LME49810 biasing procedure temp comp and simple resistor

I am looking to bias the lme49810 to 1.5Amps with temp compensation circuit given in the application note. The formula is as follows..
VBIAS = VCE = RB1 * I + Vbe(V)

its said that Vbe is limited to 6V in the circuit with 2sc5171 as biasing transistor.

what is the general value of vbe observed or does it changes when biasing?

what should be RB1 for a bias of 1.5amps? considering
I = 1.5A
Vbias = 10V
Vbe = 5v

so the result is RB1 being 3.3 ohms? what is the general procedure to bias in this case..
 

multisync

Member
2008-01-31 10:34 pm
Carp
you do not bias the LME chip, you bias the output transistors. It is difficult to do this mathamatically, as the bias current depends on the hfe of the output transistors, the number of output stages, the resistors used, the configuration etc. You may notice nearly every bias circuit uses an adjustable resistor. Some people after adjusting the bias with an adjustable resistor replace the adjustable resistor with one of a fixed value.
 

multisync

Member
2008-01-31 10:34 pm
Carp
I have something similar at home. The boards were bought assembled.

see this thred
http://www.diyaudio.com/forums/solid-state/155173-anyone-recognise-design-based-lme49810.html

None of the 5 pairs of output transistors were matched. I set the bias to an average of 30ma each pair. Some pairs were at 20ma some at 40ma. The total for all 5 was 150ma. I used +/-65volt rails. The power dissapated was 130volts x 0.150 amps or 19.5 watts per channel. In your case you can set the bias to an average of 300ma per pair or 1.5 amp per pair. Depending on what you do, your heat sink will vary in size from very large to gigantic or large and force air cooled. Higher bias currents will not necessarly give you a lower distortion with BJT transistors. You have to find the sweet spot. Measure the voltage across one 0.22R resistor. Current will be the voltage across the resistor divided by the resistor value. 0.220 volts DC across one resistor will be about 1 amp of bias current, 0.022 volts will be about 100ma of bias current
 
I have something similar at home. The boards were bought assembled.

None of the 5 pairs of output transistors were matched. I set the bias to an average of 30ma each pair. Some pairs were at 20ma some at 40ma. The total for all 5 was 150ma. I used +/-65volt rails. The power dissapated was 130volts x 0.150 amps or 19.5 watts per channel. In your case you can set the bias to an average of 300ma per pair or 1.5 amp per pair. Depending on what you do, your heat sink will vary in size from very large to gigantic or large and force air cooled. Higher bias currents will not necessarly give you a lower distortion with BJT transistors. You have to find the sweet spot. Measure the voltage across one 0.22R resistor. Current will be the voltage across the resistor divided by the resistor value. 0.220 volts DC across one resistor will be about 1 amp of bias current, 0.022 volts will be about 100ma of bias current

I got this from the source http://connexelectronic.com/product_info.php/cPath/39_40/products_id/78
I think this guy is a forum member and seemed a bit reliable than one on ebay. Like you said the design seems interesting.

I also bought lm4702 with a pair of 5200 1943 transistors and they sound pretty good no complains here..

300ma per pair makes sense but 1.5 amps per pair seems impractical I think you wanted to tell on the total its 1.5 amps.

Like you said there wouldnt be much difference in THD after certain level. Its there in the National app note that after 500ma there is no perceived difference but I think its for one pair.

regarding Heat issues once "andrew said that he tried at 500ma and cant put fingers on the heatsink imagine how how it would be at the junction"

I will be using this heatsink two of them if required each about 6 inches in length..
Imageshack - photo0306t.jpg

so the max per pair I think we can go is 500ma or ill be rather stay at 400ma max per pair..

right now i have this in BJT soon will be with Toshiba mosfet or Hexfets.

but all this is done varying the pot on it?

thank you very much for the kind help... :)
 
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+/- 35 volts will give you about 50-60 watts @8ohm.


1.5 amps of bias is about 18 watts at 8 ohms class A and then the amplifier will go into class AB. This will be quite loud with most speakers.

There are lots of DIY members who are happy listening to amplifiers in the 1-5 watt range


Ti/national make a special version for fets LME49830, the bias voltage can go higher for the fets

yes, just adjust the pot, measure and wait to see how hot the heatsink gets remeasure and adjust again. There is a bit of space between the bias transistor that senses the heat and the output transistors, therefore there will be a significant delay between the furthest transistor getting hot and the bias transistor detecting that heat and lowering the bias. The final case you put the amplifier in, other heat sources, sunlight, summer temperature where you live will all affect the final design. Before I was happy I let my amplifier sit for days and I would check and adjust and check and adjust till I was satisfied things would not get too hot to blow up.

The amplifier is stable if the transistors are not matched, some transistors will get hotter than others, that is why you must measure each transistor bias for worst case. You did not read all of the link I gave you.
 
+/- 35 volts will give you about 50-60 watts @8ohm.


1.5 amps of bias is about 18 watts at 8 ohms class A and then the amplifier will go into class AB. This will be quite loud with most speakers.

There are lots of DIY members who are happy listening to amplifiers in the 1-5 watt range


Ti/national make a special version for fets LME49830, the bias voltage can go higher for the fets

yes, just adjust the pot, measure and wait to see how hot the heatsink gets remeasure and adjust again. There is a bit of space between the bias transistor that senses the heat and the output transistors, therefore there will be a significant delay between the furthest transistor getting hot and the bias transistor detecting that heat and lowering the bias. The final case you put the amplifier in, other heat sources, sunlight, summer temperature where you live will all affect the final design. Before I was happy I let my amplifier sit for days and I would check and adjust and check and adjust till I was satisfied things would not get too hot to blow up.

The amplifier is stable if the transistors are not matched, some transistors will get hotter than others, that is why you must measure each transistor bias for worst case. You did not read all of the link I gave you.

here the amp bias is 1.5Amps peak so the final output in class A to be

P RMS = [ (Ipeak x 2)^2 * 8ohms ] / sqrt(2)
P RMS = Sqrt ( 3^2 * 8 ) /sqrt(2)
P RMS = 72/1.414
P RMS = 50.91

yeah it takes time to stabilise the bias.

I read the link that you gave to me...

I am looking forward for working on two pole compensation improving the HF distortion.

so I have a doubt that the way you recommended to measure the Bias current is continuous not peak is it?

the above calculation is nelsons method of calculating class A.. he first calculates everything in peak and the divides it by 1.414 to get the final power...
 
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When you measure bias current you measure DC current flowing through the output transistors with the input shorted. There is no peak current, just DC.

The current into your speakers will vary with frequency as nearly all speakers are a reactive load with varying phase angles between voltage and current and varying impeadeance with frequency. A good capacitor will nearly dissapate 0 watts of heat while having tens of amps of current flowing through it and having hundreds of volts across it. Same with a good inductor.

Think about where the currents flow in a class A output stage, with no load, with no load and signal, with a load and no signal, with a load and signal.
 
according to your way it seems correct as its all about the I^2 x R which is right but how come nelson is measuring peak but why is he going that way rather than direct I^2 x R...
strange... does anybody has an explanation? why is nelson considering peak first?

but how is he calculating peak when one can measure the DC directly..
 
what will be the Vce for the above amp is it some value less than the psu like if my psu is 78V so Vce could be 70V?

what about Ic?

Im just calculating the max transient current that can be drawn for a duration of 100ms and 10ms so that the psu can be optimised...

so how to calculate the Vce?
 
I don't understand your last few posts. Are you making a new design? I thought you are you using a prebuilt board.

Doug Self has discussed the value of Re in his book. Look for his articles on "Distortion in amplifiers"

maybee you shouldn't as you will find other problems to loose sleep over

The power transformer is usually chosen depending on desired power output. If you will have a lot of bias current add that to the transformer ratting and then a bit more for piece of mind.

example stereo 100 watt/channel power amplifier @8ohms

transformer = 200-300 VA

same amplifier but with 500 ma bias.

to get 100 watt @8ohms the rails will be about +/- 50 volts DC

500ma bias x 100VDC = 50 watts per channel in standby ie no signal

so add another 100 va to get 300-400 va transformer.

Some people here use KVA transformers for 50 watt amplifiers.

Most music has a 10-1 dynamic range.

You will almost never run your amplifier constantly to clipping, maybe sometimes when you have had too many drinks and there is a loud party.

just about every transformer can run 200% rating for short times. Ie you can pull double the VA out of it with no real problems.

For capacitors in the power supply 2000uf per amp of output current is enough. There are tens of thousands of amplifiers that have less and they sound good. There is a long thread here somewhere that discusses this to death.

for 100 watts per channel you need about 3.5 amps @ 8 ohm, no speaker is exactly 8 ohms so lets say 5 amps. so we get 5 x 2 x 2000, = 20,000uf per rail.

I just took apart an old kenwood amplifier and I reused the chasis for a project. The kenwood amplifier was rated at 130 watts /channel. The capacitors were 7,500 uf in the old Kenwood. I have 2 other similar kenwood amplifiers that I use and they sound fine.

see post 2024 in the folowing thread
http://www.diyaudio.com/forums/soli...ifier-based-lme49830-lateral-mosfets-203.html


the transient current will depend on the load and the resistance of all the wiring and all the series parts. Could be hundreds of amps esp if the output is shorted, I have done it and parts dissapear only to leave a black smudge and smoke and melted wires.

Just build your amp and have fun. Add some changes later, does it sound better or different or worse if you have 1,000 or 100,000 uf capacitors.

some people don't like to have relay contacs as the say the relay contacts change the sound. Some people use solid copper wire some use braided wire some silver wire. Some use gold connectors some brass etc etc. Some people like to align the amplifier so the earth's magenetic field only crosses signal wires at 90 degrees.

Hey maybe I should start a new thred on that idea.
 
i had a strange HF rolloff after 14khz so i suspected that its becoz of the RL network on the output so I just connected the output straight before the RL netwoked and all worked super good and even the transparency increased alot.

what RL values are recommended at the output.

what problems one would face if its completely removed?
 
i had a strange HF rolloff after 14khz so i suspected that its becoz of the RL network on the output so I just connected the output straight before the RL netwoked and all worked super good and even the transparency increased alot.

what RL values are recommended at the output.

what problems one would face if its completely removed?

Did you measure it? What R and L were used?