What am i doing wrong, why wont my JLH operate in class A / get warm? - diyAudio
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Old 7th October 2011, 04:49 PM   #1
Shamron is offline Shamron  Norway
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Default What am i doing wrong, why wont my JLH operate in class A / get warm?

Hi there people.

A year ago i rediscovered electronics after 15 years of just selling them. in 95 i studied electronics to become a radio/tv repairman but being a rebelling student not too focused on school and then not giving electronics any attention for 15 years haven't given me the most knowledge or experience with fixing and understanding it.

I found a JLH 2x10W amplifier pcb kit with components on ebay and started working from there. I buildt the JLH in 92-93 at school, but we kinda never got it working as it should. We then used 2n3055, and this one from ebay also came with those.

I replaced the outputs with MJE15003's. The two other transistors is a A970 and a c2240. I dont understand the workings about transistor schematics / datasheets and to know what to look for to understand if these are gonna work good with this circuit, but i trusted the guy putting the kit together. I've measured all the resistors and can confirm that their values are correct compared to the schematic on the original circuit at http://www.tcaas.btinternet.co.uk/jlh1969.pdf

Now if these values still are correct with the "new" transistors is something else. That's where my knowledge comes short.

The problem is...

My amplifier doesn't get warm at all. I can feel only a slight difference between it being off or on, but i remember the one we buildt at school became HOT. I DO have slightly better heatsinks, but i first buildt it with comparatively small heatsinks and still almost nothing.

R1 and R2 are 100 and 560 Ohm, since i'm using 8 Ohm speakers. Also i replaced all the electrolytics with Elna Silmic II. The sound is very nice, but being a class A amplifier, it SHOULD generate heat and since it doesn't, i assume the quiescent current is too low.

I've given it a rather large power supply, regulated it at 27V exactly and thrown in 1F capacitors for each channel.

How to solve this? Do i mess with the R1/R2 or is will that screw up the voltages over the rest of the transistors as well?

This problem haunts me and i would really, really appreciate help.

Again, i know i should know this, but i don't. :-P

EDIT: I forgot to mention, in a desperate move, i replaced the c2240 with a 2sc3421 which i use in the 2x40w JLH (which DOES get hot and sounds beautiful) but it didn't change anything.

Ben Cato Malkenes.

Last edited by Shamron; 7th October 2011 at 04:51 PM. Reason: Forgot something...
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Old 7th October 2011, 08:13 PM   #2
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Throw a small resistor in series with your 27V,
so you can measure the total current drawn
as a small voltage drop across that resistor.
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Old 7th October 2011, 08:51 PM   #3
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You will get a lot more help if you post these questions in the very large thread on the JLH Class A 10watt amp. This is somewhere else on this site and all the talent there should help.
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Old 7th October 2011, 10:33 PM   #4
Shamron is offline Shamron  Norway
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Quote:
Originally Posted by kenpeter View Post
Throw a small resistor in series with your 27V,
so you can measure the total current drawn
as a small voltage drop across that resistor.
I just used my multimeter and measured the currents drawn from the 27V directly, it's 0,59A. This should be 1.2A for a 8 Ohm load, right?

I have seen some drawings with this amp where the 560 Ohm resistor is replaced with a 1.2K pot, but all i got are those multi-turn pots, isn't it too high a current for those? :-P

Thanks for helping

Ben.-
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Old 8th October 2011, 01:51 AM   #5
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Linear class A into 8 ohms would require 850mA
quiescent to swing rail-to-rail from 0 to 27V....

Considering JLH currents are non-linear in a way
that runs far hotter than Linear A when its doing
nothing, you would probably need at least 1A.

JLH takes more base current for one transistor
to reach the rail than it does for both to stay
perfectly idle in the center. A slightly excessive
bias becomes the necessary evil of that curve.

Got workaround for that, but its a whole other
topic. Let's fix your more serious problem first.

Last edited by kenpeter; 8th October 2011 at 01:54 AM.
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Old 8th October 2011, 02:15 AM   #6
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I am in disagreement with JLH over R1 R2 not paired of the same value.
Mismatch increases parasitic loss within the bootstrap for same bias...

I don't have a clue how much base current an MJE15003 would need
to conduct 1.7A peaks? But that requirement would dictate how much
bootstrap current you must have, at a bare minimum.

Last edited by kenpeter; 8th October 2011 at 02:18 AM.
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Old 8th October 2011, 02:27 AM   #7
AKSA is offline AKSA  Australia
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KP,

Assume a beta of around 20 at 1.7A and 25C, slightly better at higher temperatures.

That would mean 85mA min base current, say 100mA for a decent margin. These are old high Vceo switching devices, and beta droop is pretty bad.....

Hugh
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Last edited by AKSA; 8th October 2011 at 02:51 AM.
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Old 8th October 2011, 03:49 AM   #8
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For 100mA base current max, boot resistors might be about 68R apiece?

I'm figuring 135R for dropping 13.5V @ 100mA in the quiescent state.
And bootstrap wants to tap right in the middle for best performance.
So this splits to a pair of 68R or maybe 75R. Gotta figure Beta will be
riding a hotter curve than 25C... Dude, its JLH! Get me a potholder...

Last edited by kenpeter; 8th October 2011 at 03:54 AM.
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Old 8th October 2011, 07:55 AM   #9
Shamron is offline Shamron  Norway
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Hi again and again thanks for helping.

R1 and R2 together make up the bootstrap, right? I ask since as i said in the initial post, i've forgotten so much, consider me a beginner who knows Ohm's law and how to solder. :-D

I lost my old school books and today's books is not nearly as thorough as they were in 91-94. I've found some pdf's of old books and i got JLH's book of amplifiers and some other analogue / amplifier books, but it's hard to re-understand it all by yourself. (i think).

So if R1 and R2 is what's called the bootstrap, are you suggesting i lower the values of both? Won't that also raise the collector voltage on TR3? or is that needed to also in the process raise the collector current of TR1?

When it comes to R/C impedances in the circuit, i'm lost. That's way above me i think. :-/

Also: Jonathan, i feel i owe you a reply. I did post in the other JLH thread some half year or so ago. I cannot remember if it was about this specific issue, but i didn't get it solved even though Geoff and some other people with high knowledge replied (again, can't remember if it was about this specific issue) but i thought if i made a topic with this specific issue, it would be easier for others with the same problem to also find it in the future. I might be wrong but that's why i chose to make another topic. :-)

Ben.-
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Old 8th October 2011, 02:07 PM   #10
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Quote:
Originally Posted by Shamron View Post
Won't that also raise the collector voltage on TR3?
Feedback from TR4 raises current through TR3 to restore voltage balance.
The collector you asked, voltage behavior won't be changed, only current...

Last edited by kenpeter; 8th October 2011 at 02:09 PM.
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