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Old 1st June 2006, 06:13 AM   #21
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Default Swamping in Output Circuits?

This has been a fun thread, but I'm out of here.

The output circuits of most power amps are not common collector. That would require the speaker to be connected to one of the high or low voltage rails. It isn't, it's grounded. And the approximate value of the drive at the output is also at ground level.

The output circuit is an emitter follower, which is a common emitter circuit with the load on the emitter side, instead of the collector side.

Also, I've never seen an output circuit that was "swamped", unless you have a different definition than the one I was taught in school. In order for the output devices to be "swamped", the emitter resistors would have to have values of 2 to 4 ohms. That's not my experience. Most of the time, the emitter resistors are fractional values between 0.12 ohms and 0.47 ohms. Essentially, they are a dead short. That's because they are only intended to be safety devices and current limiters.

There is a reason why the output cannot be "swamped" at the emitters in an emitter follower output. The reason is damping factor. Most solid state amplifiers brag about that. Damping factor is the ratio of the speaker impedance to the impedance of the amplifier output. For example, an amplifier with a damping factor of 100 into an 8 ohm speaker has an effective output impedance of 8/100ths of an ohm.

There's no way to achieve that if the emitters are "swamped" with a 2 to 4 ohm resistor. No, not even with some major negative feedback.

Now, about this matching business. I was pretty young in 1964, when I started doing stereo. The first couple of times that I replaced output devices, I got lucky and everything worked OK. It was beginners luck and fate favoring the foolish.

It took me a while to figure out what those little dabs of paint on the factory original output devices were all about. The brand new transistors/Darlingtons/MOSFETs that I was buying didn't come with those dabs from the OEM supplier. So how do they get there?

Well, on the manufacturing line, there is a tech who marks the output devices for their matching tolerance. For example, let's say that a certain manufacturer expects the output devices to cover a range of hFEs from 1200 to 2200. The tech will pre-test the devices and mark them with a dab of paint to indicate the hFE code: 1200 to 1300 is black, 1300 to 1400 is brown, 1400 to 1500 is red, etc.

Then the installer on the line simply picks two devices with the same coloured dab of paint and installs them, knowing that the hFEs will match within 100. Pretty slick, huh?

You know, come to think of it, there actually was a solid state amplifier output circuit that was "swamped". Bob Carver built it to win a bet. The bet was that he had to produce a solid state amplifier with the same transfer function as a Futterman direct coupled tube amplifier. He won the bet by lowering the damping factor of the transistor amplifier down to the same damping factor as the tube amplifier. And he did that by increasing the value of the emitter resistors until the output was "swamped". I think the designer was Bob Carver. Is that right?
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Old 1st June 2006, 07:21 AM   #22
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Default Re: Swamping in Output Circuits?

Quote:
Originally posted by chewrock
This has been a fun thread, but I'm out of here.

The output circuits of most power amps are not common collector. That would require the speaker to be connected to one of the high or low voltage rails. It isn't, it's grounded. And the approximate value of the drive at the output is also at ground level.

The output circuit is an emitter follower, which is a common emitter circuit with the load on the emitter side, instead of the collector side.
Excuse me, but that's not what I've been taught. In my world an emitter follower is a common collector circuit.
An amplifier can be built using either common collector or common emitter circuits as outputs, but common collector is by far the most common.

Now, any amp that's designed such that the minimum gain of the output transistors is even close to what the drivers can reliably drive is going to have substantial problems with a variant load, such as a loudspeaker.

However, I'm sure amplifiers from way back sometimes had "creative" ciruits that could go bust if mismatched devices were used, due to cost constraints.

Rune
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Old 1st June 2006, 07:30 AM   #23
AndrewT is online now AndrewT  Scotland
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Hi Chewrock,
Quote:
Also, I've never seen an output circuit that was "swamped", unless you have a different definition than the one I was taught in school. In order for the output devices to be "swamped", the emitter resistors would have to have values of 2 to 4 ohms. That's not my experience. Most of the time, the emitter resistors are fractional values between 0.12 ohms and 0.47 ohms. Essentially, they are a dead short. That's because they are only intended to be safety devices and current limiters.
Something was lacking in the way electronics knowledge was imparted by your school teacher!

With regard to emitter resistors.
They are not swamped, that is something to do with water.
They do not need to be 2r to 4r.
They are not a dead short.
They are not intended to be safety devices and current limiters.

Their main purpose is to maintain stability in the output device current and secondary (I think) to define the current gain more accurately.

They can be used as tapping points for a current or VI protection detector system, but this is optional and the correct circuit operation does not depend on the protection circuit.
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Old 1st June 2006, 09:18 AM   #24
ilimzn is offline ilimzn  Croatia
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Default Re: Swamping in Output Circuits?

Quote:
Originally posted by chewrock
This has been a fun thread, but I'm out of here...
I was pretty young in 1964, when I started doing stereo...
Maybe it's better that you are out of here... apparently you still don't understand the operation of the circuits you use since 1964?
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Old 1st June 2006, 01:21 PM   #25
anatech is offline anatech  Canada
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Hi chewrock,
Quote:
I think the designer was Bob Carver. Is that right?
That is very basically correct. Those amplifiers did a couple other things, but there was a resistor in series with the output outside the feedback network. Notice there were no non-linear fuses, just a "linear" resistor (I think it was wire wound). The TFM series adopted these concepts.

-Chris
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Old 1st June 2006, 08:53 PM   #26
elaar is offline elaar  United Kingdom
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Hi again,

Thanks very much for all of the replies, they've been very helpful.

I finally got my new batch of sanken sap15s and 100ohm trimmer pot replacements through today. I put the trimmers in the halfway 50ohm position, everything was installed correctly with mylar insulators, good thermal paste and all the old damaged components were replaced.

Powered it up and it works, sound and everything!

With no manufacturer circuit specs I'm a bit lost (I tried forums and ringing the company with no luck).
The board has just one trimmer per channel. I've been examining the sap15 datasheet as that's my only source for now. Like the datasheet, the amp circuit has the trimmer between the diode outputs of the pair of transistors, the reason for this is to adjust so that the current between these points measures 2.5mA. This will also have an affect on the idling current which should measure 40mA where the emittor points meet.

Still with me?

Well, i tried adjusting a trimmer on a channel, this seemed to have absolutely no effect on either the DC offset OR the forward current. The voltage across the trimmer increases or decreases as it's changed so that the current across the trimmer is always about 3.6mA (not 2.5mA).
The DC offsets by the way are 14mV on the left channel and 23mV on the right.

So.... any ideas? As you can see i'm not really that clued up

Cheers,
Andy
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Old 1st June 2006, 09:06 PM   #27
elaar is offline elaar  United Kingdom
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Sorry, me again.

The emittor resistors are included inside the transistor package, and are 0.22ohms, how would you measure the voltage across this to get the quiescent current? Because both emittors in the pair meet (creating the amplified output), this means that the resistors are in parallel (2*0.22ohms), correct? How would this effect the quiescent current?

Cheers again,
Andy
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Old 1st June 2006, 09:09 PM   #28
anatech is offline anatech  Canada
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Hi Andy,
You live a charmed life. You could have instantly lost the new parts.

It sounds as though there are some other defective parts as well. Don't turn the amp on again until you have checked everything. Try to use some way of limiting the current in case of a fault.

The bias transistor (if there is one) may be partially shorted, resistor open (check the low value parts). You may have a predriver blown and leaky, that took a while for me to find once.

So test all the other transistors in that channel. Take a picture first so you get everything back properly aligned.

A schematic would really come in handy right now.

-Chris
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Old 1st June 2006, 09:14 PM   #29
anatech is offline anatech  Canada
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Hi Andy,
Measure across the S and E pins as seenhere (thanks Farnell). Your voltage reading / 0.22 = the current.

-Chris
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Old 1st June 2006, 10:54 PM   #30
elaar is offline elaar  United Kingdom
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Hi anatech

"You live a charmed life. You could have instantly lost the new parts.", hehe, well i'd run out of options

I spent many painfull hours last week checking EVERY single transistor on the board, of course all damaged resistors and caps were replaced, there isn't much else on the board, so there was a fairly low chance of something major going wrong

Basically i was pretty certain there were no other problems on the board.

Thanks for the reply, I measured and adjusted and as the datasheet recommends, set each emitter current to 40ma. Without any service manuals I can't think what else to do, it's not a hugely expensive amplifier so it's not worth getting it professionally repaired. The DC offset doesn't seem too bad, I'll have to find some cheap speakers to test out the sound.

Thanks for the help,
Andy
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