Filament Drop Resistor consumption - How to calculate ?

[kagliostro]

This thing is confusing me

Starting from a 9v Ac I wan to reduce it to 6.3v for 3 tubes, total consumption of the tubes is 0.9A

If I'm not wrong to drop 9v to 6.3v I must use a 3ohm resistor

9v - 6.3v = 2.7v

2.7v / 0.9A = 3ohm

2.7v * 0.9A = 2.43W

Then I lost me ..........

which is the formula to calculate the current consumption of the 3ohm resistor ?

Sure I'm missing something ............. 

Thanks

K

[stratele52]

Current in the 3 ohms resistor  ?

Same as tube and you write it ; 0.9A

In serie current is the same.

 

[kagliostro]

So I need a 1.8A 9v AC supply (at least) ?

K

[stratele52]

So I need a 1.8A 9v AC supply (at least) ?

K

Why 1.8 A ? your circuit need 0.9 A

It is the same and only 0.9 A in the 3 ohms resistor and in the tube (s)

I write ; in serie current is the same ; mean total current , one time 0.9 A

[kagliostro]

OK

but if the current needed by the tube is 0.9A the in series added resistor haven't a consumption ?

this thing is confusing me

K

[sluckey]

In a simple series circuit such as you have, there is only one path for current to flow. The current flowing thru the filament is the SAME current that flows thru the resistor. No need to calculate the current thru the resistor since it must be the same as the current flowing thru the filament.

[jjasilli]

I think you need to balance the resistance you calculated by putting a series resistor of 1/2 that value on ea leg of the AC filament supply.  Otherwise the 60 cycle waveforms will not fully cancel, thereby allowing hum.  Alternatively, if an artificial CT is needed anyway, you might try a 1/3 voltage divider.

[HotBluePlates]

Right. Split the 3 ohm resistor into two 1.5 ohm resistors, one in each leg of the filament circuit.

1.5 ohm * 0.9A = 1.35w, so use 1.5Ω 3w resistors.

[Zipslack]

Think of series current as traffic going down a one-lane road.  How fast does it go?  As fast as the slowest car.  How fast are the rest going?  As fast as that slowest car - they all end up going the same speed (current flow).  With multiple lanes (parallel paths), the cars can go around and go different speeds in each lane and you get more overall traffic (current flow).

[PRR]

Think of series current as water going down a ditch. How much water flows? Say you want 0.9 quarts/liters per second. How far does it fall? Total 9 feet (or CM). Say you need to turn a water-wheel 6.3 feet high. You need to drop the other 2.7 feet with a chute or pipe.

The chute-pipe uses/wastes 2.7 feet at 0.9 quarts per second. Same water current as what flows in the water-wheel.

The friction in the chute/pipe must be 3 feet/quart-second to get the right flow in the water-wheel.

The useful energy to the water-wheel is 6.3*0.9= 5.67 feet * quart-seconds.

The wasted energy in the chute/pipe is 2.7*0.9= 2.43 feet * quart-seconds.

Yes, you waste 1/3rd of the available water power. If you had a 9 foot water-wheel you would have no waste. If you only needed the same 5.67f*qs of output you would only have to supply 0.63 quarts per second of water. But you already know that you can only get 6V heaters and have accepted the waste of electrical power.

BTW there are several 9V heater tubes. In the US types:

9A8  tp
9AH9  tp
9AU7  tt
9BW6  P
9CL8  tq  
9DZ8  tP
9EF6  P
9KZ8  tp
9P9  P

[kagliostro]

Many thanks to all

I was thinking about the resistor like a mode to drop the voltage, but also I was considering that the resistor has its own consumption

the consumption of the drop resistor is 2.43W but I would like to know his consumption in A as to know if a 1A 9v winding is enough to supply the 3 tubes (3 x 0.3A = 0.9A) + the consumption of the drop resistor = ??

My confusion was about the formula to establish the correspondence between the 2.43W (of wasted power) and the current consumption

K

[HotBluePlates]

... I was considering that the resistor has its own consumption ...

A resistor does not consume current. It slows the rate of current and as a result it dissipates heat.

[kagliostro]

A resistor does not consume current.

 

the 2.43W dissipated don't mean there is a consumption by the resistor ?


K

[phsyconoodler]

 Voltage is pressure
The resistor slows the voltage or pressure down and turns it into heat.

[stratele52]

... I was considering that the resistor has its own consumption ...

A resistor does not consume current. It slows the rate of current and as a result it dissipates heat.

Yes it consume current , that's why a resitor can be hot  

[PRR]

> the correspondence between the 2.43W (of wasted power) and the current consumption

You already computed 2.7 Volts across it, and 2.43W of heat.

2.43W/2.7V= 0.9 Amperes.

[kagliostro]

Yes PRR

Now I think I've understand

the total consumption of the circuit isn't 0.9A (the tubes) + 0.9A (the resistor)

because the 0.9A of the resistors consumption refers to 2.7v

As to recognize if the transformer has enough current to supply the tubes + the resistor I must use the W not the A rating

so the (three 0.3A filament) tubes consumption is 5.67W and the resistor consumption is 2.43W

total consumption is 5.67 + 2.43 = 8.1W

the transformer is rated at 9v & 10VA and assuming that VA = W (also if not exact) I've 1.9W more than that is required

Is this consideration correct ?

Thanks

K

[HotBluePlates]

The problem we have is the word "consumption" as though the resistors "eat" current, which then somehow disappears.

They don't.

It's a series circuit, so 1 current goes around the loop formed by the circuit. The exact same current must go into and out of all the components of the circuit, because there's no where else for the current to go. A parallel circuit would provide one or more "other places" for current to go on its way around the loop.

Maybe NEETS will help. I'm sorry it's only in English, but look at Module 1, starting at Chapter 3.

If anyone else is reviewing this thread and doesn't have a strong basic electronics background, Modules 1-9, 16, 19 and 21 are most useful to guys like us.

[kagliostro]

Hi HBP

Thanks for the link, I'm very curious about and I'll study chapter 3 with attention

My lack of education in electronics weighs on me more and more

My hope is that my english is enough good

K

[Zipslack]

The NEETS books can be a little intimidating for newcomers...here's a series of free textbooks that aren't too bad
www.allaboutcircuits.com

{edit - untangled URL code -- PRR}

[kagliostro]

I've just finish to print the chapter three and I'm starting to study 

K

[TIMBO]

Hi K, Good Topic. I have a similar problem of using a separate transformer to power the tubes heaters. In my last build i had the tubes heaters in seriers and used a 12v tap to do this but when checking the voltage with tubes in i was only getting about 11.3v to the tubes and although this may be ok i would like to have had the voltage a bit higher (most tubes says that 10% voltage either way is OK).

So i was going to do a similar thing for dropping 15v down to 12.6v.
My thoughts were (and i am assuming that the resistors that you are talking about are only on the AC leads not the artificial tap resistors)could the artificial tap resistors be used to decrease the voltage???

Thanks

[HotBluePlates]

The NEETS books can be a little intimidating for newcomers...

Maybe, but mainly the ones on waveguides, wave transmission, radar, oscillators (when it gets beyond the basic R-C oscillator we know from trem circuits), etc.

But those are topics we don't need to mess with, so just don't read those modules.

Also, skip the synchros and servos, unless you need a low-muscle-power way to turn your battleship's turrets. 

[kagliostro]

I've read something like 25-30 pages of chapter 3 and I must say that is very well explained  

Thanks again HBP


TIMBO I was considering to use a pair of drop resistor (one for each branch - 9v AC) and than an Humdinger pot

The consumption in current of a pair of 100ohm resistors used as artificial ground is around 30mA, and I don't thing that this resistors can be used as drop resistor, you must have also the drop resistors in your circuit

K

[kagliostro]

Hi Tubeswell

even if we were talking about a much more simple argument, about  to reduce the voltage intended to the filaments , your explanation is very interesting and deserves a great attention

Thanks

K

EDIT: In the attached image the schematic that I want to use

[kagliostro]

I've read the chapter 3 and now I'm sure about which was my mistake

I was considering current (A) as a meter of consumption like if it was power (W)

Many thanks again for the link

K

[tubeswell]

Hi Tubeswell

even if we were talking about a much more simple argument, about  to reduce the voltage intended to the filaments , your explanation is very interesting and deserves a great attention

Thanks

K

EDIT: In the attached image the schematic that I want to use

Sorry I was on a completely different planet in the wrong thread (post removed)