LME49810 - a new cousin for LM4702

panson_hk said:
For current-limited regulator, you proposed a shunt regulator? The chip's Iq is 11/13 mA. If the CCS is set at 50 mA, a shunt transistor will consume about 39 mA x 60 V = 2.34 W in idle mode. Will it be too much?
Hi,
using a transistor with 10W 100MHz to 200MHz hFE100 to 150 would dissipate your 2.34W using a 10C/W sink at ambients upto 30degC. and Tj still below 90degC.
I have not yet implimented either series or shunt front end regulation so have no opinion which sounds better.
The inherent current limit of CCS + shunt is attractive.
The few papers I've read seem to suggest that noise suppression is better with a good shunt.
 
LME49810 Audio AMP (Samples)

LME49810 Audio AMP (Samples)
 

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Interesting circuit! And I like your design monkey. Nice indeed. What kind of output transistors do you think would be best to use?

Im new on class-a and have some questions on the output stage. How do you increase power. Is it correct as I did in the attached file? (Take no notice on the transistors I only picked a random one.)

Also, how do you callculate output power? I guess it has to do with the MOSFETs or?

Best regards...
 

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Your distortion plot seems to indicate that it actually already IS current limited. :D

If we assume it is current limited at 50mA it would be able to provide about 35mA RMS to the load before clipping and high distortion. The 100 ohm load curve shows distortion coming in slightly below 4 volts. 35mA*100R = 3.5V. The 600R load shows clipping setting in at around 20V. 35mA*600R = 21V. :)
 
CJ900RR said:
Interesting circuit! And I like your design monkey. Nice indeed. What kind of output transistors do you think would be best to use?

Im new on class-a and have some questions on the output stage. How do you increase power. Is it correct as I did in the attached file? (Take no notice on the transistors I only picked a random one.)

Also, how do you callculate output power? I guess it has to do with the MOSFETs or?

Best regards...

I think that bipolar transistors and MOSFETs are all good choices for different applications.

Increased output power can be used to Darlington transistors, more than the parallel right.

BTL structure and improving Voltage of the output stage is also raising output power method.

Thanks and best regards!
 
Increased output power can be used to Darlington transistors, more than the parallel right.
integrated darlingtons have the driver (and it's resistor) in the same package. That limits flexibility. I believe this is more suited to mass production where it costs money to place each component on the PCB.

If you want best performance then use separated driver and output.

Now on to power.
A single darlington will match a single driver and output device for power output if the two devices are rated the same.
But there are many more separate output devices to choose from and many will exceed the power and SOA capability of the restricted range of integrated darlingtons.

Doubling up the output pairs will roughly double the output current ability when operated from the same voltages. This applies to both darlington and separate outputs.
Alternatively twin output pairs can be run from higher supply voltages and stay within their SOA limits at a higher output current than a single. Again this applies equally to separate and darlington. However the extra SOA available in separate devices allows even more output current if one chooses the appropriate device for the duty.

In summary, I disagree, darlingtons are cheap and produce less power than doubled up separate output devices.
 
AndrewT said:

In summary, I disagree, darlingtons are cheap and produce less power than doubled up separate output devices.

I agree with Andrew's conclusion. Darlington is just a high current gain component/approach. It doesn't mean high output power. To have higher power W = V x I, increase either I or V, or both. Then design a suitable output stage: parallel increasing current output capability or serial for higher voltage or bridge. But not darlington.
 
Thank you for your answers. I have done reading and learned alot. But I am still a bit confused over one thing.

Monkey29 claims that he uses DC-voltage of +-50V and +-45V, and that his construction has 3 Fairchild FJL4215, and 3 Fairchild FJL4315 as output transistors. He also claims that hi's construction delivers 300W into 8 ohm.

But I also found a schematic on a russian website using +-100V at the chip, and +-75V at the output transistors (3 * MJL21196 and 3 * MJL21195) and the output power is to be 250W into 8 ohm and 500W into 4 ohm...

I dont say that anyone is lying but how can this be? Increased voltage should increase output power? Or?

Best regards...
 
Hi,
simple arithmetic shows that 300W into 8r0 needs an AC voltage of sqrt(300*8)=49Vac.
the peak output voltage is sqrt(300*8*2)=69.3Vpk.
one cannot get that level of output from either +-45Vdc nor +-50Vdc supply rails.

I would doubt one could get 300W into 8ohms from <+-73Vdc supplies. More likely the supplies would be near +-80Vdc.
 
Thank you for your answers.

AndrewT said:
Hi,
simple arithmetic shows that 300W into 8r0 needs an AC voltage of sqrt(300*8)=49Vac.
the peak output voltage is sqrt(300*8*2)=69.3Vpk.
one cannot get that level of output from either +-45Vdc nor +-50Vdc supply rails.

I would doubt one could get 300W into 8ohms from <+-73Vdc supplies. More likely the supplies would be near +-80Vdc.

I thought it where strange. And your callculation example states it.

panson_hk said:
My LEACH amp generates 150 W/8 Ohm from +/- 63 V DC rails. It would be a magic if +/- 50 V DC rails can deliver 300 W/8 Ohm.

You need to consider V drop across output BJTs' C and E. Again, I recommed you visiting LEACH amp web page

http://users.ece.gatech.edu/~mleach/audiothings.html

You can increase output power by bridge. Pls read NS app note AN-1192 for LM3886.

Thank you for your link panson. Now I have even more to read and learn.

Im just wondering why monkey says on his webpage that his module with the LME49810 outputs 300W.
 
about "300W+300W *(8¦¸)"

Website nominal "300W+300W *(8¦¸)" is that 49810 AMP module is capable of 300 watts output power at 8 Ohm load, when in proper working voltage.

DC45V voltage is a recommended value. This does not mean that the DC45 voltage can 300w output Power.

The letter was not sufficiently detailed expression caused by misunderstanding, expressed regret. And thanks CJ900RR pointed out mistakes, thanks AndrewT the calculation and deduced.