Discussion in 'General Hardware' started by Derek12, Aug 20, 2011.
see, i told you so.
my formula is totally crisp
The stripe goes towards negitive. Not every diode in a PSU is suitable. There could be zener diodes in there. You need something like a 1n4002 (cheap at radio shack). Look for the black ones in the PSU. Don't use the glass ones as they could be zeners or just low current signal diodes.
You were using a red and black for 5v right? With the added drop of a diode, 4.2v might not be enough to start a fan.
To have the best of everything, use a 12v rail, a series resistor (to drop the voltage), and parallel a capasitor across the resistor. 1000-2200uf 16v would do it. This will give the fan a kick when it's powered up to assist it starting from a dead stop.
Many thanks, I found a 36 ohm 1/2W resistor (I think it's even more powerful) from a broken appliance (a TV) and effectively the fan voltage is now very close to 5V (4.70V) and it runs fine, the resistor isn't as warm as the previous one so I think it will be fine, anyway I will look for a fan controller like the Scaminatrix one because my "project" is very botched: the resistor is hanging (though there is no risk for my computer innards
Lazzer408 this fan seems to run fine directly from the circuit without a capacitor, must I add one just in case? thanks
no. i dont see how it helps.
also the capacitor will also be connected to the PSU which isnt advisable. you poop the cap you poop your PSU as well.
and i dont think having a cap across the resistor will give it a kick. the cap will dissipate energy through the fan as soon as the PSU is turned off.
Many thanks maybe then I will leave it as is unless I see problems, then I would donate a capacitor to there
what i said will work. no questions asked
Correct it works like a charm, many thanks to you specially for the circuit and formula
no probs man. glad to help
The capacitor isn't needed if things are working. Glad the OP has it working now.
The cap is in parallel with the resistor for one so no it won't disapate through the fan. How does a capacitor help? A fully discharged capasitor is essentially 0 ohms. If your series resistor drops 12v to 5v that means there's a 7v drop across the resistor. The capasitor will have to charge to that voltage when the circuit is powered up. During that time, the voltage at the fan will ramp from 12 to 5v as the voltage across the resistor climbs from 0-7v. This causes a 12v kick to the fan to help get it moving. Some fans can't start at low voltages but will run at lower voltages once you give them a push.
The OP mentioned something about the fan not starting so I thought it appropriate to mention the capacitor "trick" incase anyone can benefit from it. Not argue about it. There's no harm to a powersupply even if the cap did fail (which is highly unlikely) and if you want I'll go blow some up tonight!
actually a capacitor is always infinite ohms in DC. as far as i know.
Not when it's discharged. When it's charged to a voltage equal to the supply voltage it will be, for all practical purpose, infinite ohms.
Anyways, go try it. You might learn something. Got a scope? If not, put a small 100uf across a current limiting resistor for an LED. You will see the LED flash when you power it up.
Ups, A capacitor is basically *short circuit* almost fully conductor when discharged, its resistace begins to increase until very near to infinite when it's fully charge
You beat me
a capacitor lets current flow into it but not afterwards. therefore its not short circuited!!
thats how we are taught in engineering.
lets see you connect an LED in series with a discharged cap to a battery. if you are right the LED will glow
Sorry I used the wrong term "short circuit", I wanted to say "almost fully conductor"
The LED would light at first then it would dim quickly until is fully off (unless the cap is bad)
Here you go. I'll assume a 100ohm fan and a 100ohm limiting resistor to keep the math easy. When the switch was closed, it produced a 12v spike that settled in at 6v because of the capacitor.
No the LED will flash brightly then dim to nothing as the capacitor charges.
no fair you are using a POLAR cap.
I don't think that matters in his circuit, he is using a DC generator I think though I don't know why it has a sine wave
Doesn't matter. A cap is a cap but if a polar is used it's polarity must be noted. It's unlikely the OP (or anyone else who could benefit from the trick) would have a 100uf non-polarized cap in the parts bin. A 100uf of higher 16v cap can be found all over the place in used power supplies or motherboards.
also once again his graph is only possible as the cap already has some charge in it when the switch is connected. start the circuit on an empty cap and it WILL show a zero voltage spike.
The cap is at 0v when powered on. The 100ohm resistor keeps it discharged and the emulator also assumes caps are at 0v when the emulator is started. I started it with the switch open then closed and reopened the switch to produce the wave form shown.
If I place the scope across the resistor/cap you will see the exact opposite waveform as the cap charges from 0v to 6v.
EDIT - Here is the charge curve on the cap as the switch is closed then opened. It starts at 0, charges to 6, then discharges back to 0 by the 100ohm resistor in parallel with it.
The oscilloscope is in parallel to R2 as I see. The capacitor, being empty at first, is like a normal cable, so all current flows through it. and not from R1 so the oscilloscope gets the supply voltage. When the cap is being charged, current begins to flow through it and R1 and voltage begin to drop. After the cap is fully charged, the current flows solely though R1 and voltage stabilizes.
Correct me if I wrong Lazzer, please
What youre saying would be correct if oscilloscope would be in parallel to the capacitor.
EDIT: Once again, I am too slow
I ish Ninja!
Yes Derek. Electricity takes path of least resistance. When the capacitor is discharged it's resistance is 0 and current flow will completely bypass the resistor until the capacitor begins to charge. Once the charge equals the voltage drop across the resistor (6v) it then becomes infinite resistance in this circuit.
Here is a video showing a situation where the cap is required. My resistor box is set to 470ohm. I'm not about to crunch the math on this because that's not the point. The point is that the resistor was required to run the fan slower then intended (to reduce noise usually). 470ohm gave the builder the required level of noise with adequate airflow but "OH NO" :O the fan won't start from a dead stop. Here is when the capacitor comes to the rescue. In parallel with the dropping resistor, it will provide the kick needed to start the fan! Yay capacitor! -.-
Holy hijack batman.
I just wanted to show a more typical/industrial application of an RC circuit that's not oscillator related (RC is commonly used in timer circuits). This circuit is a simple switched mode power supply with it's control circuit omitted. Lets say our 12v supply has unlimited current or maybe a few 100 amps in an automotive application. The goal is to protect the expensive transformer from high currents in case a switching mosfet failed in a short-circuit condition. Under normal operation, the pulse width is shorter then the charge/discharge time of the capacitor and the resistor will flow very little current. For whatever reason, a mosfet just failed. :O The capacitor will charge up to the supply voltage and all the current must flow through the (wattage undersized) resistor causing the resistor to blow thus opening the circuit and protecting the transformer. A few pennys spent for one resistor and one cap saved a $5 transformer which would have gone up in flames. You could say "Well thats what the main fuse is for" but to save on the cost of multiple fuse holders, only one fuse was used. The power supply we are protecting has other circuits that comunicate faults and must stay powered.
i think we were taught BS at college.
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