Quick Biasing Question for FastRedPonyCar's Nitro

[Rushtallica]

Hi. Just trying to help a cool poster. Here's his original post from HC:

Despite it's name, I'm not sure if the Nitro's bias is set totally correct or not using the weber bias rite.

I'm switching from EL34's to KT88's. I'm supposed to set it to 60mA per scott's recommendation.

I put all 4 tubes in and used one of the inner pair under the bias rite tool, set it's little switch to cathode current (as opposed to plate volts), plugged it into my voltage meter and set it to the 200m on the DC voltage setting, plugged in the cab and then flipped it from standby to on. It started off reading 70.0 so I turned the screw till it measured 60.

The amp sounds incredible so I'm thinking I did this right but I only played for probably 10 seconds just to verify it worked and now here I am.

So, biasing gurus, is this the correct setting to use on the voltage meter and the right type of measurement to have the weber probe set to?


and though nothing blew up in the process and the amp sounded great when I plugged in and tested it, I just want to verify everything is set to where it needs to be set before playing a full 3~4 hour practice with the band.

Thanks for any help.

 

[Loudness250]

Tell him that if he checks his plate voltage he can multiply the plate voltage with the current like this.

500V x
.060 = 30. watts.

So that would be two thirds of the tubes supposedly max rating of 40watts. But I rate the kt88 as more of a 35 watt tube "for life of the tube concerns". So I would bias the tube at 25 watts. But that's just me. Long story short, he needs to do the math for his amp.

 

[Marshall Law]

Tell him that if he checks his plate voltage he can multiply the plate voltage with the current like this.

500V x
.060 = 30. watts.

So that would be two thirds of the tubes supposedly max rating of 40watts. But I rate the kt88 as more of a 35 watt tube "for life of the tube concerns". So I would bias the tube at 25 watts. But that's just me. Long story short, he needs to do the math for his amp.

the way you do the math is you take the wattage 30 and divide by the plate voltage you say 500 so
30 divided by 500 = .06 or 60mA but thats at 100% dissipation , to hot now, you take .06 and times it by .7 thats point .7 for 70 percent which gives you .042 or 42mA

30 / 500 = .06 or 60mA at 100% dissipation
.06 x .7 = .042 or 42 mA thats at 70% dissipation = Safe
I wouldn't run it at 60 mA or you may be asking for trouble

at 35 watts a tube it would still only be 49mA

thats what it should be at the numbers you gave at 70% you can go a little hotter or colder

 

[FastRedPonyCar]

thanks guys. The weber bias tool has a switch that tells you either plate voltage or cathode current.

I'm using this volt meter

http://www.partsexpress.com/pe/showdetl ... er=390-550

the volt meter set to 2m over on the right side and the weber tool's switch set to cathode current reads 62.0 is that saying 62mA?

If I switch it back over to DC volts and set it to 200m, and set the weber to plate volts, it reads 46.2~.3


following weber's instructions for the BR-AH bias tool, it reads

.046 plate volts
.062 cathode current

multiply those to get .002852

Is that to be read as 28.5 watts?

If so, divided by 500 = .057... so 57mA?

I value everyone's opinions here and Scott himself said no more than 60mA. I guess this is to be translated as the absolute maximum amount possible.

(and thanks for the thread rush!)

 

[Loudness250]

Tell him that if he checks his plate voltage he can multiply the plate voltage with the current like this.

500V x
.060 = 30. watts.

So that would be two thirds of the tubes supposedly max rating of 40watts. But I rate the kt88 as more of a 35 watt tube "for life of the tube concerns". So I would bias the tube at 25 watts. But that's just me. Long story short, he needs to do the math for his amp.

the way you do the math is you take the wattage 30 and divide by the plate voltage you say 500 so
30 divided by 500 = .06 or 60mA but thats at 100% dissipation , to hot now, you take .06 and times it by .7 thats point .7 for 70 percent which gives you .042 or 42mA

30 / 500 = .06 or 60mA at 100% dissipation
.06 x .7 = .042 or 42 mA thats at 70% dissipation = Safe
I wouldn't run it at 60 mA or you may be asking for trouble

at 35 watts a tube it would still only be 49mA

thats what it should be at the numbers you gave at 70% you can go a little hotter or colder

Actually, in my description I told him that 30watt is two thirds of 40 watts which is actually 75% dissipation which was a very vague description. I should have been more exact. But other than that my approach is just different than yours. I take the max wattage of a tube and then bias my tubes at 60 to 70% of that figure. I turn down my bias trimmer and then put the new tubes in and work my way back up to the 60 to 70% figure.

 

[Loudness250]

thanks guys. The weber bias tool has a switch that tells you either plate voltage or cathode current.

I'm using this volt meter

http://www.partsexpress.com/pe/showdetl ... er=390-550

the volt meter set to 2m over on the right side and the weber tool's switch set to cathode current reads 62.0 is that saying 62mA?

If I switch it back over to DC volts and set it to 200m, and set the weber to plate volts, it reads 46.2~.3


following weber's instructions for the BR-AH bias tool, it reads

.046 plate volts
.062 cathode current

multiply those to get .002852

Is that to be read as 28.5 watts?

If so, divided by 500 = .057... so 57mA?

I value everyone's opinions here and Scott himself said no more than 60mA. I guess this is to be translated as the absolute maximum amount possible.

(and thanks for the thread rush!)

For your plate voltage reading, your meter should be set to volts as far as I know. your volts should read before the decimal point like 500.volts. But the current part looks right. I dont know without having your plate voltage but if the guy who built it thinks 60mv is good then he's probably right in the ballpark. If it was me I would measure it exactly myself.

 

[duesentrieb]

Firstly you have to look up a KT 88 datasheet. It has a max power (a+g) of 40 Watts, a = anode/plate, g= grid. Pa is 35 Watts max.

Read your plate voltage off PIN 3 of the octal sockets (or with your bias rite if it has the VI option). DMM set to 1000V DC, red to PIN 3, black to chassis/ground. Should be something in the 450-525V area normally.

Since you are measuring on the cathode you can calculate with 40 Watts - BUT a max. tube dissipation of 55%-60%.

The calculation then goes:

say 450V plate volts, 55% max. plate dissipation:
40 * 0.55/450 = 0.048 A = 49mA
say 450V plate, 60% max. plate dissipation:
40 * 0.6/450 = 0.053A = 53mA

say 500V plate volts and 55% max d.
40*0.55/500 = 0.044A = 44mA
say 500V plate and 60%
40*0.6/500 = 0.048A = 48mA

I've made a little graph for you - and this tells you the completely safe area:
on the Y-axis you have a range of plate volts (so you need to read yours), on the X-axis there's the current for a single KT88 cathode biased between 55% (purple) and 60% (red) of the max. plate dissipation.

What you can see is that no matter which plate voltage your amp has - 48mA is safe, should sound good and keeps your tubes healthy for a long time - and also this doesn't stress your power transformer too much.

Let me know if you have further questions.

 

[Loudness250]

Yeah, it's amazing how many ways people go about arriving at the same conclusions. I bias my kt88's at aroun 48mv myself, at 535v or so , give or take a couple to three. I went on multiple websites to learn how to bias and I found what is the most simple way for me. Once you get the hang of it , it's pretty basic. So assuming a kt88 is a 40 watt max tube,

So I would take

535v x
.048 = 26 watts. and then to get the percentage "which I can already see is in the ballpark",I would,

26watts divided by 40watts=.65 or 65%. I think that's very simple.

 

[Rushtallica]

Thanks for the responses.

 

[FastRedPonyCar]

Ok so the plate volts right now are showing 460.

Assuming I'm wanting to aim at the target range of 48~50mA

And at the moment, the cathode current measurment on the meter is showing .062.


Loudness, using your formula:

460v X .062 = 28.5 watts

/ 40 watts = .7125

So right now, they're way too hot at 71%

correct?


If so, here's what I'm thinking needs to be done so again, correct me if I'm wrong.

I'm looking to lower the bias knob until it reads in the .056 range?

One other question (and I guess I'll find out after work when I get back to the amp) when I adjust the bias pot, will that also alter the plate voltage?

 

[duesentrieb]

62mA is IMO too hot.
Just set it to 50mA and you're done.
Play around a bit within the range of 45-55mA to hear what you like the best.

The plate voltage won't change (at zero volume). Just without tubes you have 20 Volts + and cranked (depending on your power transformer and the filtering) it can go down a bit.

 

[FastRedPonyCar]

if the volt meter is reading .062 cathode current, is that saying it's running at 62mA? Or is cathode current and the mA reading two different things?

If it's not the mA reading, I'm guessing I need to figure out with a 460v plate voltage, what cathode current I need be reading on the volt meter to achieve that 50~52mA target.

Correct?

 

[Loudness250]

No, it's ok, the current is the milliamp reading. All is well. Just lower that number to 50ma, it's a little much at 71%.

 

[FastRedPonyCar]

Done. 52mA it is.

Amp sounds KILLER!!!!!!!!!