How to test a Bi-Directional Triode Thyristor

[jim hardy]

Finding a 100 watt clear light bulb in today's world on the other hand is a bit hard to do..lol

Home Hardware, Krome and about N 6th... you'll love the place.

I guess the one that sold is coming to me Jim. I ordered the first one.

Ahhh, laughing. Good , it'll serve you well..
Study up on holding current, gate characteristics, and protection for dv/dt and di/dt.
Then you'll have the basics.

 

[Planobilly]

I built the triac switching circuit and tested the old triac which tested good. Well...it worked the way the the schematic said to test it.

I am starting to think that perhaps there is no gate voltage for a reason. The triac is acting as a safety device based on some fault in the amp. How that would work I have no idea at the moment.

Diode D402 was shorted which is a 1n4002. I replaced it with a 1n4004 as I did not have a 4002 on hand. I don't think it matters much as the only difference is the max voltage.
I will study the schematic a bit more and do some testing tonight.

Billy

 

[Bonkers]

1N4004 is fine, strictly you should replace 403 with the same part (variations in Vf might affect the 50:50 purity of the triac conduction, resulting in DC then saturation in the transformer)
shouldn't matter much tough the thing is "digtally" driven, the opto-coupler here:
http://www.beck-elektronik.de/fileadmin/templates/beck_folder/opto/optokoppler-opic/PC401.pdf
is a digital output part, there is no linear control loop,
the opto is essentially blanked (LED is lit) for N% of a mains half-cycle, this inhibits the triac from firing, the output transistor of the opto shorts the bridge D401 out, no gate voltage, positive or negative.
(presumably,) under no-signal conditions, there would be a 120Hz pulse on the opto, the thyristor etc, to ensure some rail volts - there are at least 3 rails, the 12v supplies should be there the whole time, so there shoudl be some phase angle through the opto.

you should be seeing some 120Hz square waves on the input to the opto.

I suspect there arent, because the "DC fault protection" is active. - again the block diagram is detailed and useful.

plan then would be to power up first teh +/-12V rails, with a lab supply, see if there is a DC fault declared.
then power the+/-30V supply with a 1A limited lab supply, see if a fault declares,

or, with only the +/- 12V, check there should be 120Hz signals at the opto input and output - this is telling the triac to pass some power. if the signal is present, then put the triac an all the mains stuff back in circuit and get back to me., ...
v then the 60V

check that with the triac out of circuit (to see if there are other sources for the rail supply) -
Then the control circuit ( the one controlling the power amp rail voltage)

 

[meBigGuy]

D402 shorted is pretty serious.

I assume you have measured all the transistors and the bridge in the power-on circuit and measured the thermal protection devices and internal transformer fuse.
Also, the opto device.
Do you see any voltage across the power switch when it is open (before turning it on)? If not, suspect the thermal protection switches.

Do you see any discoloring in that vicinity?

WOW --- that is a tricky circuit.

 

[jim hardy]

Okay some USWAGS here, Unscientific-Wild-A**-Guesses ..

what the heck is this on the transformer ? Maybe others have a better idea..

I think it's the source of gate current for the Triac.

D401 is a full wave bridge. Its AC corners are left and right, DC are the vertical ones.
Current through that bridge is allowed, disallowed, or modulated by PC401 above.
That current flows out the left corner of D401 into Q401 (or Q403 depending on which half cycle) ,
and on out through Q404 & D402(or Q402 &D403) into the gate.
Need about 50 ma of gate current to trigger the triac you linked.
http://www.datasheet.hk/view_online.php?id=1927120&file=0465\sm16gz41_6629879.pdf
So how does gate current get into that bridge's AC side ?
Here's the only way i can see...

You can squeeze 50 ma peak through R401 if it's 5.1K but not if it's 51 K.
Is it green-brown-red or is it green-brown-orange?

Must tap into a transformer winding someplace ?
Aha ! Near its bottom would be like the test circuit you used...as soon as the triac fires, you no longer need gate drive that's how they get away with only 1/2 watt R401.
The thermal gizmo that looks like two shepherd's hooks, might shut down the amp in case of overheated transformer by removing gate drive.
Power switch does same thing by shorting gate drive to ground at bridge output?
LOL a power switch that's open to apply power ? Showoffs !

Anyhow that's my USWAG. I'll wait for comments from the guys who know this stuff better than me to chime in.

Gotta help Fair Anne with some stuff .. TTFN ...

old jim

 

[Planobilly]

Replacing R409 at the moment.
I will read again and be sure I understand Bonkers post.
I have not checked the thermal switch yet Big Guy. No indication of anything burned or discolored, plus the main fuse never has blown.
I would be glad to trade this "tricky circuit" for some good old tube amp circuits!...lol

Edit: Green Brown Orange Jim

 

[jim hardy]

Edit: Green Brown Orange Jim

okay - i don't understand how they get gate drive current

<white flag of surrender icon>

thanks

 

[Planobilly]

There is a direct connection from the mains through R409 to the gate. Why would that not open the gate?? There is 120V AC to the R409 1000 ohm resistor which is connected to the gate. There is 120V AC at T1 and T2 of the triac

Getting a bit tired...need to review everything that has been said.

I just removed the 100 ohm resistor and put a 1000 ohm resister in the test circuit I built and it turns on the gate and current flows...

Perhaps I need to take a break.

Cheers,

Billy

 

[jim hardy]

i think the Triac's datasheet tells us here

it's voltage between gate and T1 that fires(triggers) the triac by pushing current into the gate
guaranteed 3 volts and 50 milliamps will fire it at 25°C
but it might fire at much less, anywhere in the shaded area.
(You'll love your new SCR manual...)

There is 120V AC at T1 and T2 of the triac

Do you mean between T1 and T2 ? Across the triac? That means it's not turning on, which you know already.
You need a very few volts between gate (terminal 1) and T1 (terminal 2). That's where R409 is connected, from gate to T1. Gate current has to come in from above, via R408 and D402 (or D403).
Now - R408 is only 4.7 ohms. That you found D402 shorted suggests it saw excessive current. If it did, that current also went through R408.
Is R408 still 4.7 ohms ? I'd not be surprised to find it swollen , cracked and open circuit.
How's D403?

I'm still puzzled by source of gate drive for this thing.
See my sketch in post 40...

@meBigGuy
I need an AC source to top of R401 to make this thing go - what am i missing ?

old jim

 

[meBigGuy]

@Jim
I wonder about how the triac is drawn. They don't label the pins.

R403 (10 ohms) to the power switch essentially breaks the path you drew, shorting the inputs to the darlingtons to neutral through the power switch.

I assume the gate current comes through R403, but I don't understand how the opto/bridge modulates the firing point.

They do call the transformer a "magnetic field coil". Maybe there is something special about it and the unlabeled pins.

 

[jim hardy]

@Jim
I wonder about how the triac is drawn. They don't label the pins.

Datasheet

Carver schematic

you are right. I assumed it's drawn T2 left, T1 and gate right. I can't make it work any other way.
T1 and T2 are same size push-on terminals and i suppose they could get inadvertently swapped... Would that put line voltage across R409 ?

I don't understand how the opto/bridge modulates the firing point.

There's a lot of auctioneering. I don't know if it's phase control or just on-off, probably capable of either.

Planobill - In your mind move the optoisolator inside the bridge and it's more apparent it controls current . That's how you control AC current with a DC device.

will look up that opto, has a part# on schematic..tlp631
http://www.digikey.com/product-detail/en/toshiba-semiconductor-and-storage/TLP631(GB,F)/TLP631GBF-ND/871364
Toshiba, it's good for 55volts, 50 miliamps so they're not blocking line voltage with it

old jim

 

[meBigGuy]

http://carvermk2.com/Docs/Carver Magnetic Field Whitepaper.pdf talks about the features in the amp, including a high level of the power supply and the use of class G power rail switching. It also a great treatment regarding the real demands of high quality amplifier power supplies and the advantages of his design.

Planobilly:
I'd start here: (essentially poke around and look for anomalies, which is tricky given 1. I am here, 2. I don't completely understand the circuit, 3. You don't understand the circuit either) So we just do the best we can.

1. put meter or scope lead on neutral (non fused side of line input)
2. Measure voltage at top of r403 with power switch open. Then with power switch closed.
3. Measure voltage at top of R408 with power switch open, then closed
4. Measure resistance of the two thermal switches with power removed
5. Measure voltage on right side of lower thermal switch with power switch open. Should be line voltage if internal transformer fuse isn't blown
(is that fuse in the transformer accessible?)

Something in that process won't make sense and will provide a clue as to the problem.

Is there anything about the transformer that might explain what the lead that is common the the temp elements are? Maybe it is just a mounting point?

 

[Tom.G]

Diode D402 was shorted...

With D402 shorted, it is very likely that Q404 is also ready for burial. Might as well check/replace Q401, 402, 403 while you're at it.

BTW, is that Power Switch momentary contact type? It looks like it supplies trigger power thru the Q401 - Q404 structure when closed. That would be how the whole thing bootstraps at turn on. If this is correct, then holding the Power Switch closed "should" energize the transformer without any of the feedback loop functional.

 

[meBigGuy]

That makes sense that it could be momentary. Then, pulling the darlingtons toward neutral fires the triac, which would occur through the 15K on the right side of the bridge. Then I can see how the opto might regulate the supply through the bridge. But, it's all conjecture at this point. I'm just guessing.

 

[jim hardy]

Glad I'm not the only one confused.

Something in that process won't make sense and will provide a clue as to the problem.

Good Troubleshooting...

Aha ! at 460% magnification, R404 is 6800 not 680K. There's how gate current gets into the four-transistor trigger circuit Q401-404, not through R401 like i drew it...
Planobilly - is R404 blue-bray-red not blue-gray-yellow?

so here's my first shot at the gate drive circuit - doubtless wrong but can be corrected and polished up.
C405 causes voltage at Q401& 403's emitters to lag that at their bases by a small fraction of a cycle..
Between zero and 90 degrees while voltage is rising,
...Q401 is held off , its eb is reverse biased.
...Q403 is turned ON but D403 blocks gate current.
After 90 degrees when voltage starts decreasing from peak the situation reverses.
...Q401 is now forward biased and Q403 reverse...
...Q401 and 402 deliver gate current to the triac

So it CAN bootstrap up
AND we have achieved phase control because
Current from the right, controlled by optocoupler, advances or retards the firing angle.
Closing power switch sinks all the current that can get in through R404 removing gate drive.

So - R404, C405 and D401/PC301 are the muscle of phase control. Brains are in that column of opamps above.
R404 sees full voltage when system is off. Likely candidate to be open.R401 and R402 drop voltage to something within the photocoupler's 55 volt rating?

Sound plausible guys? You were doubtless there already...

old jim

Edit back to drawing board
R404 is 680K per partslist

so i do not see how that circuit as drawn can deliver trigger current more than peak volts/R401,
170v/51k = about 3 milliamps.

Does it use the charge stored in C405 for a trigger pulse?
dv/dt = I/C , it could deliver 50 ma for 10 microseconds with a 3.3 volt drop...

?
How to check C405 in circuit ?

Check my arithmetic, not to mention the logic.

 

[Planobilly]

Out until tonight.

 

[Salvador]

I apologize before that I will not be able to be helpful here with my post but I just wanted to say a few things, first of all , looks like the triac isn't much of a " regulated psu" after all than a mere safety thing implemented in such a rather puzzling way but maybe it's really useful and has good protection I think Planobilly will see after he solves this mystery.
second of all personally I find the schematic rather hard to read maybe the copy is bad quality but without extreme magnification the traces seem to connectet everywere , although I've seen worse so maybe isn't that bad except for the labeling , I struggled with Onkyo integra myself until i finished it and now it works , although it had a more classic approach when the outputs were blown it simply disconnected the speaker out and when powering up i was able to measure and see the faults easier due to the DC voltage being at the output and also input stages so I frankly knew then were to look for faults.

They do call the transformer a "magnetic field coil". Maybe there is something special about it and the unlabeled pins.

Big Guy said

Well I think it's bla bla radio GA GA , the HI FI high end stuff has many unorthodox names simply for appearance and or any other magic that folks who buy them believe in or like.
For example what Carver meant with magnetic field amplifier as he calls some of his models like the one in question ?

 

[meBigGuy]

I guess you did not read the Carver white paper I linked to.
Too bad.

 

[Planobilly]

Update
After replacing almost everything around the triac and taking a good look at all the transistors without any change I jumped the triac. I lowered the input voltage to around 50V with my variac. One of the filter caps smoked so I removed the secondary connection and am checking the transformer which I think is most likely ok. Nothing got hot except the cap at 50V. I also had a light bulb in the circuit which came on and went off when the cap failed.

I need to order a couple of filter caps and there are a couple of diodes that I want to change. The idea is to find the fault in the amp and then go back to the input of the power supply to resolve the gate issue.

This is for sure not a easy amp to troubleshoot.

This is a link to the patent for the power supply. I really did not understand much of it...https://www.google.com/patents/US4218660

No matter, I am not designing the amp...so I guess I replace parts till get it right...lol

Onward through the frog...or fog or something like that..lol

Cheers,

Billy

 

[Tom.G]

This is a link to the patent for the power supply. I really did not understand much of it...https://www.google.com/patents/US4218660

That patent describes high frequency switching to the power xformer primary. For that to happen, the Triac (TR401) would have to be a Gate Turn Off (GTO) device. That makes sense when you notice that the polarity of D402 and D403 supply Positive and Negative voltage to TR401.

https://en.wikipedia.org/wiki/Gate_turn-off_thyristor

(Sounds like more confusion for this project!)

 

[Salvador]

I read the patent too and quite frankly I don't get were is the big fuss. Carver explains quite reasonably that bigh power HI FI home amps are heavy because of big heatsinks and big transformer.Ok that's true.He also says that heat in heatsinks is a waste product which again is true.
Ok I get his point of jerking the transformer primary to higher and lower power with respect to the input signal waveform with the help of a circuit that " reads" the input and then controls the primary current accordingly by the means of that triac.
But here's the point I don't get , his main points are the size and weight of such amps. with this circuit he hasn't got rid of much the weight only maybe of some of the waste heat in the heatsinks since they now have lower currents at lower signals levels and higher when they are needed. Ok that goes for smaller heatsinks , but the mains transformer being a iron core mains frequency transformer still needs to be as big as the max possible or predicted load of the amplifier will be , in other words it has to be as heavy and big as to be able to supply the amp when it's driven at it's max or near that.
since when a triac is fully open it becomes just a wire and so it simply let's through the whole mains sinus which is no more than 60hz

So where's the gain in that ?

 

[meBigGuy]

Gate Turn Off (GTO) device.

There is a link to the triac spec sheet in the OP.

 

[jim hardy]

Sorry for my gaffes above. I've often said it - I'm a plodder.

There is a link to the triac spec sheet in the OP.

and a link to a Carver brag-sheet in one of meBigGuy's posts.
In that document Mr Carver himself describes the power supply , albeit briefly.
It's just a phase control like a lamp dimmer but instead of a knob it uses that D401/PC401 to control firing angle.
Observe PC401 is under control of the opamp just above.
That opamp's inputs include power supply voltages, and via the heavy dark line a measure of the output.
So i'll correct my last sketch

Good work PlanoBill

 

[Salvador]

Oh trust me Jim , I have added less to this thread than you could ever and I hope Planobilly isn't angry at me for that , I do wanted to say some stuff , I just read the white paper which Big guy desperately wanted me to read.
Much of it is talk-talk although true but still smart advertising then some of it's actual physics and electronics , the thing that still haunts me is were he talks about that a unregulated transformer passes current through only at the peaks of a sinus , isn't that BS? a transformer has the same sinus at it's secondary output the current may lag voltage but it can't simply be there when the voltage has risen to it's max.And with all the benefits at lower power levels and etc I still asked the question which big guy largely ignored dismissing me as a fool , is at full power when the psu is under it's heaviest load , what can the TRIAC help there but to be fully open and acting simply as a wire , so at full power the transformer in this design still becomes the very " dumb" unregulated transformer Carver himself talks about in the paper.So as any transformer it's power rating must be sufficient to deliver the load without dropping much voltage.
I wonder what does the TRIAC and all the fancy name calling (magnetic field etc) help in this case? My guess is nothing.

Also he talks about that a transformer " resists" the ac sine wave 4/5th's of it's time , that would probably be most of the rise and fall time of each cycle , but then how does his TRIAC solve this problem , the TRIAC turns on near the peak of each cycle ad that means that the rise that the transformer primary sees is even steeper almost vertical , ok I see that uses only the most powerful parts of each half cycle , but without that triac those parts are still there with the rest of the rise and fall of the half wave so at near max power again what reduces the very size vs power here ?the frequency is the same.
a genuine question of mine not a show off as some may think.
I also apologize for messing up tis thread but this question seemed very interesting to me.

 

[jim hardy]

the thing that still haunts me is were he talks about that a unregulated transformer passes current through only at the peaks of a sinus , isn't that BS?

no, it's not BS...

Think about a transformer and simple bridge rectifier with capacitor filter

Current flows only during that portion of the cycle where voltage from the transformer exceeds voltage on the filter capacitor. Rest of the cycle the diode is reverse biased. That is the narrow shaded area under the sinewave peak in Carver's drawing.

Here's an example
http://arduino.stackexchange.com/questions/873/reading-a-varying-voltage-into-arduino

That sharp pointy waveform has a high RMS heating value for its average value so it requires big wire
which requires a big core to surround it
I suspect he uses a much smaller transformer than would be required for 1000 watts continuous, and his smart amplifier overloads it only for brief intervals.. hence thermal sensing on the transformer.
But that's only a suspicion - I've never seen the inside of a Carver.

read up on "Crest Factor"

old jim

 

[Salvador]

ok right, stupid me , thanks for pointing that out.I was thinking all the time of a transformer with no bridge rectifier and filter caps at the secondary but those added they do change the picture , hence everything having capacitors also talks about power factor.

according to this graph then we could just use a simply TRIAC for every mains transformer operated psu in it's primary even without the input signal control over it , simply cut off the very starts of rise and fall on each cycle since they would rarely ever go to the cap bank , unless it's being driven so hard the caps are empty almost all the time but that would be beyond hard clipping and distortion anyway.
and so get away with smaller transformers for the same power output ,
why don't others do that ? especially considering the tiny price a thyristor has compared to copper and iron.

 

[jim hardy]

See ? You knew already... Much of learning is discovering what we know.

 

[Planobilly]

Hi guys,

First, I am not put off in the least about what anyone said. A lively bunch of guys are always going to express different thoughts about any subject.

I also did not understand the Bob Carver statements about a transformer resisting the sine wave 4/5 th's of the sine wave. I see Jim is explaining that as I write.

The Carver amp design may in fact be very good, well until one has to repair it...lol Heavy iron may cost lots of money and well, be heavy but it is a hell of a lot less trouble to understand and work on.

One thing is for sure, this project and reading everything everyone has said has forced me to learn a bunch of new stuff!

Cheers,

Billy

 

[Salvador]

now that I'm coming to terms with this approach i would say it's not as complicated as an smps and not as easy as a simple iron with wire outputs , somewhere inbetween.I guess I was a bit out of my manners when I replied to " me big guy" because when I did that I wasn't fully aware of how this thing actually works until Jim pointed out the fact I had forgotten that once you put a rectifier with a cap bank at the load of any transformer it then charges the top of the AC cycle because the rest is already there in the capacitor and this creates uneven load for the AC sine wave.
hence the PFC in the smps units and other stuff.

So Planobilly this is what Carver meant when he was writing about the 4/5ths of a cycle getting through , although I would say the white paper could have been more directly about physics than comparisons and half of it simply about manufacturers that " cheat" and the ones who don't but I guess it's business you have to say some nice words about yourself first.

 

[Tom.G]

There is a link to the triac spec sheet in the OP.

Yup, I missed that. So the patent

This is a link to the patent for the power supply. ...https://www.google.com/patents/US4218660

with its high frequency switching is not applicable here.

 

[Planobilly]

Hi Salvador,

I sort of read around the sales pitch part of the white paper. All this seems straight forward now that Jim posted the graph. Thanks Jim !

Bob Carver seem to be saying "I can build you a amp for $1500 that is just as good as one costing $10,000 and in a blind taste test you can not tell the difference"

I will be glad when the parts get here so I can finish this project.

Cheers,

Billy

 

[Planobilly]

Hi Tom,

I sort of assumed the patent was not directly related to this amp. I did not read all of it. It was more general background information for me into what direction Bob Carver was going toward.

Repairing this amp has taken on a life of it's own and has caused me to consider a good many new things. I guess if one is going to take a dip in the deep end of the pool this amp is as good as any..lol I am sure to many of you this amp design is easy to understand but I have a ways to go to be able to really understand it.

Now if I can just get Jim to show me how to draw on a schematic and get it in a post !

Cheers,

Billy

 

[jim hardy]

Now if I can just get Jim to show me how to draw on a schematic and get it in a post !

click "copy image"
open MSPAINT
click PASTE
click the pencil icon, select a color, draw

click "save as" and give it a name ,
i always save in jpg format because it makes a smaller file than png

Back at pf click UPLOAD IMAGE and start guessing where Windows hid your drawing. I made a folder named "for pf" on desktop where i put mine.

Find Microsoft's "snip" feature, it let's you select and copy part of any image.

 

[meBigGuy]

Just want to share an insight (from the white paper) about why this is a "magnetic" power supply, and can have a significantly smaller transformer and output caps compared to conventional supplies.
(also, it is a class G amplifier, which vastly reduces output stage dissipation)

When the triac switches off, it opens the primary (like an ignition coil) and causes most of the transformers magnetic field energy to dump into the output capacitors.

 

[meBigGuy]

@Planobilly
I really wish you could have made the simple measurements I suggested. That would have helped locate the issue. Who knows how much has been damaged now.

BTW, starting with low line voltage is a good way to destroy things.

Eventually you will need to make the measurements I suggested unless you get real lucky.