I want to use one potentiometer to adjust two lines simultaneously, one between 5-12V, and 0-12V, like this: 100% 12V / 12V 90% 11.3V / 10.8V 80% 10.6V / 9.6V ... 20% 6.4V / 2.4V 10% 5.7V / 1.2V 0% 5V / 0V Where each decreases by the same percentage of their voltage range simultaneously. Sorry if this isn't clear. I've thought of a few ways, but I'm sure I've missed the best ones.
Using a dual gang potentiometer would probably be the easiest method. Keep your voltage regulator circuits totally separate, with the only link being the mechanical link and your power supply.
Slight problem with regulators is that they only turn down to 1.25V, and (unless you use low drop-out) up to about 10.5V. You might get better results with a couple of emitter-follower controls, use a dual-gang 1k pot with the bottom end of one going to a rectifier diode (to start at zero) and the bottom end of the other going to a 5V6 zener (to start at 5V). Max output will be about 11.3V. Like this, with the diodes instead of R1:
I thought I could use a zener diode in some way, certainly better than the idea I originally came up with. That is perfect, thanks guys.
What current are you wanting? Below about 500mA you don't need Q2, just a 2N2222A for Q1. Otherwise, any high-gain NPN for Q1, power PNP tranny for Q2.
I would use the circuit that cpemma suggested, nice and simple. I would use Q2 in any case since the to92 case like the 2n2222 can only dissipate a watt or 2. I'd use a 2n2222 or equiv. for Q1 and a TO220 package transistor, like a MJE2955T, for Q2 so you can easily heatsink it if you need to. both of these transistors are very common.
I have an 8-segment bargraph (red LEDs), what would be the best way to run that off the 0-12V side? Considering what I'm getting at with this, are there any better ways of doing the whole thing, while still using a pot (and not a rotary switch)?
If what you're getting at is an LM3914-driven bargraph display that shows zero at 5V and full scale at 12V, you can just raise the zero point to any voltage level you like, and just need 1 variable supply. "To raise the zero point, we add a resistor R2 between ground and the "lo" pin (#4) of the IC." Make R2 a 10k preset pot and tweak to suit.
Add a diode in reverse parallel with the fan if you use a FET as they're quite easily damaged by back emf. FET transistor = Field Effect Transistor transistor
Trying to control the output with a mosfet seemed very fussy component-value wise when I looked at it, but if you have a nice circuit...
oops! (fett is a word in norwegian... means fat (like in food)... it's often used for something else aswell... but this is a family forum, right? ) cpemma: The Power MOS-FETS are voltage controlled, right? so mabye use a voltagedevider on the input?
Yes, a voltage divider on the input would work. Stick to low value resistors/pots though e.g. no higher than 10k.
If you really want to use a FET there only needs to be a few minor changes to the circuit proposed by cpemma. First off you dont need to use an additional diode, eventhough what steveyg is correct, all FET's (the MOSFET variety)ive ever seen have a diode built right in. Secondly, you probably don't want to drive it directly since fets need about 3.5V on their gates before they start to turn on as opposed to the .7V transistors need. So if you just used a voltage divider your maximum fan voltage would be about 8V or so. The best option would be to use the circuit cpemma suggested bt replace Q2 with a fet i'd suggest an IRF9Z34 and the put a 1k resistor from Q1's collector to the 12V rail. But i'd still use transistors since this circuit still won't get you any higher than the 11.3V from before since the 0.7 loss comes from Q1, not Q2.
