# Help Getting Started - Simple DC Circuit

• posted

You will want to make sure the source voltage is a greater than the led's forward voltage and the current limit resistor voltage drop. I think white Leds are about 3V and Red Leds are 1.7V, but be sure to check.

David

• posted

2 'C' cells in series should give you enough power for short periods, and 3 volts will be sufficient to light the LEDs. You could try 2 'AA' cells, but I'm guessing they won't handle 30x0.02A = .6A for very long (if at all).

Each LED&switch will need a separate resistor. Don't try to parallel them thru 1 resistor; you'll let the magic smoke out of the LED's. :(

To calculate the resistor values:

R = (Vcc - Vled) / Iled

where Vcc is the supply voltage, Vled is the forward voltage drop of the LED, and Iled is the operating current of the LED.

You can find a minimum R value for your LED's, and try larger values that still give acceptable brightness with less power consumption.

HTH

• posted

Red, yellow and green LEDs are around 2 volts, and I find they are quite bright enough with 10 mA, so, with 3 volts, you could use 100 ohms.

Blue and white LEDs want about 3.6 volts, I believe, so you'd need another cell or two for them.

Resistors under 1K are certainly not uncommon - I think I've got a few values under 10 ohms in my stock at work (although I'll admit that I don't use such low values very often.)

```--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca  ```
• posted

Typical MCD: 620 30 mA current (max.) Typical wavelength: 570mm Size: T-1-3/4 (5mm) Viewing angle: 12=B0 Sold in package of 1 Typical voltage:

2=2E1V, with a maximum of 2.8V

chances of running 30 led's from 1 nine volt are slim to none... at least not for very long... you could put 2-3 nine volt batteries in parallel... and that would divide the current draw between them...

ya know... i'm not big on radioshack prices... so i found a nice little place...:-D

can't get them all at once.. but free is a good thing...

• posted

If you need that much quantity, don't go to RadioShack. Check out

- they are WAY cheaper and have no minimum.

• posted

(I apologize for posting this basic circuit question. 25 years ago in college I had to know E=IR etc. Just having problems getting started...)

I'm trying to "invent" a mechanical device. I've been having some alignment problems with it and this afternoon it occured to me some simple electronics would help a lot. But it has been many years since I've done anything more complicated with electronics than change batteries in the household smoke detectors. ;-) I'm looking at the Jameco web site, seeing if I can find enough info to play with E=IR but there are just too many options & choices...

I want to go to my neighborhood Radio Shack and buy - 30 ea SPST NO switches - 30 ea green LEDs - 30 (?) ea resistors to limit current to the LEDs - A circuit board to solder the LEDs & resistors to - A battery case (e.g. 4 "D" cells) or 9v clip I have an electronics soldering iron and plenty of wire.

The goal: When everything is lined up right on my mechanical device, all the switches will be closed and all 30 LEDs will be glowing. Then I'll unplug the battery, as the machine is ok once all the switches are closed. (E.g., no long-term lighting requirement.)

How can I get started with this? For the battery which value to use, 1.5,

3, 6 or 9 volts? Does each LED need its own resistor or is one resistor enough? (I don't care how bright the LEDs are so long as they are visible.)

Thanks.

-- Mark

• posted
[piggybacking - for some reason Mark's reply didn't show up here...]

Since they are assorted LED's, I think you can count on using 'assorted' values for the resistors. Try finding a set of resistors from 47 ohm to maybe

150 ohm. Test each LED with the higher value resistors and step down until a reasonable brightness is achieved.

Nope, because you're dropping such a low voltage (0.5 volts) at such a low current (.015 A).

A Vled of 2.5v just seems a bit high to me, though; I'd suggest using 2v in your calcs. You can always step down to a lower ohms value if an LED is visibly dim.

• posted

Randy & David -- Thanks for the replies.

I'm just back from Radio Shack with a package of these to get started thinking:

20 assorted LEDs, 2 - 3V, 10-20 mA

(They were the only thing economical. Other LEDs were one or two per package, \$1.98 to \$4.98. I'll get another package when I go to get the battery holder & resistors.)

Am I reading the specs right to use Vcc = 3 V, Vled = 2.5 V and Iled = 15 mA? R = (3 - 2.5) / 0.015 = 33.33 Ohms

When I did electronics many years ago most of the resistors were in the kiloohm and megaohm range. 33 ohms seems wrong. Am I missing something?

Thanks again!

-- Mark

• posted

Thanks for the link. The last time I did any serious soldering the 80186 and 80286 were brand new. ;-)

-- Mark

• posted

Peter --

So I didn't make some kind of math error. Thanks.

-- Mark

• posted

Check this URL for the Vf of different colored LED's:

• posted

Just a thought - it is much easier to see one lit LED among many unlit ones than it is to see one unlit LED among a bunch of lit ones, so I'd wire the thing so that all lights are out when things are all properly aligned. Also, if you do it this way, there is no need to remove the battery while running the machine, and you will get a warning if anything does slip out of alignment when it shouldn't. You would, however, want to remove or switch off the battery while the thing is disassembled.

```--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca  ```
• posted

Radio Shack is far too expensive for toggle switches, as you discovered. You can get them for 40 cents each in lost of 10 from Allelectronics -

catalog # MTS-75PC

You can add the 31st LED and a simple circuit to tell you when the other 30 LEDs are lit. That way, you need glance at only 1 LED to see if all the others are on instead of needing to look at all 30 of them. You'll need 30 diodes - catalog #

1N914TR (100 for \$2.00) an NPN transistor (any NPN would work - catalog # PN2222A is a suggestion) and a couple of resistors. The single LED can serve as a "run" light. When it is lit, the battery must not be disconnected. When it goes out, the battery can be disconnected. (You could use it as the basis of an automatic battery disconnecting circuit if you want.) Here's the circuit: +3 ---+------------------} }----+--------------------+ | | | [LED1] [LEDn] [LED31] | | | [R1] [Rn] [R31] | ->|- | ->|- | +---[Diode1]---+ +---[DiodeN]---+ | | | | | | [Switch1] | [SwitchN] | | | | | | | Gnd | Gnd | | | | | +---} }-------------------+ | | | [4.7K] | | /c +----| NPN | \\e [100K] | | | +-----+ | Gnd

Use 100 ohm resistors for R1 through R31 and a 3 volt supply. (If you don't want to use batteries, you could use catalog # DCTX-330, which is a 3 volt 300 mA wall wart power supply.) That will limit current through each LED to about 10 mA, which is plenty bright enough for red LEDs. Catalog # LED-1 gets you 10 standard red LEDs for \$1.00. The catalog numbers for the resistors are 100-1/4 (100 ohm, 1/4 watt), 4.7K-1/4 and 100K-1/4 They cost 50 cents for 10 and must be ordered in lots of 10.

Ed

• posted

you're pushing the envelope there...

with a 3.0V supply and a 1.7V led drop (typical for green LEDs) and a 0.6V drop in the diode and in the be junction of the transiistor there's only 0.1V left through 4.7K that's about 20uA at 0.6v the 100K will pass 6uA leaving 14 to flow into the base of the transistor.

unless that pn2222 has an hfE around 500 that last LED could be pretty dim. and that's assuming 0.6V Vbe is sufficient

sticking a 1K resistor in parallel with each LEDn-Rn pair would be one way to fix that.

Bye. Jasen

• posted

Ed --

Thanks.

Wow, great idea. Thanks!

-- Mark

• posted

Peter --

Excellent point. Thanks.

-- Mark

• posted

Yes. I had a "window" of only roughly 3.25 to 2.25 volts. Your excellent suggestion (more on that later) fixes that nicely.

The trick was to get the single LED to light without pulling enough current in the base path to make a "main" LED (one of the first 30) glow even a tiny bit.

Not quite. The diode drops about .45 volts at that very low current.

I guess "pretty dim" is relative. On the breadboard, the LED glows merrily, and is easy to see. I used 2 10K in parallel in place of the 4.7 K, and 150 ohms in series with the "main" LED.

500 is way too high for hfe. As I recall, a 2N2222 is good for about 300, max. On the breadboard, without the 1K in parallel with the main LED, base current measures ~23 ua and LED current measures ~3.69 mA so hfe ~ 160. With the 1K in parallel with the main LED, base current is ~280 ua and LED current is ~ 7.52 mA, so gain is ~27

It fixes the "window" problem very nicely. The window problem was that the single LED would glow with the battery voltage no lower than 2.75. Below that it was too dim. And above roughly

3.25, the "main" LED would start to glow dimly. With the 1K resistor across the main LED, the circuit works down close to 2 volts, and will work without causing the main LED to glow above 6 volts. 3 volts is still a viable option, but he could use 6 volts if he changes the value of R1 through R31 to 220.
• posted

Here's an updated schematic to incorporate Jason's idea of adding a 1K resistor in parallel with each of the first

30 LEDs. It will allow the circuit to work as the batteries drain below 2.75 volts. It will also allow you to use a higher supply voltage if you want, like 3.6, 4.5 or 6 volts, but you would need to change the value of R1 through R31 using the values below: 3.6 volts use 150 ohms; 4.5 (or 4.8) volts, use 270 ohms; 6 volts, use 330 ohms.

Ed

+3 ---+-------+----------} }----+-------+------------+ | | | | | [LED1] [1K] [LEDn] [1K] [LED31] | | | | | +-------+ +-------+ | | | | [R1] [Rn] [R31] | ->|- | ->|- | +---[Diode1]---+ +---[DiodeN]---+ | | | | | | [Switch1] | [SwitchN] | | | | | | | Gnd | Gnd | | | | | +---} }-------------------+ | | | [4.7K] | | /c +----| NPN | \\e [100K] | | | +-----+ | Gnd
• posted

Ed --

Much appreciated!

-- Mark

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