Voltage drop resistors - Check my math please!

[frequencycentral]

I'm not a math guy, but I'm trying to get to grips with it.

I'm looking at a voltage drop resistor to power 5672 tubes from 12 volts.

5672, 1.25 volts / 50 ma heater requirement:

Voltage drop required 10.75 volts

10.75 volts / 50 ma = 215 ohm
10.75 volts x 50 ma = 0.5375 watt


And for two 5672, so 100ma:

10.75 volts / 100 ma = 107 ohm
10.75 volts x 100ma = 1.07 watt


...for 9 volts, therefore 7.75 volt drop

7.75 volts / 50 ma = 155 ohm
7.75 volts x 50 ma = 0.387 watt


.... two 5672 @ 9 volts

7.75 volts / 100 ma = 77.5 ohm
7.75 volts x 100ma = 0.775 ohm

Have I got this right?

thanks,

Rick

[Jimmy-H]

Hi Rick,


I think your math is correct.
But maybe it's a good thing to put a zener across the power supply.
So the voltage never gets higher than 12 or 9 volts and the heater never gets more than 1.25 volts.

[Minion]

yes you got that right...I hate doing math also and I suck at it but to calculate a Voltage drop resistor I actually use a LED Resistor calculator Like this one...

http://led.linear1.org/1led.wiz

You just put in the Voltage you need and the Current you need and the supply voltage then it calculates the resistor needed and the wattage of that resistor (to the closest available Value)....

Cheers

PS: you can also use a resistor + LED to make a simple 1.25v Regulator that will be more stable than just a resistor, Or a LM317 regulator....

[kurtlives]

I'm not a math guy, but I'm trying to get to grips with it.

I'm looking at a voltage drop resistor to power 5672 tubes from 12 volts.

5672, 1.25 volts / 50 ma heater requirement:

Voltage drop required 10.75 volts

10.75 volts / 50 ma = 215 ohm
10.75 volts x 50 ma = 0.5375 watt


And for two 5672, so 100ma:

10.75 volts / 100 ma = 107 ohm
10.75 volts x 100ma = 1.07 watt


...for 9 volts, therefore 7.75 volt drop

7.75 volts / 50 ma = 155 ohm
7.75 volts x 50 ma = 0.387 watt


.... two 5672 @ 9 volts

7.75 volts / 100 ma = 77.5 ohm
7.75 volts x 100ma = 0.775 ohm

Have I got this right?

thanks,

Rick

P = Voltage Drop (squared) / resistance

I thought>?

EDIT: I'm tied tonight. Both you and I are right, either way you get the same answer.

[frequencycentral]

Thanks guys, thats great.

Minion - the calculator is really useful.

PS: you can also use a resistor + LED to make a simple 1.25v Regulator that will be more stable than just a resistor, Or a LM317 regulator....

So i just use a resistor and LED in series between +ve and earth right? And tap off the voltage from where the resistor and LED connect?

[Jimmy-H]

But most LED have a voltage of 2 or more.
So that's a littlebit to high.
But with a 1N00X you have a voltage of 1.1
The spec for the heater is 12V + 20% = 1 to 1.5 volt.
So that should do the trick......just try

[Jimmy-H]

The spec for the heater is 12V + 20% = 1 to 1.5 volt.

I meant the spec. of the heater is 1.25V + 20% 

[R.G.]

Here's a thought. You have 12V available. Put the heaters in series. Instead of 1.25V at 50ma, now you need 2.5V at 25ma. Now the math says:

"I have 12Vdc. The heaters will use up 2.5V of that, leaving only 9.5Vdc left in the resistor. The resistor needs to pass 25ma, so the resistor must be R = 9.5V/0.025A, or 380R. The power is P = V*I or 9.5*0.025= 0.2375W, so a 1/4W resistor would barely work, and a 1/2W or two 1/4W resistors would be conservative."

380 is a non-standard value. You could use 390R at 1/2W, or two 750R at 1/4W in parallel. Or three 1.1K 1/4W in parallel to get 367R. All of these get you well within +/-20%.

Notice that the 1.25V nominal rating is the voltage of a carbon-zinc dry cell. These miniatures are intended to run from batteries. Two AA cells will run your two heaters for a reasonable time.

On the regulation thing. Be careful with shunt regulators if you don't know more about electronics and math. A shunt regulator like a zener, a diode string in parallel with the load, or an LED all depend on the series limiting resistor to let through the right amount of current for both the load and the regulator over the range of input voltages. The shunt regulator eats up all the excess current to keep the voltage constant. If you disconnect the load, the regulator can burn up. If the load gets too big, the voltage falls out of regulation.

[frequencycentral]

Here's a thought. You have 12V available. Put the heaters in series. Instead of 1.25V at 50ma, now you need 2.5V at 25ma.................

I had thought about running two heaters in series, as I do with 6111 heaters. The difference with 5672 is that Grid #3 is directly connected to the heater - I'm not sure how that would affect things, but it seems to me that one of the tubes Grid #3 would therefore not be directly earthed. 

[R.G.]

I had thought about running two heaters in series, as I do with 6111 heaters. The difference with 5672 is that Grid #3 is directly connected to the heater - I'm not sure how that would affect things, but it seems to me that one of the tubes Grid #3 would therefore not be directly earthed. 

That could well be a problem in some circuits, all right.

Well, never mind that on then.   

[frequencycentral]

I had thought about running two heaters in series, as I do with 6111 heaters. The difference with 5672 is that Grid #3 is directly connected to the heater - I'm not sure how that would affect things, but it seems to me that one of the tubes Grid #3 would therefore not be directly earthed. 

That could well be a problem in some circuits, all right.

Well, never mind that on then.   

R.G., you look so beautiful when you blush.........

[~arph]

I'm not a math guy, but I'm trying to get to grips with it.

5672, 1.25 volts / 50 ma heater requirement:

So does this mean that the heater resistance is about 25 ohms?

[Mike Burgundy]

Ohm's law!
Do keep in mind that that's the heaters "on" (as in hot) resistance. It's significantly higher when it's hot than cold. I've seen factor 2 to 10 difference I think, though I have no idea which tube does what exactly.
This low cold resistance is one of the reasons a soft-start is a good idea for tube longevity. Ever notice how incandescent bulbs always fail exactly when you switch hem on?

[~arph]

Yeah good ol' Ohm. That soft start doesn't sound like a bad plan either..

I'm cooking up a different way to supply the heater. With a zener so it can accept different supply voltages while te heater supply stays roughly te same.

[PRR]

> with 5672 is that Grid #3 is directly connected to the heater

Right, and G1 voltage is specced relative to "F-".

So the most logical connection is all F- to ground.

1.25v at 2*50mA is 1.25V 100mA.

> power 5672 tubes from 12 volts

Brain training:

You have 12V. You want 1.2V (roughly).

You want one-tenth.

Or conversely: you want a resistor about ten times bigger than the heater. (Actually more like 9 or 8.6, but in the ballpark of 10 will get you warm.)

One heater 1.25V/50mA = 1.25V/0.050A = 25 0hms.

Two heaters must be 12.5 Ohms.

Resistor ten times bigger is 125 Ohms. Or if you computed 8.6X, then 107.5 ohms.

Total power is 12V at 0.1 Amps or 1.2 Watts and most of this is in the resistor. Even exact math says 1.075 Watts at 12V exact, so it's gotta be over 1W rating. Shop the 2W parts.

120 ohm is the nearest very-common 2W value.

BTW: the LED calculator is a handy trick/cheat but since you need 50mA or 100mA you have to specify that as "LED" current, plus more if you expect an LED to regulate. But there are few 100mA LEDs. And it is NOT essential to regulate tube heaters. The battery jobs were intended to run 1.5V down to 0.9V.

Now stand back. You have picked a low-power battery tube, then adopted a heater supply that wastes most of the power you throw at it. Aren't there 6V and 12V tubes which can power off 12V without the waste?(*) And don't some of them have separate cathodes, no dinking-around with directly heated filaments and hot drop-resistors? And the cooling-space needed around a 1+W resistor kinda takes away from any space-savings you thought the sub-mini might give.

And BTW the battery series "tend" to be short life: tube life didn't matter because battery costs would break you long before tube costs, and working the filament hard made the most of your precious battery power.

(*) http://pw2.netcom.com/~wa2ise/radios/penciltubes.html

> higher when it's hot than cold.

That's a reason to use excess voltage and a huge resistor.... with 12V and 120 ohms the peak current can't be over 100mA, ~~50mA per tube, whereas with a D-cell battery (the "right" power source) these "50mA" tubes do draw 200mA peaks and (moreso than indirect tubes) do tend to pop-on-ON.

With indirect heat tubes and any rational heater supply (series-string TV sets excepted) heater failure is so very rare it isn't worth worrying about. Unless you pay a guy $250 to come out and change your tubes.

[Perrow]

I've just built a led driver using a 317 and a few resistors to make a current source, couldn't that be used here too?

If I read datasheets correctly a TO-92 LM317 can handle (typically) 200mA. The resistor will see 1.25V voltage drop at 100mA -> 0.125W, should be safe with a 1/4W resistor.

Edit: I only used "a few" resistors as I needed 400mA, and had no higher wattage resistors at hand.

[~arph]

Yes Rick's used te 317 solution over and over I believe.

Thing is I want to be able to power it with about any regular 9v adapter you find in te stompbox field. I have tried numerous times to get this working with the 317 and somehow always ended up with power troubles, the LM317 eating up so much current that there wasn't anything left for anything else. I was using a one spot btw..

[PRR]

> a TO-92 LM317 can handle (typically) 200mA

The TO-92 won't safely dissipate the proposed 0.1A*(12V-1.25V-1.2V)= 0.95 Watts of heat.

And since the supply voltage is steady(?) and the filament voltage is much less than supply voltage, a dumb resistor will fix the current just fine.