OK, I feel like this has to have been mentioned (shot down?) before, but I haven't found it anywhere. Typical problem. I want to lower the voltage across a fan. I don't need it to be adjustable, I just want permanetly lower voltage. I'm not goint to do the 7 volt trick, because, I think I want more voltage than that (8 or 9V), and, anyway, I want to play with circuits. So, I figure I'll just put a resistor in. No big deal. But, this fan I have is pretty heavy and seems to require a lot of voltage to start, plus, since I'm doing the mod for someone else, I really don't want it to not start on some random day. Now, I've read the capacitor jumpstart solution for the 7 volt mod, and was wondering if a similar thing would work for a resistor voltage drop. Specifically, would wiring a capacitor in parrallel with a resistor, on the ground side of the fan, acheive the same result? I.e., would the fan see the full 12 volt potential while the capacitor was discharged and therefore run the current through that, and then, once the cap charged, just run at the middle voltage through the resistor? like this -- Code: __resistor___ 12 v+ -----------Fan -----|____cap___|----ground Am I missing something or should this work? Great forums, by the way. Best, Sam
..to skin a cat.. The basic digram you set would work. The only thing you may have to play with is the size of the cap. The charge time on the cap may be too short to get the fan running. You can slightly lenghten that time by adding a smaller restistor in line with the cap. You may also look to using a zener diode in place of the mid-point dropping res. This way if you add/change fans on that load you will not change the supply voltage. The zener will need to be mounted on a small heatsink. It's never a good idea to use a passive component to regulate voltage. Using a zener diode is a cheap and easy way regulate the fan voltage under a changing load. remember ohms Law: voltage equals current(I) times resistance(R)
Welcome to Bit, sam_d The circuit will work, but with a fairly low-resistance fan you need a very high-value capacitor to get more than a split-second boost. A neater way is to by-pass the resistor with a PNP transistor switch, with its base lead connected to ground through a resistor & capacitor in series. A fairly high-value resistor can be used with a smaller capacitor. There's a schematic here but you may need a beefier transistor if the fan draws over about 500mA.
I thought a resistor would be OK because I don't plan on changing the fan (or adding any more in series). Is there any other reason not to use a resistor? As for the zener, the problem was when I was looking -- at mouser.com -- they didn't seem to have many zeners rated over 1W. this fan has a pretty high draw -- about 300mA at 12v -- which, if i use the zener to drop four volts, would bring it pretty near its tolerance. 4v * (.3A*8/12) = .8W, if my math is right. is that why you suggest the heatsink?
thanks for the welcome! and the schematic looks good too. especially since this is a low resistance fan -- around 40Ohms by my calculations.
One snag with resistors - when the fan starts up it draws more current, so the resistor (by Ohms Law) drops more voltage than planned. In effect, you get opposite to a kick-start, a soft-start. Your zener maths is correct, the 1W ones are OK for low-power fans, but you can buy them rated to several watts. There's the common 1.3W BZX85 series, along with 2W, 3W and 5W types, but the larger ones have generally been supplanted by cheap regulator ICs.
Ah, right. But if had my kickstart bypass the resistor until the fan got going I'd avoid that problem. Btw, is the increased current draw caused by the fan's inductance? Hmm. Don't know why Mouser doesn't have (or I couldn't find) those 1.3W ones.
Yes, soft-start only applies to a fixed resistor used as a voltage dropper. The increased current is while the fan motor is accelerating in speed so using more energy than when just steady-speed cruising. Same for any motor, even us.
What purpose does R2 serve in this schematic? Is it just for the capacitor to discharge over when the circuit is off?
That's right, someone came up with a neater, faster, discharge solution using a second transistor, but the resistor will work if the power has been off for an hour or so. Twas Smilodon, and tis here
Could I just use a smaller resistor at R2, say 1000 uF, to get faster discharge? I'm not sure I understand the transistor discharge method; I'll have to study the schematic some more. Thanks for the help. I love this forum.
a 1000uF resistor? The 1k2 resistor for R1 allows about 9.5mA max (11.3/1.2) to flow through it with the cap totally empty, gradually falling off as the cap charges up, but while ever the current is above 12V_Fan Current/Transistor Gain the transistor is full on and the fan gets about 11.8V. The graph line is horizontal. Below that current the transistor starts to close down and the fan voltage gradually drops as shown on the graph to 6V, set by fan resistance & dropper resistor. With a full cap and R2 = 1M, current through R1 & R2 is about 11uA, so if transistor gain =100, about 1.1mA leaks through the transistor, which has little effect on the voltage the fan gets, the dropper resistor Rd is having 99% of the effect on fan current with your fan. But as R2 gets smaller, more current leaks through Q1, reducing the effective dropper resistance of Q1 || Rd, and the fan voltage gets higher. It's suck it & see, 100k for R2 seems OK in simulation, 47k having a more obvious effect, but you can compensate if necessary with a higher-value dropper resistor.
