| joe mobile |
I Just assambled a Leach Amp version 4.5. I made the boards from Mr Leachs original layout. It works, but when I try to adjust the bias current, I get no response, the current just stays the same.
Has anyone encountered the same problem, or has anyone an explanation for this?
I searched the forum, and in this thread
http://www.diyaudio.com/forums/show...2925&highlight=
someone had the same problem, but no one responded to his post
I built everything very carefully and i`d really like to finisch the amp, but i can`t figuere out whats wrong. Please help
Thanks,
Joe |
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| acenovelty |
Removed the short circuit jumper in parallel with C12?
"Double check the diode assembly. If a diode is backward, the output transistors can blow. You can check the diode polarities one at a time with an ohmmeter. The ohmmeter will not have enough test voltage to forward bias all 4 diodes simultaneously. If there is a short circuit between a diode lead and the heat sink metal, the amplifier will not work and the output transistors can be blown." |
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| joe mobile |
Thanks for your reply Acenovelty
I checked everything and I found nothing wrong. I already installed the output transistors and the amp does plays music, just the driver transistors are getting a little hot (a little too hot I would say), And I have a little less than 100mV DC on the output.
But the thing that I really don`t understand is that absolutely nothing happens when I turn P1... I connected a signal generator to the input and observed the output waveform on a scope, and it stays as it is when i try to adjust the bias.
Here is the circuit diagram
http://users.ece.gatech.edu/~mleach...raphics/ckt.pdf
and here the board layout
http://users.ece.gatech.edu/~mleach...ics/layouts.pdf |
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| acenovelty |
Defective pot?/ 2K ohms, right? Confirm solder/traces from the pot to other components/cold joint etc. If possible measure pot resistance with DVM to confirm value.
At least you have music. The transistors should be only warm to the touch unless driven hard.
"Do the leads to the heat sinks connect to the right points on the circuit boards? I recently saw a student's amplifier where the base leads for the npn and pnp output transistors on one channel were reversed on the circuit board. The bias potentiometer would not adjust the bias current in that channel."
HTH |
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| Arius |
Hi Joe,
1) Uhh, what are you trying to do by adjusting P1 with a signal generator at the input and a scope at the output? You should set the bias current without any input and by monitoring the voltage across the output emitter resistor(s).
2) 100mV DC you say at the output? Is this without anything on the input? If so, then that's pretty high for an output DC offset. Might need to check your input stage to see if everything's properly balanced or you may have grossly mismatched input transistors.
Do note that the bias current setting affects the temp of your output devices and not the VAS stage drivers. Good luck with your amp. |
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| mikeks |
| quote: | Originally posted by joe mobile
..leach amp vbe multiplier... |
It's not a vbe multiplier.. |
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| AndrewT |
Come on Mike, please explain.
You have my permission to use more than one sentence |
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| mikeks |
| The voltage across Q7 is not a function of Q7's Vbe. |
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| darkfenriz |
| Check if Q7 is NPN for sure. |
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| audiofan |
| i build a pair of those a few years ago......and if I remember... you should use a jumper around this transistor to make previous test on the board if the jumper is still in place you will not be able to set bias it will stay to minimum and will operate in class B |
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| joe mobile |
hello everybody
To Arius: I did not try to adjust the bias by that, but if I turn P1 the waveform should change I think, thats why I mentioned that.
To Acenovelty: The pot is ok too and it is 2K, and the leads to the transistors are in the right places too.
to Darkfenriz: I checked it, it is mpsa 06, so it`s npn.
and to Audiofan: the jumper is removed.
AndrewT:
I don`t know that much about electronics and it`s my first amp, so I can`t explain that much, only what I measure and what I see that happens... But if you need to know something perticular to make an assumption what might be wrong I`ll try to explain it.
I`ll go on trying, if nothing helps i`ll just build a new one, good for practice...
Thanks to all for helping.
I`ll be away for the Weekend, so nice weekend to everybody
joe |
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| AndrewT |
Hi Mike,| quote: | | It's not a vbe multiplier.. |
| quote: | | voltage across Q7 is not a function of Q7's Vbe |
The ratio of voltage from collector to base and base to emitter works in a similar way to a resistive voltage divider.
The output voltage is a function of the base emitter voltage. |
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| sreten |
| quote: | Originally posted by mikeks
It's not a vbe multiplier..
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Erm.... it is a Vbe multiplier if A+B are connected.
:)/sreten. |
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| Ultima Thule |
The upper left 4-diode string should be connected to A & B, it's a temp dependent baker clamp.. (edit)allmost! :D
Cheers Michael |
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| jacco vermeulen |
| Can't we just call it a diode string with an adjustable zener ? :clown: |
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| Arius |
| And how does debating semantics help the thread starter who's just into his first project? |
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| MikeBettinger |
I'd be checking for a change in the voltage drop across the Re for the voltage amp or a change in the voltage across the (disputed) Vbe multiplier. Also, is the drop enough to bias on the outputs or not. There was no mention as to whether the bias was low or high (measured across the output emitter resistors) and not changing, or none existent.
A quick run through of the various stage currents might point to what's working or not.
Regards, Mike. |
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| mikeks |
| quote: | Originally posted by AndrewT
Hi Mike, The ratio of voltage from collector to base and base to emitter works in a similar way to a resistive voltage divider.
The output voltage is a function of the base emitter voltage. |
No.
| quote: | Originally posted by sreten
Erm.... it is a Vbe multiplier if A+B are connected.
:)/sreten. |
Yes. |
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| AndrewT |
Hi Mike,
did you forget to read Leach (it's what this thread is about)?
If Sreten is YES then I am YES not NO. |
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| mikeks |
Just think about it for a while, and you, in all probability, will understand just why Leach is wrong.
Here's a clue:
| quote: | Originally posted by Ultima Thule
The upper left 4-diode string should be connected to A & B.... |
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| MikeBettinger |
Hi Mike K,
I have a question for you, since you do appear to be quite knowledgable as to the reasons for and the proper way to implement most aspects of amp design.
If I examine older amp designs I have noted that the bias for the output stage has been handled many ways, from a simple resistor or diode string to some very complex signal or temperature tracking approaches.
My question is, is there a real technical advantage to using a tracking type circuit relative to a few diodes and an adjustable resistor coupled to properly designed heatsink?
I have experimented with a signficant number of approaches and sonically prefer a simple resistor (which I've found is not practical to implement) but I'm currently using 3 1N4148's thermally coupled to a heatsink that is only luke warm after 5 minutes under a blow drier. I will admit to an hour or so warmup time playing music before it all locks in.
In your experience is there a measurable improvement aligned to a more active approach to supplying the bias? Can you siuggest any further reading on the topic?
Regards, Mike. |
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| mikeks |
| quote: | Originally posted by MikeBettinger
If I examine older amp designs I have noted that the bias for the output stage has been handled many ways, from a simple resistor or diode string to some very complex signal or temperature tracking approaches. |
With a diode string, your bias voltage varies significantly with variation in the current through the diodes.
This problem, of course, is vastly worse if a simple resistor is used to set bias.
This is why i recommended this approach. |
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| AndrewT |
Hi Mike,
Leach's Vbe multiplier is not 2c or 2d either.
It is a circuit where| quote: | | The output voltage is a function of the base emitter voltage. |
The voltage across the circuit is a multiple of the Vbe of the transistor (just like a resistive multiplier), the diodes give a measure of temp compensation (just as a resistive multplier uses the transistor), the voltage across the circuit varies if the VAS current is not constant (just like a resisitive multiplier).
The voltage varies just like (not exactly like) a resistive Vbe multiplier, it just arranges the temp compensation differently.
It looks like a multiplier
It behaves like a multiplier
It fits the amplifier as a multiplier
It is a multiplier. |
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| mikeks |
Hi Andrew,
The diode string connected betwen A and B make this untrue, as the voltage across these diodes is not a linear function of the current through them.
This is why it is this string of diodes and not Q7 that is mounted on the heatsink for thermal compensation. |
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| AndrewT |
Hi Mike,
you are right,| quote: | | the voltage across these diodes is not a linear function of the current through them |
you are also right when you say| quote: | | This is why it is this string of diodes and not Q7 that is mounted on the heatsink for thermal compensation |
but| quote: | | The diode string connected betwen A and B make this untrue |
does not follow
It is still a voltage multiplier with a different arrangement of temperature compensation.
However, the current through the diode string is nearly constant over a big range of VAS current. The Vbe voltage holds the bottom leg resistor at nearly constant voltage. The base current is fairly small and variations in this change the diode current only a little.
the main variation is current coming in through the collector and this I think causes temp changes at the junction and in turn change Vbe. If this is the mechanism that causes the slope in the multiplier curve then it explains why your proposal, D.Self's solution with the added Collector R and Hagerman's all improve on the basic multiplier's voltage change with VAS current changes.
But back to where we started. The fact that any multiplier has a varying voltage with varying VAS current does not of itself stop the multiplier being (and behaving as) a multiplier. |
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| mikeks |
| quote: | Originally posted by AndrewT
The Vbe voltage holds the bottom leg resistor at nearly constant voltage. |
This is correct at a given temp.; however, with a true Vbe multiplier the bias voltage across Q7 is a function of Q7's Vbe.
This is not the case here since, as you've noted, the voltage across the diode string does not vary ''significantly with current''.
Viz. Diode current is not linearly related to diode voltage, which is why the Vbe multiplier equation is inapplicable: the diode string and not Q7 must, therefore, be mounted on the heatsink. |
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| AndrewT |
Hi,
why have you introduced this distraction?| quote: | | the diode string and not Q7 must, therefore, be mounted on the heatsink |
We all know that Leach intended the diode string be mounted on the main heatsink. He describes the mounting and the need for the two isolating resistors in detail.
We have no disagreement with this.
The disagreement is whether the Leach style of Vbe multiplier is entitled to be called a multiplier or not.| quote: | | with a true Vbe multiplier the bias voltage across Q7 is a function of Q7's Vbe |
applies to the Leach Vbe multiplier just as much as it would to a resistively (with no added diodes) loaded multiplier |
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| mikeks |
| quote: | Originally posted by AndrewT
Hi,
why have you introduced this distraction? We all know that Leach intended the diode string be mounted on the main heatsink. He describes the mounting and the need for the two isolating resistors in detail. |
He has no choice: mounting Q7 on the heatsink will not cause the bias to track the heatsink temp.
| quote: | Originally posted by AndrewT
The disagreement is whether the Leach style of Vbe multiplier is entitled to be called a multiplier or not. |
I don't think you've actually read my post.
It is called a Vbe multiplier because its bias voltage varies with Q7's Vbe.
How can this be the case if you include a diode string between Q7's base and collecor?
The diode string's voltage will vary hardly at all when Q7's Vbe varies. This should now be self-evident. |
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| jackinnj |
Pick any number and device -- do a temperature sweep -- left axis is the ratio of Vce.
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| mikeks |
Hi jackinnj,
Run Temp sweep with respect to Q1's Vbe, and plot Vce.
Similarly, run Temp sweep for the diode string alone and again plot Vce.
Further, run temp sweep for Q1's Vbe with the diode string removed, and, ounce more, plot Vce.
You'll find that Vce varies most (with the diode string in situ) when the temp of the diodes is varied, rather than when Q1's Vbe changes. |
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| jackinnj |
Perhaps I should have labeled them better -- the Vbe plots with and without the diode string sit atop each other -- the Vce plots are shifted by the number of diode drops and have different slopes. Not enough time to do everything.
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| AndrewT |
Hi,
I just spent 25minutes compiling a reply to Mike refuting his contention that the Leach version is not a Vbe multiplier.
But I lost it.
Mike,
I think this argument is not worth another 25minutes, I give up (not give in!). |
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| MikeBettinger |
| quote: | Originally posted by mikeks
Note that figure 2c and 2d are NOT Vbe multipliers. |
Hi Mike,
Basically voltage references. I thought about trying to use one but then It was easier to fit a few of diodes.
I have a few obsrvations on the suggested circuit but would like to do my homework before speaking. We'll see how the week pans out. Thanks for the input.
Regards Mike. |
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| sreten |
Hi,
Hmmm....... A Vbe multiplier can have an effective "gain" of near 1.
So its very near a Vbe reference.
Personally I'm not keen on Leach's inacurracies regarding bias.
Seems in the case shown you have a string of diodes + an adjustable
Vbe multiplier to set the actual bias voltage, thus the output voltage
is not a direct multiple of the transistors Vbe.
D.Self covers the complete issue comprehensively and far more more
accurately than Leach, who simply misunderstands optimum aB bias.
:)/sreten. |
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| Ultima Thule |
It appears the Leach's bias circuit is a hybrid between a normal Vbe multiplier and Hagermans circuit 2c, 2d.
In Leach's bias circuit Rbe(R27+P1) is 3,2 kOhm and Rbc(R25+R26) 2 kOhm plus the diode string.
The background to Leach's "deviating" bias circuit is as follow quoted from his site :| quote: | | Many amplifiers have the VBE multiplier transistor on the heat sink with the output transistors. This eliminates the need for the diodes. The wires which run from the circuit board to the transistor exhibit capacitance to ground which can affect the high-frequency response of the second stage. At worst, this could cause oscillation problems. With the diodes on the heat sink, resistors on the circuit board can be used in series with the wires to isolate this capacitance from the second stage. These resistors are R25 and R26 in Figure 10. If Q7 were mounted on the heat sink, isolation resistors could not be used because they would affect the voltage regulation between the collectors of Q8 and Q9. |
Cheers Michael |
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| acenovelty |
Are there any practical suggestions for a solution for joe mobile's
stated question? |
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| joe mobile |
I think I really started something... although I don`t understand too much of the discussion.
I just blew the amp by accidently shortening R36 so for now my problem is solved... I`ll start a new leach amp. I guess thats learning the hard way.
I guess it will take quite some time till I finish the new one since I have to etch new boards and I don`t have that much time, but at least I got all the parts.
I`ll post when I got it finished
Thanks again to all the people who helped, and I`ll go on reading everything.
Joe |
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| mikeks |
| quote: | Originally posted by Ultima Thule
From Leach's site :
| quote: | Many amplifiers have the VBE multiplier transistor on the heat sink with the output transistors. This eliminates the need for the diodes.
The wires which run from the circuit board to the transistor exhibit capacitance to ground which can affect the high-frequency response of the second stage.
At worst, this could cause oscillation problems. With the diodes on the heat sink, resistors on the circuit board can be used in series with the wires to isolate this capacitance from the second stage.
These resistors are R25 and R26 in Figure 10. If Q7 were mounted on the heat sink, isolation resistors could not be used because they would affect the voltage regulation between the collectors of Q8 and Q9. |
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Leach's explanation is rather fanciful.
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| MikeBettinger |
| quote: | Originally posted by acenovelty
Are there any practical suggestions for a solution for joe mobile's
stated question? |
I thought I offered a couple of reasonable suggestions for seeing if the bias circuit is functioning in post # 18.
Regards Mike. |
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| MikeBettinger |
| quote: | Originally posted by joe mobile
I just blew the amp by accidently shortening R36 so for now my problem is solved... I`ll start a new leach amp. I guess thats learning the hard way.
Joe |
Joe,
I guess the my last post is not needed now. To bad, troubleshooting is where we learn the most..
Just out of curiousity, why is is necessary to build a new amp from scratch? I've blown up more amps than I'd care to admit to, but I've always just troubleshot the damage and got it running again, if only to make myself feel less stupid about whatever dumb-#&% mistake I made. I'm assuming the boards are in rough shape getting it to this point in the game.
Oh well, it's always nice when the current project is listenable.
Regards, Mike |
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| acenovelty |
Unlike other commentaries, you did indeed MikeBettinger. Others seem more interested in esoteric semantic discussions. Our friend from Hamburg asked for help and got a doctorial sounding lecture from some admittedly very clever folks.
And then he presents his own solution. However, a burned 220 ohm R36 resistor adds a certain difference to the problem than that first stated.
As for me , I only know what Prof. Leach says about his creation. Uncommonly good practical advice in my experience.
| quote: | | R36 - 220 ohm - If you have a problem with the VBE multiplier or the bias diodes, the bias voltage can go too high and this resistor can smoke and possibly burn the circuit board. I recommend putting a 1/8 inch long piece of insulation stripped from a piece of hookup wire on each lead of this resistor before soldering it to the circuit board. The insulation will hold the resistor up off the circuit board in case it burns. If you wish, you can use a 1/2 watt resistor for R36. |
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| Tony |
| quote: | Originally posted by acenovelty
Unlike other commentaries, you did indeed MikeBettinger. Others seem more interested in esoteric semantic discussions. Our friend from Hamburg asked for help and got a doctorial sounding lecture from some admittedly very clever folks.
And then he presents his own solution. However, a burned 220 ohm R36 resistor adds a certain difference to the problem than that first stated.
As for me , I only know what Prof. Leach says about his creation. Uncommonly good practical advice in my experience.
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hi ace,
indeed, a very practical advice from prof. leach...i've seen japanese commercial amps even used ceramic beads(tubes if you will) to serve as standoffs for resistors that are liable to burn when something goes wrong. |
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| MikeBettinger |
| quote: | Originally posted by acenovelty
Others seem more interested in esoteric semantic discussions.
As for me , I only know what Prof. Leach says about his creation. Uncommonly good practical advice in my experience.
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The discussion was good and actually got me thinking on a different plane about my own approach which, with time, I believe will be beneficial.
Although, I couldn't help thinking that the discussion was, as you say, a semantic difference.
The Professor using a diode string for temperature sensing and calling it a Vbe multiplier. The intent being to keep the discussion of the circuit in a normal frame work (my interpertation). While technically he's not using the transistor junction as the sensing mechanism, which would be the correct implementation of the terminology.
Good discussion though. Although, I'm not so sure the improved circuit posted by Mike K will actually pan out as sonically any better than my current simpler approach.
My thoughts, since I haven't been able to build anything, tells me that a bias shift could be responsible for the difference I hear between lower and higher listening levels. (the problem is definitely in the output stage)
With my current interpertation of the posted circuit, with no component values or operational discription, I'm assuming:
Transistor one to be the sensing junction.
Transistor two provides a mild amount of gain to increase the sensitivity and more or less isolate the junction from the effects of the current through the circuit block.
Transistor three being the pass transistor (so to speak) to develop the bias voltage across and present a low impedance to the output stage. Like I said I've only been pondering this.
My reservations are based on using three junctions in series with a noise generator as an input to create a stable voltage reference. Kinda like using a sledge hammer to put in a finishing nail. Especially if, with my three 1N4148s touching the heatsink, the amount of change in current through the diodes does not produce a major shift in bias in normal operation. A few test and I'll have my answers.
Has anyone already been down this path? Is my loose description of the circuit operation fatally flawed?
Regards, Mike. |
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| Tony |
| hmm...about this Vbe thing, in my book it is still a Vbe multiplier since the base emitter voltage is multiplied by the number of series string diodes... |
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| AndrewT |
No,
the multiplier function is determined by the ratio of the upper and lower resistors. And is adjusted in exactly the same way.
The diode string adds an offset to the multiplied Vbe.
With the offset only a small multiplication ratio (of about two times) is required to arrive at a suitable output voltage, rather than the five times or so that comes from a resistive multiplier without the added offset.
Leach has arranged for the temperature compensation to be applied to the diode offset rather than the transistor and he explained why he chose this route, even if his logic is correct or flawed. |
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| acenovelty |
| Hmmm, correct or flawed? Except for the issue of drilling four holes in the heatsink, is there a more elegant and simple method to accomplish the task? Probably not. |
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| BobEllis |
With all respect to the lawyers arguing over the meaning of words, the original post seemed to be asking for help figuring out why his amp was not responding to his attempts to set the bias.
FWIW Joe, If you haven't been successful getting the amp to react properly, my guesses are:
1. Solder bridge remaining after you removed the jumper for testing.
2. incorrect/reversed transistor in the Vbe multiplier (or whatever you choose to call it.)
If you have been successful, please post what you found so we can add to the collective "mistakes that have already been made" database. |
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| joe mobile |
Thanks Bob, I already checked that, the jumper was removed. The transistor was the right type, mounted correctly. Maybe the transistor was just broken, but it`s too late to check that, because I blew the amp... all I can say that now T7 is broken.
When I blew R36 the driver transistors were blown. I replaced them, but after that I had 60 V DC on the output. So there are other things broken too.
I will rebuild the amp now, and I hope that this time the bias- setting will work correctly ( and everything else too ).
I also want to thank acenovelty for reminding everybody of my original question. Thank you. Still I think it`s nice when the discussion just goes on in some direction. I`m not sad about that.
Nice forum here, I already spent hours reading.
good night or nice day to everyone,
joe |
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| joe mobile |
to mike
thank you too, you indeed gave me some tips, but I have to admit I`m not far enough with electronics to understand all you proposed... but I read it and I`m learning.
I want to rebuild it, because I think there must be quite a lot broken. Someone gave me the tip to replace every diode and transistor, so why not build a new one. But I`m still measuring on the broken amp to see wich parts are broken.
joe |
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| jackinnj |
| the other thing to look out for is the potentiometer's wiper -- if this goes out it doesn't VBE anymore ! I have seen a resistor paralleled with the potentiometer to prevent this from happening. |
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| audiofan |
Have you build a pair of amp ?
If there is a second amp how the second one is running ? |
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| MikeBettinger |
| quote: | Originally posted by joe mobile
I`m not far enough with electronics to understand all you proposed... but I read it and I`m learning.
I want to rebuild it, because I think there must be quite a lot broken. Someone gave me the tip to replace every diode and transistor, so why not build a new one. But I`m still measuring on the broken amp to see wich parts are broken.
joe |
Joe, Once you get to the bottom of what blew up and the amp is working my suggestions were actually pretty straight forward checks you could make. Unfortunately it's easy to assume that everyone has the same level of understanding and one can tend to get lost in the terminology.
If you have 60 volts on the output it's pretty clear that something is still blown. I'm not looking at your schematic (so I'm not sure what the resistor you shorted out does) but a few minutes spent looking at the place in the schematic that you shorted might draw power from and where the short might have sent power through. This can help you to narrow done where to look for bad parts. You don't need to swap everything out.
From what clues you've supplied I'd assume that you were checking something in the bias circuit or trying to measure the bias on the outputs. If it was the bias circuit then I'd be checking the voltage amp transistors that feed it. If drivers are blown, the outputs are gone as well, even if they don't read shorted (they might have opened up internally from the overload.
Troubleshooting is like a game where you look at the clues and try to figure out what they mean. It's a good way of reconciling your knowledge with the real world.
Send me a link to the schematic your working with and I can be of some help. Also when troubleshooting it helps to wire a 60 watt light bulb in series with the AC to the amp. If you make a mistake the bulb just glows brightly and you don't blow-up a bunch of parts. Do take the time and spend a couple of euros to build a solid/ safe assembly for doing this. You don't want a hotwired power connection on your workbench. You'll need a light socket, outlet box, outlet and power cord. I can draw up a wiring diagram if you're unsure of what I'm describing. You defintely want to be safe when it comes to messing with the AC power.
Good luck troubleshooting. Regards, Mike. |
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| joe mobile |
just finished my first amp ever!
it`s a leach 4.5, but Ichanged the layout so i could use the parts i got from my suppliers here. Plays fine! |
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| joe mobile |
| from the inside |
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