Can someone clarify this when biasing tubes

[dude]

I know putting a 1 ohm resistor between the cathode and ground gives me exact millivolts. When you look at various biasing settings for tubes, it's in current, mA's. So, it's said the 1 ohm resistor reading is mV but it's really mA'a, cause it's across a 1 ohm resistor..? Can't wrap my head around this, how does volts change to current...? 


Weber's bias calculator says my AB push and pull amp with 6L6's at a 395 volts on the plates, needs 51 mA to be at 70% dissipation.


Usually in the pass I was told, forgetting plate voltage for a monument, that 6L6's should be biased around 30 to 35 mA for about 70% dissipation (estimate). I'm trying to figure out why mVs becomes mAs across that 1 ohm R ...? and why weber's bias calculator says 53.1 mA, do they want the mVs at the 1 ohm R to be at 53.1, (kind of high ...?)


http://www.tedweber.com/webervst/tubes1/calcbias.htm 


al

[sluckey]

I = E/R. If R is 1, then I = E.

[92Volts]

Usually in the pass I was told, forgetting plate voltage for a monument, that 6L6's should be biased around 30 to 35 mA for about 70% dissipation (estimate). I'm trying to figure out why mVs becomes mAs across that 1 ohm R ...? and why weber's bias calculator says 53.1 mA, do they want the mVs at the 1 ohm R to be at 53.1, (kind of high ...?)


http://www.tedweber.com/webervst/tubes1/calcbias.htm 

The equation sluckey posted is Ohm's Law-- the relationship between voltage and current is basically the definition of a resistor.

Any "rule of thumb" for bias current is based on assumptions about plate voltage, and any difference in plate voltage changes that. 35ma seems wrong, though-- that would be 70% at 600v, higher than a 6L6 "should be" (and usually is) run. Maybe that was an estimate for 6V6s, based on their lower ratings and lower voltages they're used with?

[dude]

Usually in the pass I was told, forgetting plate voltage for a monument, that 6L6's should be biased around 30 to 35 mA for about 70% dissipation (estimate). I'm trying to figure out why mVs becomes mAs across that 1 ohm R ...? and why weber's bias calculator says 53.1 mA, do they want the mVs at the 1 ohm R to be at 53.1, (kind of high ...?)


http://www.tedweber.com/webervst/tubes1/calcbias.htm 

The equation sluckey posted is Ohm's Law-- the relationship between voltage and current is basically the definition of a resistor.

Any "rule of thumb" for bias current is based on assumptions about plate voltage, and any difference in plate voltage changes that. 35ma seems wrong, though-- that would be 70% at 600v, higher than a 6L6 "should be" (and usually is) run. Maybe that was an estimate for 6V6s, based on their lower ratings and lower voltages they're used with?

No, that's what came up using Weber's calculator when I entered 6L6GC, AB class operation and 395 plate voltage, 53.1mA's Maybe the calculator's off, seems to be, that's what led me to the question in the first place. 

[dude]

I = E/R. If R is 1, then I = E.

OK, you're making me read and study Ohm's law, probably a good thing, I have no idea what "I" or "E" even stands for 
I have no electronics education, none. Studied Business Administration in school.


Ok, I'll look up Ohm's Law, read and study the theory.

[92Volts]

E = potential (usually measured in volts), sometimes written as V instead.
I = current (usually measured in amps), almost always written as I.

No, that's what came up using Weber's calculator when I entered 6L6GC, AB class operation and 395 plate voltage, 53.1mA's Maybe the calculator's off, seems to be, that's what led me to the question in the first place.

53.1mA is correct. You mentioned you’d read or somebody told you that 30-35mA was typical— that one’s wrong.

Power=voltage*current which we can use to check:
6L6GC power limit = 30W
30W * 0.7 = 21W power target
395v * 0.0531a = 20.9745W
So 53.1mA is spot on (minus a small rounding error).

An actual 6L6 or 5881 has a 23W power limit but most you can get now are 6L6GCs

[ALBATROS1234]

i am an electrician and have an associates in electronics. i will try and explain electricity .i think of it like this in a way. voltage is the potential and current is the pressure. sort of like the speed you are driving in a car is the voltage and the current is the amount of torque the engine can produce. you can have super high voltage(50,000 to a couple million volts) with super low amperage (tesla coil) you can touch the lightning bolts emitted because the current is so low but reverse that and take low voltage with massively high current and youre dead. current is the balls behind the voltage. together I x E=W or current times voltage =wattage(overall power) the only time i really think about current when building tube amps is " can my transformer handle the current the tubes will be needing to draw" other than that i think of voltage for the most part.

[PRR]

> An actual 6L6 ...has a 23W power limit

Original 6L6 was 21W for a few months but quickly backed down to 19W. Many of the later "6L6_" have this rating.

5881 was the 23W job, yes.

> current is the pressure

These electro-hydraulic analogies can be drawn either way.

The more common way for laymen is to say Voltage is Pressure. 100PSI will explode your tire. 100,000 Volts will jump through rubber.

Then current is current like a river. The Mississippi flows a million gallons a second at St Louis. A 6L6 may flow a bazillion electrons a second.

And resistance is like (not quite) the drag in water pipe. If I have a mile of 1/8" tube, I need a huge pressure to get a little current. If I have a mile-wide gully (the Mississippi at St Louis) then very small pressure (1/10th mile drop in 1,000+ miles to New Orleans) makes a very large flow. A 1V drop in a 1r resistor makes 1 Amp of current; you can work that three ways. It says 1 Ohm, you read 1mV, you know it is 1mA.

[dude]

Looks like I've probably been using ohm's law for years and didn't think about it. I wired my house, three of them from the meter to the every outlet, 120v, 240v, all single phase stuff. Knew ohm's law as amps, volts and watts. When I laid out circuits, I would estimate the watts the circuit would need, divide that by 120 volts and get max amps on that circuit for whatever gauge wire I was using, usually 14, so max. amps allowed was 15, 20 gauge wire, 20 amps, like a kitchen, etc. Yes, it's the current that kills not voltage. I've touched live 120 volt circuits and nothing cause I wasn't grounded, had rubber boots on, of course this was a mistake that could have put me on my butt.


So,  I guess Ohm's Law is basically no different then - watts divided by volts equals amps, or amps * volts = watts or etc....


Am I looking at this right, using Ohm's law and didn't even realize it..?
 

[sluckey]

Ohm's law has nothing to do with power and watts! Ohm's law deals with the relationship between voltage, current, and resistance. That's all.

There is also a relationship between power, voltage, current, and resistance, but this is not defined by Ohm's law. I've just always referred to it as the power formula.

It's good to know both basic formulas, ie, P = E x I, and E = I x R.  You can use simple algebra to transpose and substitute the two equations to come up with all kinds of useful equations. For example, P = EI, or I²R, or E²/R.

[PRR]

> using Ohm's law and didn't even realize it..?

Not really.

You are (US NEC) not supposed to "estimate" residential loads. General Lighting is 3 Watts per square foot, which generally covers all small appliances also. Special Cases: Kitchen gets *two* *20A* circuits, Laundry gets a circuit, as do Furnace, well, and the other special loads.

For most work you have two common choices. 15A fuse in #14 wire, or 20A fuse in #12 wire. Since some 20A circuits are mandated, many electricians just run everything 20A in #12.

This is about wire not Burning Up. (Actually about the insulation on the devices not softening.) While Ohms Law is a factor, you need a lot more data to know how hot the wire will get, how hot is safe, safe for a minute or for a lifetime.... NEC's guidelines spare you much mental pain.

Voltage Drop tends to not be a factor for these gauges on runs the size of a house.

I did have problem with 20A in #12 on a long run in a school. The studios were not expected to have more than a small lamp. I had about 9 Amps of computers plus coffee-maker and large laser-printer. Voltage dipped very low when the coffee cycled.

Ohm did prove the concept of resistance, and (after much debate and verification) we named it for him.

Watt worked on Power l-o-n-g before electric power. His approximate "horse power" was redefined in a way that 1V*1A= 1 Watt (more of a mouse power). The electrical units are better defined than weight and gravity, and now that horses are pets we more often measure engines with electrical dynamometers which read in electric units. If you look in European engine specs they often give power as Watts (or KW). That direct derivation from electrical units has no special name.

To use Ohms Law in your house:

You have a BIG house. The wire from fusebox to kitchen is 100 feet long. You pretend you have a 20A load at the end. You like to have less than 2% voltage drop at maximum load.

2% of 120V is 2.4V.

2.4V drop at 20 Amps is 0.12 Ohms.

The resistance of #12 Copper is 1.6 Ohms per 1,000 feet. The 100 foot run is out and back, 200 feet of wire. 1.6 * (200/1000) makes 0.32 Ohms in the 100 foot run. This is more than the 0.12 Ohms desired. 20A in 0.32 Ohms is 6.4 Volts, a 5.3% sag.

Or working backward: #12 is 1.6r per 1,000 feet. We want no more than 0.12r total. We can only have 75 feet total wire, which is 37.5 feet of cable. My kitchen cables run 24' back and 19' over, 43', so I am a hair over 2% sag. A really first-class job would run 20A in #10 wire even for my 43 feet. 100 feet suggests #8 but that won't fit a 15A/20A outlet's screws so it gets messy.

NEC does not mandate a sag, but does suggest 2%. You can certainly decide that 5% is acceptable. If instead of a general kitchen, this is dedicated to a cooker which when fed less Volts will just run longer, 5% may be totally OK. In fact calculations for well-pumps usually allow huge sag. Over 20% at start-up and maybe 7% running. However for general residential work, when you go to sell the house, if the buyer's Home Inspector notices lamp-dimming he will put it in the report with a page of Electrical Expenses which will scare the home buyer.

[shooter]

this will help spin your head;

 https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws
 

[sluckey]

Isaac Asimov's "Three Laws of Robotics"

•A robot may not injure a human being or, through inaction, allow a human being to come to harm.
•A robot must obey orders given it by human beings except where such orders would conflict with the First Law.
•A robot must protect its own existence as long as such protection does not conflict with the First or Second Law.

[Joel]

The Zero-th Law of Robotics means that there are actually 4 Laws of Robotics. But since 3 + 0 = 3, and "The Three Laws of Robotics" sounds snappier than "The Four Laws of Robotics" they're still called the 3 laws.

[jjasilli]

Everything said is true (except I'm not getting the robotics comments).  My 2 cents:



I know putting a 1 ohm resistor between the cathode and ground gives me exact millivolts.
All voltage can be expressed in mV.  E.g.: 1V = 1000mV.  I.e., there nothing magical about the expression of a value in mV, mA, etc.  E.g., the same capacitor value can be expressed in pF, nF, uF or mF, or F.  The point is that to understand amp circuits, free of mental confusion, it's important to get comfortable with the placement of the decimal point, and how this affects proper nomenclature. Failure, refusal or procrastination in learning this will keep your head spinning.

Also, whether a value in a schematic is expressed in mV or V; nF or uF; etc. is a matter of convention.  But, the formula's -- Ohm's Law, and the Power Formula, etc. --  require the values to be stated in Volts, Amps, Ohms and Farads. I.e., if you use mV and mA in Ohm's law, you'll get the right "absolute" answer, but the decimal point will be in the wrong place. 


When you look at various biasing settings for tubes, it's in current, mA's. So, it's said the 1 ohm resistor reading is mV but it's really mA'a, cause it's across a 1 ohm resistor..? Can't wrap my head around this, how does volts change to current...? 
You have to learn & get comfort with Ohm's Law or be doomed to mystery & confusion.  Voltage Drop = Resistance x Current.  Hence: Current = Voltage Drop / Resistance. 
If me make Resistance 1 Ohm, then Current = Voltage Drop / 1. I.e., Current = Voltage Drop!!! 


If a voltmeter tells us that 33mV are dropped across the 1 Ohm bias sense resistor, we now know that = 33mA. 


There is no magic in the use of a 1 Ohm resistor.  You could use ANY small (so as not to disturb the overall cathode resistance) value for the bias sense resistor, say 2.7 Ohms.  Then, Current = Voltage Drop / 2.7.  But it's quick & easy to use a 1 Ohm resistor and skip the arithmetic.


OTOH, a 10 Ohm 1/2W bias sense resistor could be used.  This shifts the meter reading by 1 decimal place.  The advantage is that the resistor serves double duty as a "fuse" to protect the power tube.  This is often seen in vintage amplifiers, but has fallen into disuse.  I mention it to illustrate the decimal point issue. (It's almost as easy to divide by ten as to divide by one.)

[shooter]

Laws of Robotics

When I was getting my degree in automated manufacturing/Robots we programmed one to "unplug" itself and display, "Compliance complete" 

[pdf64]

With regular guitar amp designs, I don't perceive the rationale of running hotter operating conditions for a 6L6GC than would be appropriate for, eg 5881, 6l6WGB.
As all 6L6 (or whatever) type tubes have the same characteristics, curves etc; that's what makes them 6L6.
And the tube doesn't have to idle at a certain percentage of its limit to follow those curves.

I suspect that Fender moved from 5881 to 6L6GC (when they became cheap enough) in order to benefit from longer tube life / lower warranty failures, rather than to take up the opportunity to increase the idle dissipation by 30%.

These are class AB amps; consider that as HT voltage goes down, for a given load, at some point idling at 70% will result in class A operation.