

I would like to know what impedance it is meant to run at and what it could drive (ie lower impedances such as 2 ohms) if the supply voltage was lowered and if other circuit mods would be required to do so.

Cheers, Junglejuice

download the National datasheet.

Yeah I have that, so how using the above circuit determine what the power would be @ 4 ohms or 2 ohms?

The math is simply. Output power (RMS) will be about (supply voltage minus min. Vce for linear operation)/(2*Load), Load = 2, 4, 8 Ohm etc.

The supply is +/- 45 VDC. For 4 Ohm, you will get around (40*40)/(2*4) = 200 W (assume min Vce = 5 V).

Ok that makes sense, on the board it has 300w listed so using the formula above for a 2 ohm load would 400w, is this too much for these transistors to handle or do I need to lower the supply voltage to 40v to give an output of 300w @ 2 ohms? 400w @ 2 ohms would be better though....
Cheers

here is some extra reading on this subject.

http://www.diyaudio.com/forums/solid...-lme49810.html

expect the output transistors to have a dissipation capacity very approximately equal to 4times the maximum output power, i.e. 300W of maximum output power requires ~1200W of devices, 400W requires ~1600W of devices.

can you please state way? is it at A class operation?
sounds like way too much dissipation..

That seems to be a bit excessive to me, I was under the impression most amps run at about 50% efficiency including power supply so that would mean @ 400w the amp and supply would draw 800w, correct me if I am wrong.....

eg.
take a pair of 130W output devices. That's a total of 260W.
Divide by 4. This results in an estimate of maximum output power of approximately 65W.

Determine what load you intend to use. Let's make that 8ohm for this example.

Determine as a first guess the supply rails for 65W into 8r0.
Vrms = sqrt( max power * load resistance) = 22.8Vac
Vpk = sqrt( max power * load resistance * 2) = 32.2Vpk.

Allow about 6V for losses and PSU sag under full power.
You need ~+-38.2Vdc to get 65W into 8r0 from one pair of 130W devices.
This should be able to drive a real 8ohm speaker reliably if normal sized transformer and normal sized smoothing capacitance and normal sized heatsink are all used.
This will not drive a 4ohm speaker. That original set of assumptions applied to an 8ohm load.

If you had 2pair of 250W devices (total 1kW) and a 4ohm load, go through the same process and determine first guess for maximum output power and PSU requirement.