Voltage Divider question - 6 from 12 Volts

[Brit01]

Guys,

Can I use a simple voltage divider circuit to get 6 Volts out of a 12 volt transformer?
I have a secondary 12V at 4 Amps.

What would be suitable resistor values (and wattage) here to get 6 Volts at 2.5 amps min. from 12 volts? I want to supply a 6080 heater.

I found this great calculator:

http://www.educalc.net/876080.page

 

[poynton]

Big Resistor needed !!!!

To drop 6V at 2.5A requires 6 / 2.5 = 2.4ohms at 15watt rating


Get a 6v transformer - less heat and probably cheaper


Andy

 

[wwenze]

How about changing your circuit to two 6080s with the 12V feeding their heaters in series instead?

 

[Brit01]

Thks Andy.

Thought it was going to take some heat.

Although I may be able to get some 20W resistors for about 2 USD here.

Iron is not cheap here. Let me find out.

 

[Brit01]

How about changing your circuit to two 6080s with the 12V feeding their heaters in series instead?

True. The other option I thought about.

If I've got 4 amps at 12V.

Then 2 x 2.5 amp heaters would be 5 amps.

But at 6 volts this would be ok yes?

 

[Rafael L]

with any value of resistor, you make a divisor EX 0.5 ohms + 0.5 ohms...
More the correct you use a Zener diode (1N5340)

 

[Brit01]

More the correct you use a Zener diode (1N5340)

sorry not sure what you mean.

A zener with each resistor? Or as a substitute to create the required resistance?

 

[Rafael L]

sorry not sure what you mean.

A zener with each resistor? Or as a substitute to create the required resistance?

You need not resitores, only one Zener diode, above 6 v leads, so does the voltage not exceed 6 v, look fig.
-his source is not symmetric, will need only one Zener diode....

 

[pjp]

Rafael.luc I hope you know what you're talking about... 1N5340 is a 5W part which can pass approx. 800mA ...

 

[Brit01]

1N5340 is a 5W part which can pass approx. 800mA .

yep. I need something they will be capable of handling 5 amps at least.

 

[Brit01]

What about connecting 8 silicon standard diodes in series (0.7V drop each).
5.6 volts drop.
6.4 V perfect.

Ones with a high current rating.

Could be an option?

 

[pacificblue]

If I've got 4 amps at 12V.

Then 2 x 2.5 amp heaters would be 5 amps.

But at 6 volts this would be ok yes?

If you use two 6080s in series you need 12 V 2,5 A. If you use 2 in parallel, you need 6 V 5 A. To get 6 V 5 A from a 12 V transformer you need a 12 V 5 A transformer and a resistor that wastes 6 V 5 A = 30 W away.

The series connection is the second best way to go, if you need two 6080s. If you don't, the unused 6080 wastes 15 W. Pretty much the same as wasting 15 W in a resistor, but the resistor might be cheaper.

Don't use the Zener. It is a waste in this project and you will still need the resistor.
Forget about the diodes either. They need to be big, which means expensive and the forward voltage may not be 0,7 V at the current you need.

The best way is to get an adequate transformer.

 

[Brit01]

Yes I'll think I'll just use 2 in series for now until I get a new iron.

I'm just going to be breadboarding initially anyway.

Cheers

 

[Rafael L]

Rafael.luc I hope you know what you're talking about... 1N5340 is a 5W part which can pass approx. 800mA ...

Yes you are correct, I had not looked at the current

But, Zener diode need not the same as the current source, who will consume the current is the load, zener can be 1/2 of the load current.

 

[akis]

Yes you are correct, I had not looked at the current

But, Zener diode need not the same as the current source, who will consume the current is the load, zener can be 1/2 of the load current.

You can amplify the zener and then it can be 1/100 of the load or more.

But no matter what, he will have to burn considerable power somewhere: on a diode, resistor, zener, transistor.

Since the load is constant a series resistor must be the most straight forward option (I presume "heater" is a heating element of some thermionic valve?)