is this any good?

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DanDini said:
Eek! I'm using the subwoofer line-out from a Sony AV amp, running into a 10k pot. Are you saying that this is a problem?

Can you suggest a suitable buffer circuit if you think it is?

Thanks,

Dan

Can be a any normal Op-Amp in buffer config.

An OPA2134 can drive 4V RMS into as low impedances as 200-500 Ohm.
3.300 Ohm is a piece of cake for an OPA2134, at gain=1.

Before such an OPA2134, at the inputs,
you are free to use almost any Potentiometer value: 10kohm - 22 - 47- 100 kohm.

I would NOT recommend use of MORE THAN 100 kohm input impedance
for any audio circuits, if can be avoided.

This is a good rule!
( There are a couple special cases exceptions from this. )


lineup
 
sith said:
found it over the internet is it worth to try?
Schematic:
http://www.diyaudio.com/forums/attachment.php?s=&postid=847622&stamp=1140169949


If I remember, this is an Elektor article from 1980-ies.
http://www.elektuur.nl/
http://www.elektor-electronics.co.uk/
http://www.elektor.de/
http://www.elektor.fr/

There has been much water running under bridges since then.
I would not call this a hifi amplifier.
It would give you some power, 100-150Watt , with this rather rudimentary circuitry
and comparatively very few ( 8 ) transistors devices.


For a Class AB amplifier at that power level, we would expect
at least 15-20 transistors. To achieve a normal hifi audio performance.
For excellent hifi, maybe twice that: 30-40 transistors/semiconductors
in the amplifier circuit alone.

-----------------

For example this version below of SymaSym, Symasym 5.3 has got only14 transistors in standard model.
Thanks to keeping it good but simple.

It will have great performance at normal powers
and this is, in my opinion, quite an achievement using only 14 transistor devices!


Symasym5_2_eagle.GIF



lineup
Lineup Audio Lab
 
Re: Re: is this any good?

lineup said:

Should be noticed!
This amplifier has got very low Input impedance,
set by 3K3 input bias resistor (and also feedback resistor is equally 3K3 to avoid offsets.)
( R3, R8 )

No, look again. The input 3.3k resistor is bootstrapped via the feedback network. Still, this is not ideal (I wonder why the used it?), it would be simple enough to use the classical scheme.

Either the zener in long tail should be changed to increase current (now it is ~ 1 mA).
Or the 680 ohm resistor ( R7 ), at one transistor input,
that sets this balance, should be adjusted to achieve
at least 48% + 52% balance, or better, of current sharing in input pair.

I would vote for increasing the zener. Higher voltage zeners tend not to have noise problems, and also, this way the noise gain of the current source is smaller. Keep in mind that any noise that enters the base will be amplified by the impedance at the collector (very high) divided by the emitter resistor, so you want to keep the latter high.
 
I was also going to state that the input impedance is not 3K3. I did not try to calculate but would presume that the feedback factor is at least 20x, which would make the bootstrapped input impedance some 60K, apart from the input resistor (100K? I can't recall).

Then the circuit diagram is not very clear on my screen; what is the value of the tail regulating zener? If this was not covered by previous post, the temperature co-efficient of the zener could present a problem if the voltage is increased much above 6V, as a constant tail current is needed for input stage balance, which is of some importance to keep distortion down. Also, I have found that zener noise could get in here - when heavily bypassing such a zener I had a mild reduction in noise.

Regards
 
Johan Potgieter said:
I was also going to state that the input impedance is not 3K3. I did not try to calculate but would presume that the feedback factor is at least 20x, which would make the bootstrapped input impedance some 60K, apart from the input resistor (100K? I can't recall).

Then the circuit diagram is not very clear on my screen; what is the value of the tail regulating zener? If this was not covered by previous post, the temperature co-efficient of the zener could present a problem if the voltage is increased much above 6V, as a constant tail current is needed for input stage balance, which is of some importance to keep distortion down. Also, I have found that zener noise could get in here - when heavily bypassing such a zener I had a mild reduction in noise.

Regards


Yes, you are right, Johan.
I missed that bootstrap in input stage. It is not much used these days!
So no problem with input impedance and NO Need for preamp to buffer.

See my attachment for better quality image of the input part.

In the Velleman K8060 manual schema we can see the ZENER is 9.1V ( not 3.1V ).
This makes the balance in input stage better. Maybe correct!

Also correct, Johan, that a zener 5V6 - 6V8 would be better from temp stability reasons.
6V2 is often used in professional circuits.

And from this zener diode voltage, f.ex. 6V2 at 10 mA, we can calculate R9 for good balance in input pair.
Will be like 5.5V across resistor R9.

lineup
 

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  • vellemank8060-part.png
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Ah, now I can finally see the diagram with parts values.
There is certainly room for improvement, I believe we covered this amp before. In short:

1) LTP biassing: the Long Tail is not very long with only 3k. The zener diode should be increased to some 5-10V below the lowest expected rail voltage this amp will see, and the resistor R9 adjusted so that (Vzener-0.7)/R9 remains the same, then adjust R11 for the same zener power dissipation. Bypassing the zener with a cap is a good idea also. The point of R9 is to approximate a current source, hence a very high impedance. 3k3 is NOT a very high impedance. Even so, this resistor cannot likely be increased over 10k, replacing the whole thing with a proper current source is the way to go.

2) Input resistor bootstrapping is really not necessary here. Increasing R8 and keeping R3 the same will keep offset reasonably small (as small as can be expected from this topology), but connect R3 to ground. 33k would be a good value. Then also increase R1 to something like 330k and you will have roughly 30k input impedance, which should be OK. To keep the gain the same, R2 should be increased to 1k. Keep C11 the same but if possible use a higher voltage part (50-63V). C2 in parallel with the input is a bit odd, it is very much undefined what kind of HF it will block.

3) BC639 for the VAS is going to be running hot. Using BD139 with a small heatisnk is a better idea. BC639 is essentially a BD139 in a small TO92 case so no other changes should be needed.

4) The output stage is asking for trouble. TIP142/147 DC SOA is limited to only about 2.3A with 40V across it, which is what will happen in a short circuit. For 80V, it is only 0.5A, and 1A for 1A for 60V, the latter can happen with reactive loads. Although this is a pessimistic view as DC SOA is the abosolute worst case, the 'protection' limits current at 8.4A!!! Attempting to drive a perfectly resistive 3 ohm load with +-40V rails will destroy the output darlingtons for sure. This will also happen for higher but reactive load impedances. This 'protection' is completely inadequate.
The same parts count could have been used to make a foldback SOA protection. Off the top of my head:
- replace diodes D1 and D2 with shorts
- put D1 and D2 antiparallel with BE junctions of T4 and T5 respectively.
- remove connection of R16 and R6 to output, and connect this conenction to ground instead.
- replace R19 and R20 with 0.33 ohms parts
- replace R16 and R6 with 3k3 (this is for 40V rails)
- make sure flyback diodes are connected (C to E of output transistors will do)
This is far from ideal (proper resistor values should be calculated) but lightyears ahead of what is used now. It is quite possible that this protection will activate too often and sound bad on difficult loads, but at least it won't blow up, assuming you have genuine output transistors. It is possible to get it working better by connecting capacitors across B-E of each T4 and T5, using dynamic SOA, but my datasheet on the TIP parts only has DC SOA so I won't speculate here.
 
I give up!

I've tried the protection diodes across e/c - it blew up. I've tried removing the current limiting all together - it blew up!

Can anyone suggest a simple amp design that will be good for a modest subwoofer arrangement? I need to be able to build it on vero using cheapish parts.

Thanks again for all your suggestions, but I think the time is coming when I need to put this one to bed - before I lose any more hair.

Thanks,

Dan
 
richie00boy said:
The positive rail has the anode of one diode, and the negative rail has the cathode of the other diode. They join in the middle at the output node.

Rather belated, I am just a little worried about that description. The diodes are connected in normally reverse mode. That would make the emitter (cathode) of one go to the positive rail and the collector (anode) of the other to the negative rail, with the junction at the output terminal etc. The way you described it above is a short circuit across the power supply. What do I misunderstand?

Then the diagram in posted in #34 by Lineup also puzzles me. The R7 in series with the load - is that all of 10 ohm??

Regards

Sorry, that was post #44.
 
Johan Potgieter said:


Then the diagram in posted in #34 by Lineup also puzzles me. The R7 in series with the load - is that all of 10 ohm??

Regards

Sorry, that was post #44.

Sorry, for the confusion, R7 is in paralell to the outputcoil, not shown in the eagle schematic, because the coil is formed by winding a wire around r7.
You see the complete schematic here.

Mike
 
Just checking: you do have T6 well thermally coupled with the heatsionk?

I think I do, but this part is not failing so I don't think it can be the cause, or am I missing something. This thing will die as soon as I crank it up, it's not even hot - anywhere.

The end result when it fries is both output transistors dying. All else appears ok as the fuse on the transformer blows before anything else bites it.

Thanks.
 
DanDini,

Yipes, you need an oscilloscope across the output to have a look - h.f. oscillation. The coupling of T6 to the heatsink is not to protect it, but it senses the heat and adjusts output transistor current to protect them - so T6 will not come to harm. However, you state that your transistors blew up before there was heat.

Fortunately you seem to have a condition (without signal) where everything is OK. If you can monitor with a scope and turn the signal up slowly and be ready to back it off, you might see what the matter is before blowing more transistors. A fuse will not protect the circuit in case of h.f. oscillation - it is too slow.

Where is your volume control? Is the feedback in order? I fear there are too many questions to assist you here on a screen. If you can get a scope that could probably tell you something.

Good luck.
 
you need an oscilloscope across the output to have a look - h.f. oscillation. The coupling of T6 to the heatsink is not to protect it, but it senses the heat and adjusts output transistor current to protect them - so T6 will not come to harm.

Ok, my friends at work have a scope, I could check for oscillation there.


However, you state that your transistors blew up before there was heat.

That's correct.

Fortunately you seem to have a condition (without signal) where everything is OK. If you can monitor with a scope and turn the signal up slowly and be ready to back it off, you might see what the matter is before blowing more transistors. A fuse will not protect the circuit in case of h.f. oscillation - it is too slow.

Yes, without input I can crank it up, so I will check.

Where is your volume control? Is the feedback in order?

I just have a 10k pot connected across the input at the moment. I've also used a 47k as suggested in the manual. Same issue. I'm not sure what you mean by 'is the feedback in order?' (sorry - non-expert here).

Thanks again.

Dan
 
Johan Potgieter said:
Rather belated, I am just a little worried about that description. The diodes are connected in normally reverse mode. That would make the emitter (cathode) of one go to the positive rail and the collector (anode) of the other to the negative rail, with the junction at the output terminal etc. The way you described it above is a short circuit across the power supply. What do I misunderstand?

You are right! In my haste I got the anodes and cathodes the wrong way round. I wanted to draw a diagram.
 
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