So let me see if I've got this right. R = Resistance (my light bulb in Ohms) C = Farrad rating of the cap Tau*5 = the cap at almost full charge So if my bulb remains the same, and I use a 4400uF cap i get Tau = RC Tau = (2.25V/.25A) * (4700/1000000) Tau = 9 * .0047 Tau = .0423 Tau * 5 = .2115 seconds fade??? PS. I've also got a power contraint. The switch for this circuit is a photocell rated at 40mW, 20V.
Here's the thing abut the calculations (which look ok by the way). the resistance of a light bulb is more dynamic that a resistor. IT will change with temperature of the filament. I don't remember offhand which direction it changes, but I know it does. So you can do the calculations, but don't get hung up on them. Go with what works. Breadboard this baby up and use that as your guide instead. Keep in maind that capacitances are additive in parallel. So you can buy a bag of assorted caps for 2-3 bucks, and futz around until you get a value that's fells good to you. Then if you want, you can order pieces at that value. We can toss around equations all day, but the real world has a way of laughing at those sometimes.
Eh, I just looked at your photocell specs. I don't know exactly how you plan to use it, but 40 mW at 20V means it can handle a whopping 2 mA current, so I hope you weren't planning on inserting that in series with your light bulb.
The photocell i'm expecting, will be the most expensive part of this. The one I put down was the cheepest. What if I switched to super bright LEDs, could I still get a 2-5 second fade with a cheep photocell? I'd need something that can handle at least 20 mA if I remember most LEDs. Thanks much for the help linear.
You gotta give me some clue how you're hooking up the photocell. And I'm not even sure waht this photocell does. Is it a voltage source? Or a photoresistor? Help me out here.
It's a photo resistor as far as i can tell. I'm going through jameco and they offer several. The cheepest one's resistance is light, 10k; dark, 100k. about this 40mW / 20V = 2mA if I'm only putting 2.25 V though the circuit, that wouldn't it be 40mW / 2.25V = 17.8mA? The photocell will be the switch to let power flow to the cap and light, in parallel. That's it. Though, there will be ~50 of these across a board and my hope is that they all interact. setting each other off, at times.
okay, I get it now. You can't just stick that thing in series and call it a switch. Your best hope is to get a cheap transistor that can handle the current (you'll want to look at collector current ratings) and use it as a switch. The photoresistor will form part of a voltage divider (so you need another resistor as well). Check out this page, it has the fundamentals spelled out pretty well, although it's more aimed at designing for continuous control rather than on/off switching. http://www.eg.bucknell.edu/~bsprunt/classes/elec_101_spring_2000/Labs/lab5/lab5.htm
I see, so use the photocell as a trigger to open the transistor, and the cap will help the bulb turn on and off in a fluid manner. Then play with resistors until an appropriate level of light is reached to trigger the photocell. Is that what you're saying?
That's the gist of it. The only design parameter you need to figure out isthe value of R that makes the circuit shown here work for you: Again, whip out the breadboard, and slap a pot in there, twiddle till it's working the way you like it, then take the pot out and meter it. That's the easy way to find R, don't mess around with the equations.
Your bulbs are drawing 250 mA, that's a pretty good size (but not huge) load. It's bigger than most of the really cheap transistors will want to handle. Also look for something that's rated for maybe double that. You may well be into something in a TO-92 package at those currents. The 2N3904 in that article won't handle more than 100 mA (from memory--YMMV). That article also didn't mention the base current at all, but you should be able to send a transistor into saturation (turn it ON) with ~ 1mA of base current, so your photocell will be okay for that if you go this route.
the best i can find in this catalog is a TO-39 NPN, Vceo = 75, Min. hFE @ Ic = 30 @ 500mA. I'm not quite sure what some of that means. They sell at .69 cents a piece. will it work?
More potentially helpful stuff on exactly how that voltage divider bit will work: http://web.mit.edu/rec/www/workshop/photoresistors.html
That will work. hFE is the ratio of collector current to base current when you use it as an amplifier. Not as interesting in your switching application, but handy to know if you're trying to figure out how much base current it takes to drive the thing into saturation. TO-39 is an inconvenient package. it's a dinky metal can. Look for a TIP31, that should be good for up to 3A collector current, and commonly available for around 60-80 cents.
three things: move the base conenection to the node between the fixed resistor and the photoresistor. One end of the photoresistor connects to ground. the supply voltage is way low, are you setting it there to drive the bulb? How are you powering this? I thought we beat this to death, but 180 uF is way small for what you want to do. You want maybe 20-50 times that size cap.
that's right, i did find a 2200 uF cap, if the supply voltage to 2.25, that should be right. as far as i knew i had to make the supply voltage less than the bulb, so i wouldn't burn it out?
