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usb mini fountain
- Thread starter netcoord
- Start date
netcoord said:
I'm assuming you just want to pull the power from USB, and not control it, correct? See here for the pinouts:
http://www.starmount.co.uk/s_usbpin.htm
You will get +5V from the USB. Depending on how this "mini-fountain" works, you'll probably have to step this voltage down a bit. Given that this fountain probably doesn't have much of a fluctuation in current draw, you could probably just use a resistor to drop the excess 2V.
LOL no. First of all a resistor is a bad idea for this. Second of all resistors aren't rated by voltages. They're rated in Ohms (resistance) and watts (power). For simplicity - you could probably just put 3 rectifier diodes in series. a 1N4007 (well anything between 1N4001 and 1N4007) Would be perfect. Current would probabaly be low enough to use small signal diodes - like the 1N914 - again you'd want 3 in series. Either should work just fine though - and I'd expect any electronics store (even ratshack) to have at least one of them, probably both. Hope this helps.netcoord said:it would be cool to control it but my main concern is power
thanks
so i would put a inline 2v resistor on the power wire in the usb and atach it to the the positive wire for the motor of the fountain. the atach the usb ground to the motor ground.
If you want to use a resistor, what you need to know is the current draw of the fountain. Then take 2V/current = size of resistor.
I diode will not necessarily be better than a resistor as far as power dissipation goes. The same amount of current will still drop the same amount of voltage, so the dissipation will still be 2V * current. The advantage of diodes is that they drop the same amount of voltage regardless of current (within a certain range), so you don't have to worry about calculating resistances.
I diode will not necessarily be better than a resistor as far as power dissipation goes. The same amount of current will still drop the same amount of voltage, so the dissipation will still be 2V * current. The advantage of diodes is that they drop the same amount of voltage regardless of current (within a certain range), so you don't have to worry about calculating resistances.
grjr said:I don't think I understood a single thing you said
Let me try to clear it up. Instead of having a resistor to drop the 2V, we instead want to use some diodes:
Each diode drops roughly 0.7V, pretty much independant of the current through them, so you don't have to do any calculations to try to figure out the size of them, like you would with a resistor. They provide a pretty steady 2.1V drop no matter what load is on them (within reason).
Ok, it sounds like a simple electronics primer is in order. We use the following symbols:
I - current (in amps). Literally it's a measure of how many culombs (a very large number of electrons) pass a point in one second
V - voltage. How much energy each electron has
R - resistance. How hard the electricity has to work to get through
P - Power. how much energy is used in a certain amount of time
The most basic equation is V = IR. Voltage equals current times resistance. You can rearrange this equation all you want, so if you have two of the values, you can find out the third.
Power: P = IV. Power equals Current (number of coulombs/sec) times Volts (energy/coulomb), and you get energy/sec.
We know that the voltage dropped across all the resistances in a circuit has to equal the voltage supplied to the circuit. In this case, the sum of the voltages across the pump and whatever you put in series (see previous post w/ schematic) has to equal the 5V that the USB port supplies. Also, the amount of current has to be the same through all components that are in series. Kinda like a waterpipe. The amount going in is the same as the amount going out.
We know that the fountain needs 3V with a certain amount of current. We have a 5V supply, so we need to lose 2 of those volts. There are a few ways to do this, but the two simplest are to use diodes or use resistors.
Resistors drop voltage depending on the amount of current going through them (remember that V=IR, so if I (the current) goes up, the voltage drop increases). So if you know that you need to get rid of 2 volts, and you know that the fountain takes, say, 200mA (.2 Amps), then we need to have a resistor that will drop those 2V at the same 200mA. Rearrange V=IR to get R=V/I, and you get 10 ohms.
Diodes can make things easier, because they drop the same amount of voltage, no matter what kind of current you put into them. At least, up to a certain point. A typical diode drops about 0.7V. So, to get around the 2V drop we need, you would use three diodes in series (like you saw before). And you don't have to worry about the current or resistance. The one disadvantage of diodes is that they are more expensive than resistors.
The one thing you need to worry about is power. The amount of power (energy/time) supplied by the usb port has to equal the amount of power dissipated by everything in the circuit. We don't need to worry about the amount of power dissipated by the fountain--it's designed to dissipate whatever power it normally does without any problem.
You need to worry about how much power will be dissipated by the resistor or diodes. P=IV, so 2V * 200mA = 400mW or 0.4 watts. Small resistors are rated at 1/4 or 1/8 watt, so they won't be able to take that kind of heat. A 1/2 watt resistor would. You'll have to adjust these calculations for whatever current you're dealing with.
If there's too much power being dissipated by the resistors, you can user two resistors with smaller values, so that each one only drops half the voltage, thus half the current.
One last thing: USB is only rated to supply 500mA to a load (like your fountain). If your fountain takes more current than that, you can either take the risk of burning something out on your motherboard, or you can find another way to power your fountain from the computer.
I - current (in amps). Literally it's a measure of how many culombs (a very large number of electrons) pass a point in one second
V - voltage. How much energy each electron has
R - resistance. How hard the electricity has to work to get through
P - Power. how much energy is used in a certain amount of time
The most basic equation is V = IR. Voltage equals current times resistance. You can rearrange this equation all you want, so if you have two of the values, you can find out the third.
Power: P = IV. Power equals Current (number of coulombs/sec) times Volts (energy/coulomb), and you get energy/sec.
We know that the voltage dropped across all the resistances in a circuit has to equal the voltage supplied to the circuit. In this case, the sum of the voltages across the pump and whatever you put in series (see previous post w/ schematic) has to equal the 5V that the USB port supplies. Also, the amount of current has to be the same through all components that are in series. Kinda like a waterpipe. The amount going in is the same as the amount going out.
We know that the fountain needs 3V with a certain amount of current. We have a 5V supply, so we need to lose 2 of those volts. There are a few ways to do this, but the two simplest are to use diodes or use resistors.
Resistors drop voltage depending on the amount of current going through them (remember that V=IR, so if I (the current) goes up, the voltage drop increases). So if you know that you need to get rid of 2 volts, and you know that the fountain takes, say, 200mA (.2 Amps), then we need to have a resistor that will drop those 2V at the same 200mA. Rearrange V=IR to get R=V/I, and you get 10 ohms.
Diodes can make things easier, because they drop the same amount of voltage, no matter what kind of current you put into them. At least, up to a certain point. A typical diode drops about 0.7V. So, to get around the 2V drop we need, you would use three diodes in series (like you saw before). And you don't have to worry about the current or resistance. The one disadvantage of diodes is that they are more expensive than resistors.
The one thing you need to worry about is power. The amount of power (energy/time) supplied by the usb port has to equal the amount of power dissipated by everything in the circuit. We don't need to worry about the amount of power dissipated by the fountain--it's designed to dissipate whatever power it normally does without any problem.
You need to worry about how much power will be dissipated by the resistor or diodes. P=IV, so 2V * 200mA = 400mW or 0.4 watts. Small resistors are rated at 1/4 or 1/8 watt, so they won't be able to take that kind of heat. A 1/2 watt resistor would. You'll have to adjust these calculations for whatever current you're dealing with.
If there's too much power being dissipated by the resistors, you can user two resistors with smaller values, so that each one only drops half the voltage, thus half the current.
One last thing: USB is only rated to supply 500mA to a load (like your fountain). If your fountain takes more current than that, you can either take the risk of burning something out on your motherboard, or you can find another way to power your fountain from the computer.
fat-tony said:Let me try to clear it up. Instead of having a resistor to drop the 2V, we instead want to use some diodes:
Each diode drops roughly 0.7V, pretty much independant of the current through them, so you don't have to do any calculations to try to figure out the size of them, like you would with a resistor. They provide a pretty steady 2.1V drop no matter what load is on them (within reason).
You know what, for some reason when I first read this thread I thought the pump ran off 2V, so I was thinking that the 2V/I would just give you the equivilent resistance of the pump but after rereading it I caught my mistake
Whatsisname
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500 mA max by specification. Anything above that and you run the risk of blowing up your usb controller.
wayne said:
GlobalFear
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IF you seem set on powering this thing on usb and simply cannot get enough power out of usb it might be an idea to run a seperate usb port for this thing running whatever voltage you need out of the psu just for power. (wouldn't want to plug anything else into that one....)
GlobalFear
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Whats the fun in that?
GlobalFear
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I wouldn't consider that fun.....Anyways look up a few posts. I posted a different idea.
Whatsisname
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Or why not get a cheapo power supply thing from http://www.bgmicro.com or http://www.allelectronics.com which will provide plenty of power for a measly 5 or 6 bucks?
kronchev said:mini fountain? more info on this. I think I want to put this in my case too
it's one of those waterfall things, but it only like 8" high.
as for why not using a power adapter, it runs on two AA batteries 3v. so the 5v of usb will more than enough. it sits on my desk next to my tower why run a plug, taking up more room on surge protector when i can just run a small usb cable
and like globalfear said "Whats the fun in that?"