johnnyz383
100 W
I have a question. I am using a 48 volt lifepo4 battery but need to plum in a fan that uses only 12 volts. How do i calculate the resistors needed to reduce the voltage down to 12 volts?
Thanks
John
Thanks
John
reducing voltage NOT using a transformer
- Thread starter johnnyz383
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bigmoose
1 MW
First you need to know the current the fan draws on 12 volts. If it varies, like startup is different than running, the resistor won't work as well and will vary the voltage to the fan. Also fans typically draw different currents as their flow is changed.
But to answer your question you are starting with 48 V (nominal-remember a charged pack is higher, and a low pack is lower) and you need 12 volts. You need a 36 volt drop (nominally). You would use ohms law to calculate the resistance that would do that at a given current.
R= E/I so R=36/current. If the current is 0.1 Amps or 100 milli Amps the equation becomes
R=36/0.1 or 360 Ohms. Now we need to decide what the power rating of the resistor needs to be.
The power dissipated in the resistor is E*I or 36 * 0.1 or 3.6 watts. A 5 watt resistor if cooled will likely get by. Notice that for any serious currents the power rating is going to be very large and a lot of power is wasted in heat in that "dropping" resistor. For example, at the above currents the fan only uses 1.2 watts but 3.6 watts is wasted to not blow up the fan.
Hope this helps, and hope I did it right!
But to answer your question you are starting with 48 V (nominal-remember a charged pack is higher, and a low pack is lower) and you need 12 volts. You need a 36 volt drop (nominally). You would use ohms law to calculate the resistance that would do that at a given current.
R= E/I so R=36/current. If the current is 0.1 Amps or 100 milli Amps the equation becomes
R=36/0.1 or 360 Ohms. Now we need to decide what the power rating of the resistor needs to be.
The power dissipated in the resistor is E*I or 36 * 0.1 or 3.6 watts. A 5 watt resistor if cooled will likely get by. Notice that for any serious currents the power rating is going to be very large and a lot of power is wasted in heat in that "dropping" resistor. For example, at the above currents the fan only uses 1.2 watts but 3.6 watts is wasted to not blow up the fan.
Hope this helps, and hope I did it right!
serisman
100 mW
You might be better off with the following circuit.
(or skip all of this, and just use a 12v switching regulator or a 12v switching power supply. You can find LM2596HV based regulators for a few bucks that will work up to 60v, and most 12v switching power supplies will work with a 48v nominal battery.)
This circuit will provide a more stable output voltage that is less dependent on either the input voltage or the output current draw (compared to a drop resistor). It can also in many cases be cheaper than a large power resistor. The heat is dissipated through the mosfet (instead of a resistor) which makes it easy to throw a heatsink on (instead of buying a larger resistor).
Here is the way this works:
1. The resistor R1 lets some current through the zener diode D1 so that voltage at the gate of the n-channel mosfet is the zener diode's breakdown voltage.
2. Once the mosfet's gate-to-source voltage (Vgs) reaches it's transfer threshold, the mosfet will act as a variable resistor and drop just enough voltage from it's drain-to-source to keep the gate-to-source voltage drop at the mosfet's transfer threshold.
3. The mosfet's source is the output. This means we can fairly precisely control the output voltage with the following equation: Vout = V(D1) - Vgs(M1)
4. The 100nF ceramic capacitors C1 and C2 are only there to help smooth and stabilize the circuit. It's possible they aren't actually needed in practice.
5. The load resistor R_LOAD is only there to enable simulation of different current draws. In your case, it is a fan instead of a resistor.
To see how this circuit works, click on the image above and run it through some simulations.
Here is how to calculate the components:
Note: I am calculating Vin using an assumed 16 cell lifepo4 battery. Charged to 3.6v per cell this will have a 57.6 maximum voltage. Discharged to 3v per cell this will have a 48v minimum voltage. Please revise the numbers based on your specific battery configuration and charge/discharge voltages.
1. Pick the n-channel mosfet (M1). The important criteria to look for are (check the datasheet if needed):
5. Pick the heatsink: A fairly standard TO-220 mosfet can usually dissipate about 1 watt without a heatsink. You can calculate how much power the mosfet will need to dissipate using the equation: P = (Vin - Vout) * I where 'I' is the output current. So if the maximum input voltage is 57.6v (48v nominal), and the load (fan) is 12v @ 100mA the mosfet would dissipate about (57.6 - 12) * 0.1 = 4.56 watts at the worst case. You would definitely need to add a heatsink in this case. Calculating a heatsink can be tricky, but you would probably be ok with something about the size of the heatsink on a motherboard chipset.
(or skip all of this, and just use a 12v switching regulator or a 12v switching power supply. You can find LM2596HV based regulators for a few bucks that will work up to 60v, and most 12v switching power supplies will work with a 48v nominal battery.)
This circuit will provide a more stable output voltage that is less dependent on either the input voltage or the output current draw (compared to a drop resistor). It can also in many cases be cheaper than a large power resistor. The heat is dissipated through the mosfet (instead of a resistor) which makes it easy to throw a heatsink on (instead of buying a larger resistor).
Here is the way this works:
1. The resistor R1 lets some current through the zener diode D1 so that voltage at the gate of the n-channel mosfet is the zener diode's breakdown voltage.
2. Once the mosfet's gate-to-source voltage (Vgs) reaches it's transfer threshold, the mosfet will act as a variable resistor and drop just enough voltage from it's drain-to-source to keep the gate-to-source voltage drop at the mosfet's transfer threshold.
3. The mosfet's source is the output. This means we can fairly precisely control the output voltage with the following equation: Vout = V(D1) - Vgs(M1)
4. The 100nF ceramic capacitors C1 and C2 are only there to help smooth and stabilize the circuit. It's possible they aren't actually needed in practice.
5. The load resistor R_LOAD is only there to enable simulation of different current draws. In your case, it is a fan instead of a resistor.
To see how this circuit works, click on the image above and run it through some simulations.
Here is how to calculate the components:
Note: I am calculating Vin using an assumed 16 cell lifepo4 battery. Charged to 3.6v per cell this will have a 57.6 maximum voltage. Discharged to 3v per cell this will have a 48v minimum voltage. Please revise the numbers based on your specific battery configuration and charge/discharge voltages.
1. Pick the n-channel mosfet (M1). The important criteria to look for are (check the datasheet if needed):
- Maximum Vds (voltage drain-source): Pick one that is more than the maximum input voltage.
- Maximum Vgs (voltage gate-source): Pick one that is more than the intended output voltage.
- Vgs(th) (gate-source threshold voltage): This will help determine what value zener diode to use to get your intended output voltage. Vout = Vgs(th) + V(diode). Note: This value may have a range in the datasheet. Worst case your output voltage will be slightly lower or higher than your intended voltage, but probably not by enough to matter to a fan.
- Breakdown voltage: Pick one based on the equation: V = Vgs(th) + Vout. In this case, V = 16, (4 + 12)
- Power: You want to run about 5mA through the zener to make sure it performs it's job properly. We can calculate the power through the resistor with the equation: P = I * V. In this case, P = 80 mW, (5mA * 16v) and a standard 1/2 W zener would work fine.
- resistance (ohms): You want to run about 5mA through the resistor (so that about 5mA is running through the zener), so pick one based on the equation: R = V / I. The resistor will drop Vin - V(D1), so, V = 32v to 41.6v, (48v - 16v to 57.6v - 16v). So, R = 6.4k to 8.32k, (32v / 5mA to 41.6v / 5mA. In this case I just rounded up to 10k. This means we will be running a little less than 5mA through the resistor and zener, but it should still be close enough to work (3.2mA to 4.16mA) properly.
- Power: We can calculate the power through the resistor with the equation: P = I * V. In this case, P = 102.4 mW to 173 mW (3.2mA * 32v to 4.16mA * 41.6v) and a standard 1/4 W or 1/2 W resistor should work fine.
5. Pick the heatsink: A fairly standard TO-220 mosfet can usually dissipate about 1 watt without a heatsink. You can calculate how much power the mosfet will need to dissipate using the equation: P = (Vin - Vout) * I where 'I' is the output current. So if the maximum input voltage is 57.6v (48v nominal), and the load (fan) is 12v @ 100mA the mosfet would dissipate about (57.6 - 12) * 0.1 = 4.56 watts at the worst case. You would definitely need to add a heatsink in this case. Calculating a heatsink can be tricky, but you would probably be ok with something about the size of the heatsink on a motherboard chipset.
Gregory
100 kW
Very informative first post serisman. Welcome to ES.
serisman
100 mW
Here is some more information on why a drop resistor does not produce the most stable output voltage. (It's output voltage is VERY dependent on the input voltage and the load current).
Again, these calculations assume a 16 cell lifepo4 battery charged to 3.6v /cell and discharged to 3v /cell.
They also assume that the load (fan) will always pull about 100mA.
To calculate the resistance we use the maximum voltage: R = V(drop) / I. R = (57.6v - 12v) / 0.1A, R = 456 ohm.
Lets round that up to a fairly standard resistance of 470 ohm.
Assuming the load (fan) will always try pull about 100mA of current, the resistor will always drop the same voltage.
V(drop) = I * R. V(drop) = 0.1A * 470, V(drop) = 47v.
So, at maximum battery voltage, the load (fan) will see 10.6v (57.6 - 47).
At minimum battery voltage, the load (fan) will see 1v (48 - 47).
That is a huge range of voltages.
If we use a 450 ohm resistor (not as standard, but available), the voltage drop across the resistor is 45 v, and the voltage range is: 3v to 12.6v. Hmmm, better, but still a large range.
Let's also calculate power (again at maximum voltage drop): P = I * V(drop). P = 0.1 * (57.6v - 12v), P = 4.56 watts.
That is actually too close for comfort for a 5watt resistor. You would probably want to step it up to a 10 watt resistor.
Final Note: I'm not entirely sure a fan will try to pull the same current with a change in input voltage. But, because a drop resistor is very dependent on both input voltage and output current, it can be fairly hard to keep the output voltage where you want it.
The circuit I posted doesn't have these issues, and the output voltage will stay pretty much consistent. It is much more forgiving of input voltage swings, and other than the amount of heat generated, it doesn't really care how much current you pull through it.
Again, these calculations assume a 16 cell lifepo4 battery charged to 3.6v /cell and discharged to 3v /cell.
They also assume that the load (fan) will always pull about 100mA.
To calculate the resistance we use the maximum voltage: R = V(drop) / I. R = (57.6v - 12v) / 0.1A, R = 456 ohm.
Lets round that up to a fairly standard resistance of 470 ohm.
Assuming the load (fan) will always try pull about 100mA of current, the resistor will always drop the same voltage.
V(drop) = I * R. V(drop) = 0.1A * 470, V(drop) = 47v.
So, at maximum battery voltage, the load (fan) will see 10.6v (57.6 - 47).
At minimum battery voltage, the load (fan) will see 1v (48 - 47).
That is a huge range of voltages.
If we use a 450 ohm resistor (not as standard, but available), the voltage drop across the resistor is 45 v, and the voltage range is: 3v to 12.6v. Hmmm, better, but still a large range.
Let's also calculate power (again at maximum voltage drop): P = I * V(drop). P = 0.1 * (57.6v - 12v), P = 4.56 watts.
That is actually too close for comfort for a 5watt resistor. You would probably want to step it up to a 10 watt resistor.
Final Note: I'm not entirely sure a fan will try to pull the same current with a change in input voltage. But, because a drop resistor is very dependent on both input voltage and output current, it can be fairly hard to keep the output voltage where you want it.
The circuit I posted doesn't have these issues, and the output voltage will stay pretty much consistent. It is much more forgiving of input voltage swings, and other than the amount of heat generated, it doesn't really care how much current you pull through it.
serisman
100 mW
Oh, by the way, I originally created that circuit above as a part of a larger circuit that I was working on for my ebike (see below).
The below circuit is a constant current led driver with variable PWM dimming that should work for an input voltage of around 12-60v. It uses a linear (i.e. heat generating) method of controlling the current through the LEDs, so you would want to match the number of LEDs used with the input voltage to limit the amount of wasted energy (and reduce the heatsink requirements).
I haven't actually built the final version of this yet, although I have each of the sub-assemblies built up on my bench and fully tested.
The below circuit is a constant current led driver with variable PWM dimming that should work for an input voltage of around 12-60v. It uses a linear (i.e. heat generating) method of controlling the current through the LEDs, so you would want to match the number of LEDs used with the input voltage to limit the amount of wasted energy (and reduce the heatsink requirements).
I haven't actually built the final version of this yet, although I have each of the sub-assemblies built up on my bench and fully tested.
hardym
100 W
I didn 't see the power requirements, but my guess is the current woudl be about 1 amp.
For about the same price as the solutoins above you can use a switching buck converter at much higher efficency.
Below is a link to DX, but these modules are commonly available on ebay.
http://dx.com/p/dc-4-5-30-to-1-25-26v-auto-step-down-converter-voltage-regulator-164313?utm_source=GoogleShoppingUS&utm_medium=CPC&utm_content=164313&utm_campaign=405&gclid=CPPZvJfAibkCFYN_QgodLQ4Atw
The resistor or linear regular solutions in this thread are not efficient, and would not be good for battery powered solutions.
Mark.
For about the same price as the solutoins above you can use a switching buck converter at much higher efficency.
Below is a link to DX, but these modules are commonly available on ebay.
http://dx.com/p/dc-4-5-30-to-1-25-26v-auto-step-down-converter-voltage-regulator-164313?utm_source=GoogleShoppingUS&utm_medium=CPC&utm_content=164313&utm_campaign=405&gclid=CPPZvJfAibkCFYN_QgodLQ4Atw
The resistor or linear regular solutions in this thread are not efficient, and would not be good for battery powered solutions.
Mark.
Wow, nice explanations, I'll subb this thread for further reference. I would add a more basic solution that has the disadvantage that it may unbalance the pack a bit: just get the voltage between the ground of the battery and the 3rd or 4th cell (depending on chemistry). A standard PC 120mm fan on 12V takes no more than 3-4 watts and can take 14V as well as 10V with a corresponding linear variation in rotation speed - so this type of fan would use no more than 0.4Amps from the battery.
Depending on the current you're taking from the pack this can be insignificant (say for 30Amp load current, 0.3A is 1%). It is likely the energy lost even in a nice dc/dc would be close to what you'd use with this arrangement to power the fan. You have the disadvantage of ever-so-slightly unbalancing the pack, so this setup would work better for high power applications where 3W are within the margin of error. It also depends a lot on the fan. You could set it in series with a thermistor to turn the fan as fast as needed as a function of the temperature.
Depending on the current you're taking from the pack this can be insignificant (say for 30Amp load current, 0.3A is 1%). It is likely the energy lost even in a nice dc/dc would be close to what you'd use with this arrangement to power the fan. You have the disadvantage of ever-so-slightly unbalancing the pack, so this setup would work better for high power applications where 3W are within the margin of error. It also depends a lot on the fan. You could set it in series with a thermistor to turn the fan as fast as needed as a function of the temperature.
serisman
100 mW
Be careful which LM2596 based switching regulator you get.
The one hardym linked to has an input range of 4.5-30V and is not suitable for a 48V battery pack.
You want one that uses the high voltage part LM2596HV that is good for up to a 60V input.
Something like this:
http://www.ebay.com/itm/261201614159
or this (encased in a metal heatsink):
http://www.ebay.com/itm/370864302106
The one hardym linked to has an input range of 4.5-30V and is not suitable for a 48V battery pack.
You want one that uses the high voltage part LM2596HV that is good for up to a 60V input.
Something like this:
http://www.ebay.com/itm/261201614159
or this (encased in a metal heatsink):
http://www.ebay.com/itm/370864302106
serisman
100 mW
I also agree that if your fan draws anywhere near 1A, you really should not be using a linear based regulator, and need to get a switching regulator.
To supply 12V @ 1A with a linear regulator from a 57.4V (max) battery pack, you are dissipating (wasting) 45.4 watts through the linear regulator (resistor or mosfet) while only supplying the fan with 12 watts. That is about 20.9% efficiency which is absolutely horrible, not to mention it would be really hard to find a good way to dump 45.4 watts of heat anywhere.
For smaller loads, a linear regulator can be a cheap/easy solution, but definitely not the best choice for something with moderate to heavy current requirements.
To supply 12V @ 1A with a linear regulator from a 57.4V (max) battery pack, you are dissipating (wasting) 45.4 watts through the linear regulator (resistor or mosfet) while only supplying the fan with 12 watts. That is about 20.9% efficiency which is absolutely horrible, not to mention it would be really hard to find a good way to dump 45.4 watts of heat anywhere.
For smaller loads, a linear regulator can be a cheap/easy solution, but definitely not the best choice for something with moderate to heavy current requirements.
dnmun
1 PW
you can find a pwm DC motor speed controller on ebay that will let you dial in the speed you wanna run the fan.
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