will wiring 2 9 volt batteries in series be too much for a 12 volt fan? do i need resistors before the fan? how many?
same question goes for the spark circuit in a stun gun.
how do i figure out how many ohms of resistance i need before an LED wired in parallel with the fan circuit to keep from overpowering it?
sorry for the questions but everything i have tried to read up on this gave too much info (making it sound confusing) for a wiring novice such as myself.
beginer wiring questions
Ohms law:
I=V/R
where I is current in amps, V is voltage, and R is resistance in ohms. For a standard LED, you don't want to go above .060A, .040A is good. The fans are rated .120A, so its still a good idea to put resistors in series.
Keep in mind 9v batteries die really quickly, so you dont need to assume a full 18v, but ~14v assumption is good.
I=V/R
where I is current in amps, V is voltage, and R is resistance in ohms. For a standard LED, you don't want to go above .060A, .040A is good. The fans are rated .120A, so its still a good idea to put resistors in series.
Keep in mind 9v batteries die really quickly, so you dont need to assume a full 18v, but ~14v assumption is good.
Last edited by rp181 on Tue Apr 06, 2010 10:01 pm, edited 1 time in total.
-
jimmy101
- Sergeant Major
- Posts: 3199
- Joined: Wed Mar 28, 2007 9:48 am
- Location: Greenwood, Indiana
- Has thanked: 5 times
- Been thanked: 17 times
- Contact:
rp181's current is a bit high, most red LEDs won't last long at that current.
To figure the current limiting resistor for a red LED;
(Power supply voltage - diode drop of LED)/(LED current) = resistor needed
Most red LEDs drop about 1.7 V (other colors have higher voltage drops, blue LEDs can approach a 5V drop). 9V battery minus 1.7V drop of the LED gives 7.3V across the resistor. You generally want about 20mA through the resistor (and the LED since they are in series) so Ohm's law says V/I=R.
(9-1.7)/0.02= 365 ohms. So, any resistor from 300 to 400 ohms will be fine.
Non-solid state circuitry (like an incandescent light bulb or a fan motor) is much more forgiving of voltage and current since they tend to be self limiting. So a 12V fan is fine with 18V though the fan won't last the many years that it will at the correct voltage. It'll still probably run continuously for months without problems.
The stun gun is trickier. In general, a solid state circuit designed to operate at an extreme, and costing less than $10 to make, will be pretty finicky about its input voltage. If you double the input voltage you may well double the output voltage. It is unlikely that a cheaply made device has the needed "safety factor" to survive. You may well get sparks jumping between components inside the case. And, in the extreme, the "magic smoke" may escaped from the circuity.
Edit:
RP said: "The fans are rated .120mA, so its still a good idea to put resistors in series. "
He probably meant 120mA not 0.120mA. The resistor really isn't needed since a motor is self limiting. That 120mA fan could be hooked up directly to a 12V car battery (that'll source several hundred amps) without any problem. A resistor could be used to drop the voltage but to go from 18V down to 12V will take a resistor that has a pretty small value. (Ignoring the impedance of the fan and just treating it like a resistor; 12/0.12A=100 ohms) So about a 50 ohm resistor would be needed to drop the 18V to 12V. The resistor will be absorbing half the power that the fan is. The fan is absorbing (12v)(0.12A)=1.44 watts. So the resistor need to dissipate about 0.72 watts. Most resistors are only rated to 0.25 to 0.5 watts. So you need a power resistor to dissipate the heat. (Or, mount the resistor in the fans air flow and have the fan cool the resistor.) Overall the resistor really isn't needed.
To figure the current limiting resistor for a red LED;
(Power supply voltage - diode drop of LED)/(LED current) = resistor needed
Most red LEDs drop about 1.7 V (other colors have higher voltage drops, blue LEDs can approach a 5V drop). 9V battery minus 1.7V drop of the LED gives 7.3V across the resistor. You generally want about 20mA through the resistor (and the LED since they are in series) so Ohm's law says V/I=R.
(9-1.7)/0.02= 365 ohms. So, any resistor from 300 to 400 ohms will be fine.
Non-solid state circuitry (like an incandescent light bulb or a fan motor) is much more forgiving of voltage and current since they tend to be self limiting. So a 12V fan is fine with 18V though the fan won't last the many years that it will at the correct voltage. It'll still probably run continuously for months without problems.
The stun gun is trickier. In general, a solid state circuit designed to operate at an extreme, and costing less than $10 to make, will be pretty finicky about its input voltage. If you double the input voltage you may well double the output voltage. It is unlikely that a cheaply made device has the needed "safety factor" to survive. You may well get sparks jumping between components inside the case. And, in the extreme, the "magic smoke" may escaped from the circuity.
Edit:
RP said: "The fans are rated .120mA, so its still a good idea to put resistors in series. "
He probably meant 120mA not 0.120mA. The resistor really isn't needed since a motor is self limiting. That 120mA fan could be hooked up directly to a 12V car battery (that'll source several hundred amps) without any problem. A resistor could be used to drop the voltage but to go from 18V down to 12V will take a resistor that has a pretty small value. (Ignoring the impedance of the fan and just treating it like a resistor; 12/0.12A=100 ohms) So about a 50 ohm resistor would be needed to drop the 18V to 12V. The resistor will be absorbing half the power that the fan is. The fan is absorbing (12v)(0.12A)=1.44 watts. So the resistor need to dissipate about 0.72 watts. Most resistors are only rated to 0.25 to 0.5 watts. So you need a power resistor to dissipate the heat. (Or, mount the resistor in the fans air flow and have the fan cool the resistor.) Overall the resistor really isn't needed.
-
Technician1002
- Captain
- Posts: 5189
- Joined: Sat Apr 04, 2009 11:10 am
Most 12 volt fans work fine on 9 volts.
As an alternative, you can use a portable mini-fan, which only requires 2 AA batteries. I find these all the time, at Walmart.
Difficult to find an exact picture of them on the internet, a quick Google search came up with this:
<img src="http://www.travelbugbooks.ca/graphics/minifan.jpg">
The ones at Walmart are made by a company called Ozark Trail and are usually less than 3 bucks a piece.
Minor trimming is required, for the blades to fit inside of 4" SCH40 PVC and you'll have to fabricate a mount... but man, do they work well!
I was so happy when I found these, because it solved the problem of having to mount a project box, along with 8AA batteries and a switch, to the outside of the launcher.
I usually choose to simply flip the switch, inside of the launcher, which is difficult, if you have large hands. But they do make for an aesthetically-pleasing addition, as there's nothing on the outside of the launher, save for a screw head, which is for the mounting system that I use.
Difficult to find an exact picture of them on the internet, a quick Google search came up with this:
<img src="http://www.travelbugbooks.ca/graphics/minifan.jpg">
The ones at Walmart are made by a company called Ozark Trail and are usually less than 3 bucks a piece.
Minor trimming is required, for the blades to fit inside of 4" SCH40 PVC and you'll have to fabricate a mount... but man, do they work well!
I was so happy when I found these, because it solved the problem of having to mount a project box, along with 8AA batteries and a switch, to the outside of the launcher.
I usually choose to simply flip the switch, inside of the launcher, which is difficult, if you have large hands. But they do make for an aesthetically-pleasing addition, as there's nothing on the outside of the launher, save for a screw head, which is for the mounting system that I use.
Oops, sorry about those mistakes, I was thinking about some IR LED's I was using.
I have never tried a 12v fan on 18V, so I guess its fine.
I have never tried a 12v fan on 18V, so I guess its fine.
Ohm's law comes in handy much more often then you would think. Memorize that equation. If you have a set voltage (like from any power source you'd likely be using e.g. a battery) and a target amperage, you can figure out exactly (well, if you factor in the resistance and capacitance of the wires and stuff) what value resistor you need to put in there. If you don't do this, the electric motor, led, etc. will draw as much amperage as there is (which could be any value).rp181 wrote:Ohms law:
I=V/R
where I is current in amps, V is voltage, and R is resistance in ohms. For a standard LED, you don't want to go above .060A, .040A is good. The fans are rated .120A, so its still a good idea to put resistors in series.
Keep in mind 9v batteries die really quickly, so you dont need to assume a full 18v, but ~14v assumption is good.
-
Technician1002
- Captain
- Posts: 5189
- Joined: Sat Apr 04, 2009 11:10 am
To use the math for the above fan example we will convert mA into amps by moving the decimal point over 3 places. 120 mA is 0.12 Amps.
The items in a series circuit have all the component voltages add up to the source voltage. In our example we want 12 volts on the motor, the source is 9V X 2 or 18 volts. We want to drop 6 volts on the resistor. We now know the current and voltage. All that is left to fine is resistance.
I=V/R. I is on the wrong side. Using basic algebra, the R= will become R= V/I. Plugging in values gives R=6/0.12. R = 50 ohms.
To prevent smoke the resistor will need to be of a high enough wattage to safely radiate the heat generated without overheating.
Power is I squared times R. I = 0.12. I squared = 0.0144. I squared times R = 0.0144 times 50. This is 0.72 watts of heat. A 1/4 watt and 1/2 watt resistor are both too small for the job. A 1 or 2 watt resistor is recommended.
@ bluetooth, regarding your sig.. I moved from that to inside the chamber..
Current build is 2.5 inch designed for high power. Unfortunately the cost is a little over $50 bucks, but not by much.
The items in a series circuit have all the component voltages add up to the source voltage. In our example we want 12 volts on the motor, the source is 9V X 2 or 18 volts. We want to drop 6 volts on the resistor. We now know the current and voltage. All that is left to fine is resistance.
I=V/R. I is on the wrong side. Using basic algebra, the R= will become R= V/I. Plugging in values gives R=6/0.12. R = 50 ohms.
To prevent smoke the resistor will need to be of a high enough wattage to safely radiate the heat generated without overheating.
Power is I squared times R. I = 0.12. I squared = 0.0144. I squared times R = 0.0144 times 50. This is 0.72 watts of heat. A 1/4 watt and 1/2 watt resistor are both too small for the job. A 1 or 2 watt resistor is recommended.
@ bluetooth, regarding your sig.. I moved from that to inside the chamber..
Current build is 2.5 inch designed for high power. Unfortunately the cost is a little over $50 bucks, but not by much.