how to reduce 12v 7Ah to 6V?

[mamun2a]

**broken link removed**

i try to reduce the 12v7ah sla battery to run 8" fan which required 6V, but only 1-15 ohms 2watt registor in series is work but the R is may burn within a minit. what should i do to down 12 v to 6 volt

 

[transistor495]

You can use an LM150/350 3A adjustable regulators for high current drain.

 

[Boncuk]

Why not just use two of those fans connected in series.

Bangladesh air is hot enough to have an extra fan.

 

[mamun2a]

boncuk u r ri8, its 30-34 Centigrade now.

but why by using R it cant down

 

[Grossel]

Why not just use two of those fans connected in series.

One motor tends to go faster than the other when connected in series.

In my opinion, a PWM controller should be the best chioce as it produces less loss of power than other voltage stabiliser circuits. I'm pretty sure other people in this forum can advice you an apropriate pwm controller ic.

 

[mneary]

A small switching regulator would be very effective. If ICs are a problem for you, try a Black Regulator. What types of small BJTs can you get?

 

[Boncuk]

One motor tends to go faster than the other when connected in series.

so what?

Boncuk

 

[Grossel]

so what?

Boncuk

What so what?

 

[Reloadron]

Why not just use two of those fans connected in series.

Bangladesh air is hot enough to have an extra fan.

I agree and who cares if one turns a little faster than the other.

Another option is remove and replace 12 Volt battery with 6 Volt battery.

Yet another option is since a 12 volt battery really has six two volt batteries in it perform surgery on the 12 volt battery. Carefully drill a hole at the junction of the 3rd and 4th cells. You now have two 6 volt batteries.

Disclaimer: This is posted in humor. I really do not suggest opening a "sealed" lead acid battery! But it would work!

Ron

 

[Hero999]

Yes, either buy a 12V fan or use a 6V battery.

Using a linear regulator or resistor to drop the voltage wastes half the power so you might as well use a 6V 7Ah battery which will be half the size or a 6V 14Ah battery which will be the same size but last for twice as long.

 

[MikeMl]

Why doesn't the OP stick shis resistor in the air stream in front of the fan. That will keep the resistor from burning up, and it will keep herm warm, too.

 

[transistor495]

Why doesn't the OP stick shis resistor in the air stream in front of the fan. That will keep the resistor from burning up, and it will keep herm warm, too.

LOL, mike you've got a good humorsense, rarely though

 

[blueroomelectronics]

Or get a 6V battery.

 

[tcmtech]

Just use a 555 timer IC to drive a power Mosfet or transistor at 50% duty cycle. Its a simple circuit and the fan motor will care less about the being driven that way. Just make the 555 timer run at a frequency above 20KHz so no one can hear it.

 

[Hero999]

Except you need a duty cycle of 25% to mimick half the voltage because the RMS voltage of a squarewave is Vpeak√duty.

 

[Thunderchild]

use a SMPS circuit or even simpler a PWM control, you can make a decent one from a 555 IC circuit or if you want to make something more of it use a pic to generate the PWM drive

 

[Hero999]

How much current does the fan use?

A pretty efficient regulator can be made from a couple of transistors. If more current is required a MOSFET can be used for the output transistor.

2-transistor Black Regulator

 

[Grossel]

Except you need a duty cycle of 25% to mimick half the voltage because the RMS voltage of a squarewave is Vpeak√duty.

No, that sounds wrong. 50% sounds right.

 

[tcmtech]

Except you need a duty cycle of 25% to mimick half the voltage because the RMS voltage of a squarewave is Vpeak√duty.

?
Where are you getting the V peak and √ duty from?

I assumed that being a 6 volt DC fan its a battery powered. So a 50% PWM duty cycle of 12 volts gives a 6 volt average as far as a DC motor is conserned.

 

[Hero999]

It's basic Ohm's law!

You want the RMS voltage, not the average voltage.

A square wave with a 50% duty cycle has an RMS voltage of √0.5 of the peak value.

Here's an example using Ohm's law:

Consider a 3Ω resistor connected to a 6V power supply:

Calculate the power:
P = V²/R = 6²/3 = 36/3 = 12W

The same resistor 3Ω connected to a 12V squarewave with a duty cycle of 50%.

P = V²/R×0.5 = 12²/3×0.5 = 144/3 = 48×0.5 = 24W

The power dissipated in the 3Ω resistor is double 12V square wave.

Calulate the power with a 12V squarewave with a duty cycle of 25%:

P = 48×0.25 = 12W

The same as it was when run off the 6V supply, therefore you need a 12V supply with a duty cycle of 25% to give the same RMS voltage as a 6VDC supply.

EDIT:

Strangly enough the power dissipation in a 3Ω resistor is also 24W when connected to a DC voltage of 12√0.5 = 8.48V.