I am planning to light up 178 LEDs (3.4V-3.8V, 25MA) using a 18V/1A Adapter (DC radio shack transformer) and I need a little help.
Power Adapter says that is 18 volt but when I test it with volt meter it shows 22V. When I connect a string of 6 LEDs (plus small resistor) with it voltage is dropping to 21.4 (and shows 9.4ma current)
I don't have problem calculating my resistors (I am using
formatting link
but it seems my 18V (or 22V) Adapter is not really what is says (it drops).
My question is what should I put in that web site as my Source Voltage? (22V? 18V? or something else).
I am planning to light up 178 LEDs (3.4V-3.8V, 25MA) using a 18V/1A Adapter (DC radio shack transformer) and I need a little help.
Power Adapter says that is 18 volt but when I test it with volt meter it shows 22V. When I connect a string of 6 LEDs (plus small resistor) with it voltage is dropping to 21.4 (and shows 9.4ma current)
I don't have problem calculating my resistors (I am using
formatting link
but it seems my 18V (or 22V) Adapter is not really what is says (it drops).
My question is what should I put in that web site as my Source Voltage? (22V? 18V? or something else).
A fair approximation for your supply (based on both the rated voltage and your experimental results) would be a 22 volt source that includes a 4 ohm series resistor. When you connect an 18 ohm resistor across it, the 4 ohm internal resistance will drop 4 volts and the 18 ohm resistor resistor will drop 18 volts.
Include that internal 4 ohm resistance in your design calculations.
#1 Newsgroup Service in the World!
120,000+ Newsgroups
Thanks but what do you mean by unregulated adapters? And also if the rating is the voltage at full load then how do we explain the Voltage Drop with 6 LEDs (that is far less than 1A for sure)?
#1 Newsgroup Service in the World!
120,000+ Newsgroups
Some "wall warts" contain internal voltage regulators that keep the output voltage very close to the stated voltage on the adapter. If yours were regulated, it would deliver 18V, regardless of load, up to its rated current output of
1 ampere.
The unregulated variety have a non-negligible internal resistance that causes the output voltage to drop as the load increases. In a rare gesture of honesty, the manufacturer actually promises the rated voltage at full load.
Well, the greater the external load, the greater the voltage that will be developed across the internal resistance. This voltage will be subtracted from the output voltage.
The voltage drop follows Ohm's law, using the internal resistance and the load current.
John has calculated the internal resistance to be four ohms using (22V -
18V)/1A. That assumes a linear internal resistance but the 0.6V drop with a current of 0.0094A suggests an internal resistance of ~63 ohms which would never allow 18 volts output at 1 A. So maybe it is non-linear (at least over part of the output current range) or defective.
It can get complicated to work with an unregulated supply. For less than a dollar, you can add a simple IC voltage regulator and some capacitors and have your output voltage stay within 50mV of
18V at any load up to 1A.
Chuck
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
formatting link
The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Circa 17 May 2007 06:37:52 -0700 recorded as looks like Homer sounds like:
Snipped the discussion about the unregulated voltage adapter, as that question seems to have been well covered. I have a different concern with your project.
If you take the 25mA as a typical 'on' value for each LED, and you want to use 178 of them, then you must calculate the total typical power that will be consumed by the circuit. That number is 178 * 3.8V * 0.025A = 16.91W. You are on the ragged edge of the ability of your supply (18W rated) to deliver the load, and have not yet factored in the power that will be consumed by the current limiting resistors, one of which will be required in each parallel leg of the circuit.
Furthermore, the design is limited to using no more than 40 parallel legs, i.e. 0.025A * 40 = 1A. 178 / 40 = 4.45, meaning that you will have to have half of your parallel legs with four LED's and half with five. 5 * 3.5V =
17.5V, so again you are topping out your design specs without counting the voltage drops of the limiting resistors.
In short, I think you need more source power or less load consumption. You've got no "wiggle room."
Circa 17 May 2007 06:37:52 -0700 recorded as looks like Homer sounds like:
Oops. You do have a problem. See my other post (not crossposted) in s.e.b:
Message-ID:
First, the "wizard" neglects to take into account your power source max current in its automatic design. With 44 legs at typical 25mA, you've exceeded your 1A limit by ten percent.
To give it credit, the "wizard" did give you this information at the end (did you read it?):
# together, all resistors dissipate 3593.75 mW # together, the diodes dissipate 16910 mW # total power dissipated by the array is 20503.75 mW # the array draws current of 1125 mA from the source.
So, by a couple means of calculation, you see that your 1A source just doesn't cut the mustard.
Thanks All for your inputs. Hearing all those bad things about not- regulated power supply, I decided to use a regulated (old Laptop Adapter) this time. It's 24V/1.7A and it should be sufficient for my project. I checked it already and volt meter shows exactly 24.1V. I will put 24V into wizard and go from there.
You need a different supply, or you need to run the LEDs below 25 mA. Try an experiment: put 5 LEDs in series with a 51 ohm resistor and put that in parallel with a 30 ohm resistance made from six 5 ohm, 5 watt resistors in series. Connect that circuit to your supply, measure the voltage, and observe the brightness. That experiment will approximate
35 strings of 5 LEDs + 51 ohm resistor for each string, and an additional string of 3 LEDs and a 390 ohm series resistor.
If the LEDs light up to your satisfaction, you're home free. Wire them up in strings of 5 in series, each string getting its own 51 ohm resistor in series. The last string of 3 is wired in series with a 390 ohm resistor. The 30 ohm resistance is not used - it is replaced by the other LED strings. Each string draws about 20 mA, for a total of .72 amps, so you should be good to go with that supply.
If you _must_ have ~25 mA, you need a different supply. I'd recommend you use a cat # 15737 PS ($7.95) and 15447-CB ($1.25) from MPJA, and qty 25 plus qty 5 of part # 512-LM317LZ from Mouser.
formatting link
and
formatting link
In lots of 25 they're 20 cents each, for $5.00, and the other 5 are 23 cents each for $1.15 Then do this for each string:
You'll do 29 strings of 6 LEDs in series, with 4 LEDs left over. The 4 LED string is wired exactly the same as the 6 LED strings, except that it is 4 LEDs, not 6.
The LM317 and 51 ohm resistor regulate the current to
Ed, I am lost. Are you commenting on not-regulated idea (my initial question) or 24V/1.7A regulated adapter? If it's the second one,
formatting link
is suggesting that I have more than enough current:
The wizard says: In solution 0:
each 8.2 ohm resistor dissipates 5.125 mW * the wizard thinks 1/4W resistors are fine for your application Help * the 560 ohm resistor dissipates 350 mW * the wizard thinks 1W resistors are needed for your application Help * together, all resistors dissipate 478.125 mW * together, the diodes dissipate 15130 mW * total power dissipated by the array is 15608.125 mW * the array draws current of 650 mA from the source.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.