ohm's law in practice

[cminor9]

I am trying to wrap my mind around something. Well, I am stuck.

I have a 6.3v @ 3a supply. I want 5v @ 3a. I don't know what value resistor I need.

So, I take I = V/R and plug in the numbers.
3a is the current, 1.3 is the voltage drop I want. Thus 3 = 1.3/n, n = .43. So a .43 a resistor @ 3.9w (read: 5W resistor.) Great.

But...how do I know that resistor is going to drop voltage and not current, other than my politely telling it to do so and perhaps buying it a nice dinner before soldering it in? How do I know this won't happen instead:
2a = n/.43, n = 0.86. Since 0.86 is the voltage drop, I end up with 5.44 v @ 2a.

This is a bit of a mystery to me. Also, that seems like a low value/beefy resistor for what I need. What am I missing? It seems like this would be a rather common question, but I could not find an answer. Lots of threads that touch on this, and some say to use zeners in such an application. I know I can use zeners, but that's not the point. I want to understand what should be a very basic thing.

[Enzo]

Ohm's Law describes the relationship between voltage, current, and resistance. They are all intimately and forever linked together.

Resistors don;t drop current, they drop voltage when current flows through them. V=I x R. The more current flowing, the more voltage is dropped. You pull less current through it, you get less voltage drops. You reduce your current by a third, reduce the voltage drop by a third by doing so. A resistor is not a voltage regulator. So if you have it set up to drop 1.3v @3A, then that is exactly what it will do... at 3A. Pull only 2A through it, the voltage rises. You see this all the time on B+ windings. Pull more current through, the voltage sags.

VOltage drop is not something we usually have just standing alone. How long is a piece of string? it is in some context, the application of it is what matters.

I am going way out on a limb here, and guessing you want to use a 6.3v transformer winding for a 5v rectifier tube. How close did I come? You could also be trying to make a 5v relay supply I guess, but 3A of it?

A 5v rectifier tube heater is a pretty steady draw, it doesn;t move all over the current meter during use. SO figure out what the tube draws, calculate your resistor and stick it in. The 3A 5U4 won;t magically draw 2A tomorrow. On the other hand, if you plan to stick a 3A 5U4 in there today, a 2A 5Y3 tomorrow, and a 1.9A 5AR4 the next day, then your simple resistor doesn;t do the job. And you can plug the numbers into Ohm's law just as you did and see what the results would be.

So the application is "I want to power a 5U4 heater from a 6.3v transformer winding." That you can do with a resistor. If your application is "I want to make 5v from a 6.3v winding regardless of current," then a series resistor is not what you want.

And remember the winding resistance. Just like the high voltage winding that sags, the difference between 2A and 3A will cause a differing voltage drop in the transformer winding wire. Not much, of course, but when we are looking at tenths of volts, it might as well be included in the thinking.

And if I had 3.9w to dissipate, I wouldnt dissipate it from a 5w resistor, myself. I would pick something larger, like a 10w. 3.9w is 3.9w, but the larger resistor will spread it out, so the hot spot will be less concentrated in the amp. besides, I like a bit more than a 25% margin on specs like that. My own rule of thumb is about a factor of 2. I want about twice the resistor dissipation rating that I need.

[Jag]

I'll try this.

You don't want a single resistor, you want to build a voltage divider.

Going with rough estimates, 5v is about 80% of 6.3v so you want two resistors where one is about 4 times the value of other, or 80% of their combined value.

To make the math easy I'll use 100k and a 25k as an example. Their total series value is 125k, the 100k is 80% of the total.

Connect the two resistors in series.

Connect one side of your 6.3v supply to the free end of the 25k resistor and connect the free end of the 100k resistor to the same ground as your 6.3v supply.

Checking the voltage across the 100k resistor and you should find it's about 5v. This is where you would tap off your 5v supply.

The current being drawn through the 100k resistor is neglegible (0.00005a) so you don't need large resistors (1/2 watt would be plenty).

Anybody else, please feel free to correct me, I'm sure there's more to this. You could (should ?) probably use smaller values for the resistors.

[Jag]

Sorry Enzo, cross posts. I was hoping to finish typing before you got in so you could correct both of us.

As I read it, he still wants to pull 3A from his 5v supply. Would the simple voltage divider I suggested work?

[Jag]

Thinking about it a little more, I guess the 25k resistor should probably be a 10w.

[Enzo]

Use Ohm's Law and see what happens to your voltage when you try to draw 3A through that 25k resistor. Hint, calculate what the internal resistance must be of a 5v 3A thing.

COntext is everything, can;t forget the current. if you just wanted a voltgae signal, then 100k works fine, just as it does in the signal path between tubes. But the current there is minimal.

[Jag]

I end up with 1.67 ohm "thing" in parallell with a 100k ohm resistor and have a 25k resistor in series with that and it wants 75,000 volts?

I knew I should have gone to bed an hour ago, totally disregarded the "thing"!

[Bruce / Mission Amps]

I wouldn't do any of the resistor stuff, too much waste of power and heat... get your self some high current power supply diodes and run them in series with your 6.3v wires.
Put pairs on each leg, back to back until you get down to 5V.
Each set will drop about .6v or so.... that should be hard to figure out how many you need.

[earthtonesaudio]

I wouldn't do any of the resistor stuff, too much waste of power and heat... get your self some high current power supply diodes and run them in series with your 6.3v wires.
Put pairs on each leg, back to back until you get down to 5V.
Each set will drop about .6v or so.... that should be hard to figure out how many you need.

This is what I would do too.

[R.G.]

I wouldn't do any of the resistor stuff, too much waste of power and heat... get your self some high current power supply diodes and run them in series with your 6.3v wires.
Put pairs on each leg, back to back until you get down to 5V.
Each set will drop about .6v or so.... that should be hard to figure out how many you need.

Unfortunately, using diodes won't save you any power or heat. The power dissipated by any component is always the voltage across it times the current through it (*with some qualifiers for AC voltage and phase in reactive parts).
So if you drop 1.4V at three amps with a resistor, you get heat out equal to
1.4V*3A = 4.2W. If you do the same with three amps through two 0.7V diodes, you get power = 2* (0.7V*3A) = 2*(2.1W) = 4.2W.

Diodes have the advantage of dropping nearly the same voltage at any current, though. They're handy that way, all right. But they don't save you any power in similar applications.

[rathmann]

Not so much an answer - but funny and relevant: xkcd - A Webcomic - Ohm

[twin-reverb]

You don’t mention what kind of tolerance you need on your 5V so I’ll assume you’ll be happy with following the 6.3V voltage fluctuations. The 0.43ohm you mention is correct if you’re sourcing exactly 3Amax. If your load sometimes draws less, then your output voltage will increase inversely. A divider would work although the resistor values would be very low. Assume resistor R1 is your 0.43ohm in series between your 6V input and your 5V output. Add another resistor called R2 on the output of R1 (your 5V output) and attach the other end of R2 to your reference (usually ground). This is a typical voltage divider. To solve for R2 = ((Vin x R1 )/ Vout) - R1 or with numbers R2 = ((6.3V x 0.43ohm) – 0.43ohm) which the calculator spits out a value for R2 as 0.086ohm. Then you’ll need to calculate the power of the 0.43ohm resistor at P = R x ( I squared) or P = 0.43ohm x 3A x 3A) = 0.43ohm x 9A = 3.9Watts. You’re unlikely to find these exact values of resistors though so you would have to recalculate with what’s available. You would want at least a 5W resistor but a 10W is probably a better choice. It will get hot!

I think the diode method mentioned above is probably a better way to go. A couple of other options may be using a 1.3V zener diode in series between the 6.3Vin and 5Vout. For closer tolerance on the 5V you could use a LDO (low drop out) linear regulator. But these are going to have power/heat issues too, again in the >4W range.

Do you really need 3A at 5V? I don’t know your application but if you’re powering a few op amps for instance they would only draw a few milliamps at most. This would make the divider much easier to implement using much higher value and physically smaller resistors and it would almost negate power/heat concerns from your supply.

Here’s another thing to consider; if your 6.3A supply can only supply 3A and you have anything else attached besides your 3A 5V output; Kirchoff’s sum of currents law says it won’t work. For example if your 6.3V supply is supplying current to tube heaters at 1A (arbitrary number here) and you want to draw 3A from your 5V supply, you’ll need the 6.3V supply to be capable of sourcing 4A.

Anyway enough of my ramblings; good luck.

[cminor9]

Not so much an answer - but funny and relevant: xkcd - A Webcomic - Ohm

you beat me to it! I laughed out loud at that one! And you know, had I thought about that in context of this discussion, it might have helped.

[cminor9]

I wouldn't do any of the resistor stuff, too much waste of power and heat... get your self some high current power supply diodes and run them in series with your 6.3v wires.
Put pairs on each leg, back to back until you get down to 5V.
Each set will drop about .6v or so.... that should be hard to figure out how many you need.

Yep, appreciate the dose of pragmatism here. This post wasn't so much about how to drop voltage, though...it was more about understanding the functioning of a resistor in a simple but real life circuit.

I have built a few amps now, and they sound great. But any deviations I have made from the actual circuit have been based on intuition, trial and error, or cookbook type material I have found online. I feel like I need a better understanding of some basics.

[cminor9]

I am going way out on a limb here, and guessing you want to use a 6.3v transformer winding for a 5v rectifier tube. How close did I come?

Good guess. Well, I don't actually have it, but was considering buying a transformer with two 6.3 v secondaries (I have no idea why not a 6.3v and a 5v.) It dawned on me that I would have no idea how to get me the voltage I need. Then it dawned on me that I really don't understand Ohm's law like I thought it did. Note to self: being able to recite a formula or a law doesn't equate to understanding it. Duh. Anyhow, if I had to dissipate 1.3v at 3A (for a 5U4GB) I'd need approximately .43 ohms. I guess that is right, in the case of a 5U4GB. For a 5Y3, I'd need .68 ohms, and for a 5AR4, I'd need .63 ohms. Not that I'd find those resistor values, but this is theoretical after all.

The more current flowing, the more voltage is dropped.

SO figure out what the tube draws, calculate your resistor and stick it in.

Yep, put in this context I get it now. One thing I did not take into account is the fact that different rectifiers will draw different current. Plugging the *rectifier* current into the formula makes sense...now. I was thinking in terms of transformer current.

And remember the winding resistance.

Now you lost me. Winding resistance? As in, the transformer has resistance?

My own rule of thumb is about a factor of 2. I want about twice the resistor dissipation rating that I need.

Interesting. Offtopic: By that logic, though, why would I bother using 1/2 watt resistors everywhere in a guitar amp? There are plenty of places that don't dissipate that much, heck, even a 220k plate resistor on a 12ax7 that drops 100v (like on my 6G3) wouldn't dissipate more than 45mW. Why am I buying 1/2 watt resistors when I could be using 1/4 watt? (I know there's not a great price differential, just theoretically speaking)