Express success! Ish......

[MikeR670]

Ok so I got it fired up. Just hooked a little 8 ohm car speaker to it. It squealed and squawked loud, like a rancid pig in a poke with a lame duck, but I realised I had forgotten the NFB loop. Didn't fix much but I swapped the OT leads and it calmed down.

Still had some buzz but I played guitar through it. Sounds rough but clean if that makes sense....

After about two minutes of playing I noticed smoke it was the 1k 5 watt resistors between the b+2 and output tubes screens. Burning up. I didn't have any 1ks, so I put two 470 ohm 5 watt in series, for each lead.

I also didn't get time to measure currents etc. Tubes all glowing nice orange, no red visible.

Any ideas of the cause of this?

[MikeR670]

Oh, I forgot to mention - this amp is the Kelly 90 express layout, except with 6550C output tubes - as my b+ too high for 6v6s.

[shipwreck]

We need pics/details

[pompeiisneaks]

Lets do some ohms law math:

V = I/R right?

P = V*I

We don't know the exact voltage, but in theory we know the wattage of the resistor at 1k is enough. If it is, we can then calculate for 5w Power and resistance of 1000 ohms as:

I = SQRT(P * R)

which in this case = SQRT(5 * 1000) which expects max amperage to handle this is 70.7 amps

IN that case lets put in max of 70.7 amps back into the power calculation and solve for what wattage we want at 470 ohms

P = I^2 / R

P = 70.7^2 * 470 = 10.6 watts.

That means that you're already over if it's at max, now they do design for overage, usually giving say 10-20% room for error, overages etc. The only problem is that the two resistors aren't likely perfectly balanced and therefore one is maybe 465 and the other 471 or some randomness. This also means the lower of the two is going to take the brunt of the work, and would likely smoke anyway, but on top of it, you may have slightly higher voltages than expected in the circuit.

two 5 watters don't always exactly equal 10 watts of power handling. (Remember the lower the voltage goes, the more current it handles)

That's just spitballing, and I could be wrong, but I think you're likely going to want 10 watters for the 470's just to make sure they've got plenty of room to handle the current.

[MikeR670]

Lets do some ohms law math:

V = I/R right?

P = V*I

We don't know the exact voltage, but in theory we know the wattage of the resistor at 1k is enough. If it is, we can then calculate for 5w Power and resistance of 1000 ohms as:

I = SQRT(P * R)

which in this case = SQRT(5 * 1000) which expects max amperage to handle this is 70.7 amps

IN that case lets put in max of 70.7 amps back into the power calculation and solve for what wattage we want at 470 ohms

P = I^2 / R

P = 70.7^2 * 470 = 10.6 watts.

That means that you're already over if it's at max, now they do design for overage, usually giving say 10-20% room for error, overages etc. The only problem is that the two resistors aren't likely perfectly balanced and therefore one is maybe 465 and the other 471 or some randomness. This also means the lower of the two is going to take the brunt of the work, and would likely smoke anyway, but on top of it, you may have slightly higher voltages than expected in the circuit.

two 5 watters don't always exactly equal 10 watts of power handling. (Remember the lower the voltage goes, the more current it handles)

That's just spitballing, and I could be wrong, but I think you're likely going to want 10 watters for the 470's just to make sure they've got plenty of room to handle the current.

Great thanks! I'll try to follow this formula - it's about time I got fully acquainted with ohm's law

Yes, I did notice that the first resistor on each side was getting more damaged than the second one, i.e.the ones on the b+ side.

Looking at the Ron Worley BOM, I saw that it only called for 1k 5 watt - although that was for a lower powered 6v6 power section with b+2 at say 410 vdc. I'm at 535.

But the old pa I used for this build had the same (or very close) b+ levels, used 6550s, and only a 470 ohm 5 watt screen resistor to each tube. Which I salvaged and used.

So I shouldn't need 10 watt in there I'm thinking.....I'll check the speaker group to make sure I didn't wire the wrong impedance line there.

It only burned up when playing through it, it idled with no apparent issue, is there a way to check current draws for some imbalance or should I just put 1k 10 watters in and try again?

[pompeiisneaks]

you can get the current draw of the tubes at idle, but it won't help what max will be unless you hit it hard at max volume. Usually it's safer to assume max current and go from there.

Edit: I forgot I was going to do that math for you. If it was at max dissipation at the full 535 volts (which it probably would never be)

I = V/R = 535 / 1000 = .535 or 535mA. Then multiply that by voltage for power: P = IV = 535 * .535 = 286 watts... the reality, though, is that power tubes never drop the full voltage. You can design based off the "maximum dissipation" on the datasheet, that's the dissipation level that would smoke the tube via redplating and make sure you have that. Use the same formula, but instead of guessing the amps it may run at, get that milliamp rating and use that last formula

P = IV, and you'll get the best estimate of maximum wattage needed.

~Phil

[MikeR670]

We need pics/details

It was a big old pa running four 6550s or 66ca7 s:

rps20170425_103434_494.jpg

rps20170425_103511_586.jpg

I wanted something compact so I stripped it and hacked up the chassis for prototyping. Ended up deciding to try the express preamp with two 6550s for the power section.

I worked from the Kelly 90 schematic:

Screenshot_20170424-165706.png

And the Todd Hepler turret layout, as I've never transposed a schematic to a board before:

Screenshot_20170424-165907.png

Ok, no guffaws at my redneck build
[attachment=0]rps20170425_104131_469.jpg[/attachment

I'll post the gut shot below, it won't fit in here.]


My plan is to get it to a reasonable sounding state, ignoring wire cross hum etc - and then build it all properly into a new chassis. I made the turret boards real quick from fibreglass and banged some cheap eyelets in them.

The final build would place the tubes better, more spacing perhaps. At least it now only measures 14 1/2" length, so about 16" overall in a nice case I'll build later.

It also contains a bunch of old stranded car wire in it right now, as I can't get any solid core 18 or 2016 awg around here. I'll order some with right colors online, to rebuild.

Any suggestions or pointers much appreciated!

[MikeR670]

Here's the guts

rps20170425_105628_431.jpg

[MikeR670]

you can get the current draw of the tubes at idle, but it won't help what max will be unless you hit it hard at max volume. Usually it's safer to assume max current and go from there.

Edit: I forgot I was going to do that math for you. If it was at max dissipation at the full 535 volts (which it probably would never be)

I = V/R = 535 / 1000 = .535 or 535mA. Then multiply that by voltage for power: P = IV = 535 * .535 = 286 watts... the reality, though, is that power tubes never drop the full voltage. You can design based off the "maximum dissipation" on the datasheet, that's the dissipation level that would smoke the tube via redplating and make sure you have that. Use the same formula, but instead of guessing the amps it may run at, get that milliamp rating and use that last formula

P = IV, and you'll get the best estimate of maximum wattage needed.

~Phil

Ok perfect, I'll work this through and report my outcome....

Thanks Phil!

[pompeiisneaks]

according to one datasheet for the 6550 (tung sol) it says max cathode current should be 190mA (that's across the cathode meaning the tube conducts it, if I get it right)

Therefore .190 * some voltage drop at the resistor

but you'd need to measure hte voltage drop across that resistor to know for sure. THe best way to be 100% sure is to just do two measurements.

1. measure the exact resistor value, say you got 1001 ohms
2. measure the voltage across the resistor (measure with your probes above and below the resistor. Say that way 50 V then you'd do:

P = (V^2) / R = (50^2) / 1001 = 2.5 watts. (this isn't going to be exact for your amp, you'd need to measure it). Then over engineer that by say 50% or more. so say 3.5 or so watts, but the next whole watt number for resistors should be 5 watts. Never hurts to be safe. (I made up those values, so be wary).

Also realize this is at idle, not being played so the voltage drop will increase as current is drawn.

The normal calulations don't work here because the resistor won't ever see all 535 volts, just the drop from the 535 down to whatever the tube draws across it.

[pompeiisneaks]

I should clarify, I was showing 'max possible' which can never happen, see my latest update.

~Phil

[MikeR670]

Ok got it, I'll dig into this and see if I can get a bit more scientific with the power section. Time to stop winging it me thinks.....

Really appreciate your time and input Phil

[JazzGuitarGimp]

When you know the resistance and power rating for a resistor, and want to calculate for maximum current to obtain the resistors power rating, use this formula:

P = I^2 * R

Power (in watts) = I^2 (current squared) * R (resistance)

The formula needs to be re-arranged to make it useful. By dividing both sides by the resistance, the formula becomes:

P/R = I^2

Now take the square root of both sides and we have.
sq rt(P/R) = I

Now plug the numbers in. We solve for P/R first: 5/1000 = 0.005

Now take the square root of 0.005 = 0.0707A or 71mA

Keep in mind that we do not run resistors at their maximum power rating. A safe operating level is about 1/2 of rated power. Some folks will derate even further to 1/4 of rated power. Suffice it to say under all operating conditions, there should never be more than 35mA following through a 1K 5W resistor. Now let's use ohms law to figure out what the voltage drop is across a 1K 5W resistor with 35mA flowing through it:

E (voltage) = I (current) * R or,

E = 0.035 * 1000 = 35V

If you have the 6550's biased to 70% with a 500V B+, the anode current will be:
Maximum anode dissipation for a 6550 is 42W, so 70% is 29.4W, so:

29.4/500 = 58.8mA

Screen current at idle is typically 10% of anode current at idle, so about 6mA per tube of screen current at idle. Noe let's see what the voltage drop across the 1K screen resistor will be at idle (this is approximate, but should easily be +/-50%)

E = 0.006 x 1000 = 6V

[MikeR670]

When you know the resistance and power rating for a resistor, and want to calculate for maximum current to obtain the resistors power rating, use this formula:

P = I^2 * R

Power (in watts) = I^2 (current squared) * R (resistance)

The formula needs to be re-arranged to make it useful. By dividing both sides by the resistance, the formula becomes:

P/R = I^2

Now take the square root of both sides and we have.
sq rt(P/R) = I

Now plug the numbers in. We solve for P/R first: 5/1000 = 0.005

Now take the square root of 0.005 = 0.0707A or 71mA

Keep in mind that we do not run resistors at their maximum power rating. A safe operating level is about 1/2 of rated power. Some folks will derate even further to 1/4 of rated power. Suffice it to say under all operating conditions, there should never be more than 35mA following through a 1K 5W resistor. Now let's use ohms law to figure out what the voltage drop is across a 1K 5W resistor with 35mA flowing through it:

E (voltage) = I (current) * R or,

E = 0.035 * 1000 = 35V

If you have the 6550's biased to 70% with a 500V B+, the anode current will be:
Maximum anode dissipation for a 6550 is 42W, so 70% is 29.4W, so:

29.4/500 = 58.8mA

Screen current at idle is typically 10% of anode current at idle, so about 6mA per tube of screen current at idle. Noe let's see what the voltage drop across the 1K screen resistor will be at idle (this is approximate, but should easily be +/-50%)

E = 0.006 x 1000 = 6V

Awesome info, thank you! In gonna get to the bottom of this

Meanwhile I plugged into my home practice 2x12 cab and the resistors stopped smoking