Hi there!

My knowledge of electronics is pretty vague to say the least and i need some basic help.

Firstly I want to know if i connect a device which requires 12V at 320mA to an 18 V battery, will the device eventually burn out??

Following that, will adjusting the voltage just cause the device (cpu fan) to speed up and slow down?

Im having difficulty getting my head around the relationship between voltage and current. Ive looked at books, but they loose me when they go off at tangents!!

My last question is if i connect this fan to a transformer with an output of 9v at 1amp, will the fan burn out? and if so how do i calculate what resistor i need to drop the current??

Would appreciate it a lot of anyone could help me out as i really do need it
Thanks again!
bren

 

Firstly I want to know if i connect a device which requires 12V at 320mA to an 18 V battery, will the device eventually burn out??

Following that, will adjusting the voltage just cause the device (cpu fan) to speed up and slow down?

First let me tell you that I'm an EE STUDENT....not a full engineer, but I will do my best here.


The 18V Battery will most likely destroy the fan quickly. Overvoltage isn't good for that device....but the battery will put out WAY more than 320mA into the load(fan) causing overheating of at least the wires.

You correctly deduced that varying the voltage would slow down the fan, but you have to set the upper limit @ 12 Volts.

Transformers WILL NOT FUNCTION with DC(except pulsed DC, but that's a different discussion )

E=I X R....Voltage(Electromotive Force) in a DC Circuit equals the current times the resistance

Using Algebra, we get:

I= E/R....Current Equals Voltage divided by resistance

we can also get:

R=E/I....Resistance equals Voltage divided by current

To make an 18 Volt battery into 12 Volts, you could connect a 6Volt Battery(or 4 AA, C, or D cells in series) BACKWARDS(with reversed polarity) IN SERIES with the 18Volt.....this causes the 6 volts to "fight against" the 18 volts to create 12.

Now, if we have 18 volts.....we want to place a resistance between it and the load to reduce the current to 320mA, so we solve for current and start plugging in resistor values

Current= 18Volts/56Ω= 0.32142857...{repeating}....Amps.....

I got 56Ω by trying a few numbers out until I got close enough to start making educated guesses.

If we do the same for 12Volts we get:

Current= 12Volts/38Ω=0.31578947368421052631578947368421 Amps

You can safely round to 316mA!

Lets switch this around a bit, so that you can get some Ohm's Law Practice.

E(Voltage)= 38 X .316
12.008 Volts

R(Resistance)= 12/.316
37.9746....Ω

(the slight disparity in numbers is due to rounding error)

I hope that this helped you in some way!!!!

Jimmy

 

Originally posted by Jimmy Lane@Dec 30 2005, 02:21 PM
Firstly I want to know if i connect a device which requires 12V at 320mA to an 18 V battery, will the device eventually burn out??

Following that, will adjusting the voltage just cause the device (cpu fan) to speed up and slow down?

First let me tell you that I'm an EE STUDENT....not a full engineer, but I will do my best here.
The 18V Battery will most likely destroy the fan quickly. Overvoltage isn't good for that device....but the battery will put out WAY more than 320mA into the load(fan) causing overheating of at least the wires.

You correctly deduced that varying the voltage would slow down the fan, but you have to set the upper limit @ 12 Volts.

Transformers WILL NOT FUNCTION with DC(except pulsed DC, but that's a different discussion )

E=I X R....Voltage(Electromotive Force) in a DC Circuit equals the current times the resistance

Using Algebra, we get:

I= E/R....Current Equals Voltage divided by resistance

we can also get:

R=E/I....Resistance equals Voltage divided by current

To make an 18 Volt battery into 12 Volts, you could connect a 6Volt Battery(or 4 AA, C, or D cells in series) BACKWARDS(with reversed polarity) IN SERIES with the 18Volt.....this causes the 6 volts to "fight against" the 18 volts to create 12.

Now, if we have 18 volts.....we want to place a resistance between it and the load to reduce the current to 320mA, so we solve for current and start plugging in resistor values

Current= 18Volts/56Ω= 0.32142857...{repeating}....Amps.....

I got 56Ω by trying a few numbers out until I got close enough to start making educated guesses.

If we do the same for 12Volts we get:

Current= 12Volts/38Ω=0.31578947368421052631578947368421 Amps

You can safely round to 316mA!

Lets switch this around a bit, so that you can get some Ohm's Law Practice.

E(Voltage)= 38 X .316
12.008 Volts

R(Resistance)= 12/.316
37.9746....Ω

(the slight disparity in numbers is due to rounding error)

I hope that this helped you in some way!!!!

Jimmy

[post=12803]Quoted post[/post]​

Hi I don't like your TITLE --electronic dummies-that apart-1-you cannot connect the fan directly to the trans former, but can use 1 diode for halfwave rectification and it should be sufficient. 2 ,they are fixed regulators eg 7812-these would regulate at 12volts capable of 1amp on heat sink 15 watts dissipation,this would hold the voltage constant/constant speed,with sufficient current.

 

Originally posted by brendyboy@Dec 30 2005, 12:42 PM
Hi there!

My knowledge of electronics is pretty vague to say the least and i need some basic help.

Firstly I want to know if i connect a device which requires 12V at 320mA to an 18 V battery, will the device eventually burn out??

Following that, will adjusting the voltage just cause the device (cpu fan) to speed up and slow down?

Im having difficulty getting my head around the relationship between voltage and current. Ive looked at books, but they loose me when they go off at tangents!!

My last question is if i connect this fan to a transformer with an output of 9v at 1amp, will the fan burn out? and if so how do i calculate what resistor i need to drop the current??

Would appreciate it a lot of anyone could help me out as i really do need it
Thanks again!
bren

[post=12801]Quoted post[/post]​

If the fan is DC and the transformer you're talking about is an AC to DC adapter type, then the fan will most likely run at 9 V but would run slower than at 12V.
If you try the 18 V battery just use an ammeter hooked up between the fan and a terminal of the battery to see how much current is being drawn by the fan, if it stays around 300-400 mA then you're probably okay, except the fan would run faster and the bearings would go bad sooner than they would at 12V.

 

[attachmentid=1074]

Originally posted by brendyboy@Dec 30 2005, 10:42 AM
Hi there!

My knowledge of electronics is pretty vague to say the least and i need some basic help.

Firstly I want to know if i connect a device which requires 12V at 320mA to an 18 V battery, will the device eventually burn out??

Following that, will adjusting the voltage just cause the device (cpu fan) to speed up and slow down?

Im having difficulty getting my head around the relationship between voltage and current. Ive looked at books, but they loose me when they go off at tangents!!

My last question is if i connect this fan to a transformer with an output of 9v at 1amp, will the fan burn out? and if so how do i calculate what resistor i need to drop the current??

Would appreciate it a lot of anyone could help me out as i really do need it
Thanks again!
bren

[post=12801]Quoted post[/post]​

Hi Brendy! Welcome to AOC.

"Electronics For Dummies" by Gordon McComb, Earl Boysen - ISBN: 0-7645-7660-7 :lol:

The relationship between voltage, impedance, and current is actually pretty straighforeward.

For a given voltage, the greater the impedence, the lower the current. A given voltage can only push so many electrons per second through a given impedence.

If we want twice as much current to flow, we must either double the voltage or halv the impedence. (Or some combination that works out to the same math)


Please look at the triangle in the attachment.

To calulate E, we multiply I times R. To calculate I, we divide E by R. To calculate R, we divide E by I.

So, in your fan:

The impedence is, for practical purposes, fixed. The current through it will be directly proportional to the voltage. 3x the voltage means 3x the current. 1/5 the voltage means 1/5 the current. The speed of the fan is based on the power used by the fan. The power used by the fan is equal to current squared times impedence. Since impedence is fixed, power used by the fan is proportional to the square of the current. (And, since current is proportional to voltage, Power is also proportional to square of voltage!) 3x the voltage or current = 9x the power. 1/5 the current or voltage = 1/25 the power.


Now, about that "9 volt transformer." Is this one of those ubiquitous blocks that plugs into the wall and gives you 9vdc at a max of 1amp? If so, then it will indeed work. The salsepeople really need to stop calling those power supplies "transformers." It just causes confusion!! Note that the 1amp is a maximum capacity of the supply. If our impedence is small enough to draw more than one amp from the 9v, then our poor little power supply will be over-worked. It will drop the output voltage below 9v to avoid going over one amp.

 

Ive actually ordered that book from amazon

Thanks for your help! I need advice on one more point (should really be in the projects forum)

I aim to use this fan in conjunction with some sort of thermostat or temperature dependant switch so that when temp goes above a certain point, the fan kicks in. It will be powered by a 12v battery prefereably around 6 amp hours (if thats possible) I just want to know exactly the best circuit kit and rechargable battery to go for. Ive found a range of kits on the net, but just wondered if anyone could advise me on the best type of kit to go for!


thanks

bren