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Old 24th February 2009, 09:34 AM   #1
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Default Help needed with bias circuit

Hi Everyone,

I have no doubt I am making some basic mistake but I am struggling to work out what is going on

I am in the building stage of an 845 amp and I have just completed the bias power supply.

I have tested it without any load.

Bench test at 20v, all OK with appropriate bias voltages for low input voltage.

Bench test at 200v. Perfect, with bias voltages within tolerances and appropriately adjustable but both resistor networks (but especially the bridge between the positive poles of the duel cap) becoming unacceptably (as in will burn out) hot within about 30 seconds.

I have triple checked my wiring, all OK as far as I can see. Furthermore, apart from the heat, the supply works perfectly.

The resistors are all of appropriate value and power rating. I measured the current in the circuit at 100mA - too many watts to dissipate!

Does anyone know what is going on? Obviously, I am measuring the supply in isolation and unloaded but I would have thought that any load would increase the current through the resistor network and make things worse?

I am sure I have made some fundamental error, can anyone help?

I have attached the bias supply schematic and link to the whole schematic.

Cheers,

Rob

http://www.audiodesignguide.com/New845/New845v2.html
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Old 24th February 2009, 05:19 PM   #2
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Hi Rob,
I think you have basically discovered the answer on your own. Too much power is being dissipated (wasted) through the resistor network. Simple ohms law (I=EśR and P=IE) will tell you this. For the values given with an estimated assumed voltage of 250V (200V X 1.41 loaded) you're dissipating over 15 watts quiescently. This puts the combined resistor wattage at or below this level, which is simply not enough. Resistors should always be chosen for two or three times (or more) their actual operating wattage.

I don't know what the author was thinking, but IMO his numbers are poorly chosen. Why run such a high quiescent current? If it is to help with regulation, I would use a 25 or 50 watt bleeder at the output of the supply. And 20mA would be sufficient, not 60 or more. Then I would increase the value of the bias pots and their series resistors so as to reduce their operating dissipation. Multi turn controls are expensive and do burn out from excessive current flow and heat.

If the intent was to provide a low impedance DC source to the 845 grids, this is all well and good. But the source does not have to be this low. I've tested enought 845s to know that a properly operating tube will only draw quiescent grid current in the microamp range. And operating at power this will increase somewhat, but never to a milliamp unless the tube is very gassy.

Victor
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Old 24th February 2009, 07:01 PM   #3
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Thanks Victor,

That's what I thought (in fact I thought that it was more like 30W when I measured current draw through the resistor bridge) and what you say makes perfect sense.

This is what I don't understand - The author of the circuit, Andrea Ciuffoli has made several versions of this amplifier. Other members of the forum have written to me and said they have built this design. No one has reported any problems. Am I the first person to notice that the bias resistors burn out within a couple of minutes???

I would love Andrea to comment but I am not sure how to identify this thread to him? I am wondering whether or not he has transcribed the schematic correctly but in looking at the photo's of his design the resistors seems to be of the stated values.

In any case, the problem seems to exist. I have my multiturn pots now and they were expensive! Is there a simpler way to modify this circuit without replacing the pots and resistors?

Rob
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Old 24th February 2009, 09:21 PM   #4
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Quote:
Originally posted by Rob11966
Is there a simpler way to modify this circuit without replacing the pots and resistors?
Sorry, not that I can think of. I went and looked at all the pictures too of the supplies. The three 18K resistors in parallel (6 total) look more like 1600 ohms to me based on the colors I see. Perhaps his camera or my monitor. Never the less, ohms law doesn't lie. If you have 100 mA of current flowing in the bias supply, that's crazy. All I can think of is a leaky filter cap or more then 200V AC going in.
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Old 25th February 2009, 01:57 AM   #5
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Hi again Victor,

I have redone my calculations and on second review, I think the schematic is OK.

According to what I have worked out, theoretically the whole circuit (as tested, without any bias load) should draw 40mA. If we just look at the 4700 resistor bridge this would equate to approximately 5W accross the network or about 0.83 W/resistor which should be fine.

However they are getting hot and I have measured the current draw at over 100mA.

If my above calculations are correct then in real life I must have a problem with the circuit - which is odd as the voltages are OK. I will quadrouple check.

Aleaky cap may explain the situation but I would also be reasurred if you could have a look at my theoretical calcs to see if I am correct concerning the current draw and power dissipation. My first calcs were based on the measured current draw (100mA) rather than the calculated (40mA)

Cheers,

Rob
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Old 25th February 2009, 02:15 AM   #6
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What is the DC voltage (from + to -) measured across each of the capacitors with everything in place?
Have you actually measured all of the resistors you are using?
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Old 25th February 2009, 02:47 AM   #7
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Hi Victor,

It is a Jensen duel cap. I will measure the voltages tonight. They should be around 280v I would guess.

I am thinking leaky cap. A current leak through the capacitor would explain the problem. I might substitute a different cap to see if it makes a difference.

As far as the resistors go, I must admit, because they were new, I only tested one in each pack.

Do you have any comment on my theoretical calcs for current draw? It would reassure me to see that the schematic was OK in principle - I could then be much more certain that I was dealing with a faulty component or incorrect hookup.

Thanks for helping,

Rob
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Old 25th February 2009, 03:00 AM   #8
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I can't do any figuring without knowing what voltages are present. Before I was estimating.
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Old 25th February 2009, 03:26 AM   #9
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Rob..
I came up with..
14.87694122 Watts total
52.886388 mA
5318.94889 Ohms total
Whats the voltage divider for the second terminal for??
Yeah I would up the values with some Five or ten watt wirewounds for the PWR resistors to keep them operating relatively cool..don't want any thermal variations to foul things up.
Check your 'Q' with the inductance of wirewounds tho' ...don't want to make a 'tank' circuit on accident.
_____________________________________________Rick. .......
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Old 25th February 2009, 05:28 AM   #10
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Hi Richard,

I come up with a slightly different number for the second resistor group (which I have labelled R2).

You came up with 2185 and I have come up with 3475. I could be wrong but I can't see where? Please correct me if you come up with your figures again.

If you use 3475 (and as I mentioned I could be wrong), then you get a total R of 6608 with a current of 42mA and a total power 11.65W Total.

The combined power ratings of the resistors are 24W and on the face of it this would seem to be OK.

So, I am getting happier that the schematic is probably OK but my made up circuit is not. I take your point regarding the wirewounds and this may be a good idea but I am measuring over 100mA in the circuit which implies a current draw not identified on the schematic. Even 5W wirewounds won't cope with this sort of current.

With respect to your question concerning the voltage devider, I think that it is just a way of arranging commonly available resistors to get the total resistance and power rating required?

I will measure voltages accross both sides of the duel cap tonight and post results. I will also check the current flow through the capacitor, this may give some more information.

I think that a possible explanation for the difference between calculated (42mA) and measured (>100mA) is most likely to be explained by a faulty cap with a current leak through it. I will replace the capacitor and see if it makes a difference. I hope it's not the cap however. It is new, from denmak and expensive!!


Thanks for the help so far, the problem is slowly becoming clearer.

Cheers,

Rob
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