Hi everyone. my name is john and i like electronics. I looked for an easy-to-build amplifier and came across the famous 2000 version jlh as the diagram below:
after assembling it I calibrated the two trimmers and heard some music on 8 ohm speakers. the music feels good and it really left me speechless given the power of only 10 watts. however I am having problems with adjusting the bias. I have a 30V power supply so the bias should be set to about 15V but for some reason turning the rv2 trimmer to full scale I only get to 10V and it doesn't go on, with consequent premature clipping. what do you think could be the problem?
p.s .: I apologize for incorrect grammar but I don't speak English very well
after assembling it I calibrated the two trimmers and heard some music on 8 ohm speakers. the music feels good and it really left me speechless given the power of only 10 watts. however I am having problems with adjusting the bias. I have a 30V power supply so the bias should be set to about 15V but for some reason turning the rv2 trimmer to full scale I only get to 10V and it doesn't go on, with consequent premature clipping. what do you think could be the problem?
p.s .: I apologize for incorrect grammar but I don't speak English very well
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Check for proper inclusion Q4 (b-c-e).
Try to reduce R3 to 47k, and R2 to increase to 150-200k.
You may find the topic jlh-10-watt-class-amplifier
JLH 10 Watt class A amplifier
Try to reduce R3 to 47k, and R2 to increase to 150-200k.
You may find the topic jlh-10-watt-class-amplifier
JLH 10 Watt class A amplifier
Let's say that the adjustment you're looking for is "DC offset" rather than bias. The bias adjustment is for the current passing through Q1 and Q2 from collector to emitter but unfortunately, you can't measure it directly because there are no emitter resistors to use for a reference in this this version.
If you also want to measure and adjust bias current, you can calculate that by fitting a small value (say, 0.2 - 0.5 Ohms) 5W resistor in series with the +36VDC power lead. Measure voltage across the resistor then I=E/R and you have the total current drawn by the ampliier, which is only a fraction more than the output stage bias.
If you also want to measure and adjust bias current, you can calculate that by fitting a small value (say, 0.2 - 0.5 Ohms) 5W resistor in series with the +36VDC power lead. Measure voltage across the resistor then I=E/R and you have the total current drawn by the ampliier, which is only a fraction more than the output stage bias.
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RV2 sets the DC output offset which--in this circuit ( blocking capacitor
at output) affects the amplifier balance as you witness in clipping .
Many cheap semi variable resistors do not accurately function at their full values sometimes clipping off the bottom and top ends .
Measure at maximum adjustment for true value with ohmeter .
JLH,s circuit regarding the offset and bias controls has been changed -
The Class-A Amplifier Site - JLH Class-A Update
Even the circuit you display isn't the original circuit of 1969 see-
jlh1969.pdf
Yours is the 1996 version .-
JLH Class-A amplifier – Circuit Wiring Diagrams
at output) affects the amplifier balance as you witness in clipping .
Many cheap semi variable resistors do not accurately function at their full values sometimes clipping off the bottom and top ends .
Measure at maximum adjustment for true value with ohmeter .
JLH,s circuit regarding the offset and bias controls has been changed -
The Class-A Amplifier Site - JLH Class-A Update
Even the circuit you display isn't the original circuit of 1969 see-
jlh1969.pdf
Yours is the 1996 version .-
JLH Class-A amplifier – Circuit Wiring Diagrams
In the lower position RV2 there will be a constant voltage at the output decreases, in the upper one it increases ..
The difference between the versions of JLH1969 and 2000
Help: what's the difference between the input stages of JLH1996 & JLH2000? | Electronics Forum (Circuits, Projects and Microcontrollers)
Help: what's the difference between the input stages of JLH1996 & JLH2000? | Electronics Forum (Circuits, Projects and Microcontrollers)
a note that I forgot: for the current measurement I am using a multimeter with its shunt in series and at the moment the current is around 1.3A, acceptable. as for the capacitor not having a 64uF in the house I used a 100uF, can it have conditioned the regulation? Finally, for the bc212 I don't know what pinout to refer to, some datasheets give it for e-b-c others b-c-e ... how do I recognize the emitter and collector in a pnp?
the fact is that I built two bases like this scheme and both give me the same problem, even if it sounds perfectly, the higher Q1 transistor definitely heats up more because of the imbalance, and I'm afraid that in the long run it could be affected.
the fact is that I built two bases like this scheme and both give me the same problem, even if it sounds perfectly, the higher Q1 transistor definitely heats up more because of the imbalance, and I'm afraid that in the long run it could be affected.
If its the British variety its -- E-B-C ,the BC range is very British used by many audio designers as well as myself .
Foreign varieties of the same stamped BJT showing a different connection are not standard in any equivalent book I have , even the well respected German equivalent books --VRT (www.eca,de) show the same as the British books.
Using an Analogue meter put the positive lead to the centre leg of the BC 212 then using the negative lead tap onto the other two legs an equal low ohmage should result .
If you only have a digital meter then reverse the leads when testing and you should have a "diode " type readout.
Foreign varieties of the same stamped BJT showing a different connection are not standard in any equivalent book I have , even the well respected German equivalent books --VRT (www.eca,de) show the same as the British books.
Using an Analogue meter put the positive lead to the centre leg of the BC 212 then using the negative lead tap onto the other two legs an equal low ohmage should result .
If you only have a digital meter then reverse the leads when testing and you should have a "diode " type readout.
...the higher Q1 transistor definitely heats up more because of the imbalance
V/2. For symmetrical cutoff of the output signal, the voltage may be slightly less than V/ 2.
Check BC212 - two diodes will be detected from the base. Check the different connections when measuring the transistor with a multimeter.
V/2. For symmetrical cutoff of the output signal, the voltage may be slightly less than V/ 2.
Check BC212 - two diodes will be detected from the base. Check the different connections when measuring the transistor with a multimeter.
here it is:
Https://ibb.co/6XVcR8h
the way it is inserted in the circuit at the moment is, looking towards the writing, e-b-c
Https://ibb.co/6XVcR8h
the way it is inserted in the circuit at the moment is, looking towards the writing, e-b-c
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What DC voltage do you see on the base of Q4?
The midpoint voltage is around 1.5 volts higher than this value. You should also see this same voltage less just a fraction on the preset. In other words very little voltage is dropped across R3
If the voltage is low then try removing Q4 altogether and recheck the voltage on the preset. If its still low then you have an error in the bias network, if it comes higher then there is a problem with Q4.
Is C4 fitted correctly (if using an electrolytic)?
The midpoint voltage is around 1.5 volts higher than this value. You should also see this same voltage less just a fraction on the preset. In other words very little voltage is dropped across R3
If the voltage is low then try removing Q4 altogether and recheck the voltage on the preset. If its still low then you have an error in the bias network, if it comes higher then there is a problem with Q4.
Is C4 fitted correctly (if using an electrolytic)?
as you said, I detect 8.90V on the midpoint between q1 and q2, and on the base of Q4 only 7.30V and as said before I do not go very far from this point. for c4 I used a 470nF NP and therefore it is not a polarity problem.
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If its ITT that's an old British company and all of them conform to the books I have ,on the other hand -Central Semiconductor Corp (USA) shows
1-C
2-B
3-E
and it looks like that applies from the front.
https://www.mouser.co.uk/datasheet/2/68/bc212_a_b-1149793.pdf
1-C
2-B
3-E
and it looks like that applies from the front.
https://www.mouser.co.uk/datasheet/2/68/bc212_a_b-1149793.pdf
You have 2.4 volts difference between the midpoint and base of Q4 which suggests a problem.
Make 100% certain you have the correct resistors fitted for R9 and R6 and that C2 is correctly fitted.
You can either remove Q4 or just lift one end of R3 to isolate it and then measure the DC voltage on the preset. We need to know that value and whether Q4 is pulling it down somehow (it should not).
Make 100% certain you have the correct resistors fitted for R9 and R6 and that C2 is correctly fitted.
You can either remove Q4 or just lift one end of R3 to isolate it and then measure the DC voltage on the preset. We need to know that value and whether Q4 is pulling it down somehow (it should not).
I lifted to the end of r3 and now the maximum voltage at which I arrive is about 11V, so not much better, at this point it seems that the problem lies in the bias network ...
another thing that leaves me perplexed is that despite having raised an end of r3, on the alleged base of Q4 there are still 8Vdc ...