LED Project... Is the diagram correct?

Hello this is the final Diagram, is there anything in wrong place or missing? Audio i also need to know what is the vaule are the resistors and capcitors in this diagram. I guess that might depend on the type of LED's i get i will provide the specs for those shortly. But just check this out if its correct.
Thank you
Max

I dont think you need those current limiting resistors for the LED's.

jrz126 said:

I dont think you need those current limiting resistors for the LED's.

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why not? wouldn't the LEDD's burn out quick

Hi!

The LM3914 have internal resistances included. You have a pin to select the amount of current you want for the leds.

JFDuval said:

Hi!

The LM3914 have internal resistances included. You have a pin to select the amount of current you want for the leds.

Click to expand...

I want about 20mA to 25mA but nothing less then 20mA
what pin is it to select the resistances?
i am not good at electronics, i am a total beginner. The person who made this schematic for me is AUDIOGURU.... he give me value for the capacitor for 555timer and resistors.
here are specs for the LED's
Emitted Colour : WHITE
Size (mm) : 5mm T1 3/4
Lens Colour : Water Clear
Peak Wave Length (nm) : N/A
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 10000
Life Rating : 100,000 Hours
Viewing Angle : 20 ~ 25 Degree

Absolute Maximum Ratings (Ta=25°C)

Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
Lead Soldering Temperature : 240°C (<5Sec)
Operating Temperature Range : -25°C ~ +85°C
Preservative Temperature Range : -30°C ~ +100°C
thanks
Max

looking at the datasheet, it says that Iled is approx. equal to 12.5/R1 where R1 is your 620 ohm resistor. this comes out to about 20 mA

Hi!

I can't tell you wich pin by heart but just read the datasheet and you will find your answer.

Hi Guys,
Of course you need those current-limiting resistors. Without them, the LM3914 will be thermally stressed with 20mA thru 6 LEDS at the end of each scan, and a supply voltage of 14.4V. The chip's temperature will go from cool to very hot for each scan, which isn't good. The current-limiting resistors aren't needed for just "dot" mode.

Max, I am sorry that I didn't finish the 555 schematic, but its pin 7 (discharge) should be used to discharge the 1uF cap, not your diode from its output which does not go anywhere near ground.
I'll arrange it so that the discharge happens quickly then you won't see the LEDs run back down.

Hello!

I didn't saw the high voltage supply. Sry,

audioguru said:

Hi Guys,
Of course you need those current-limiting resistors. Without them, the LM3914 will be thermally stressed with 20mA thru 6 LEDS at the end of each scan, and a supply voltage of 14.4V. The chip's temperature will go from cool to very hot for each scan, which isn't good. The current-limiting resistors aren't needed for just "dot" mode.

Max, I am sorry that I didn't finish the 555 schematic, but its pin 7 (discharge) should be used to discharge the 1uF cap, not your diode from its output which does not go anywhere near ground.
I'll arrange it so that the discharge happens quickly then you won't see the LEDs run back down.

Click to expand...

would that work? Should pin 7 be used, how much current can it take?
Does it depend on who makes the IC555?
and if it does what make should i use for this?
here is the updated

Hi Max,
Good, now you have a circuit that will basically scan the LEDs linearly when the power is applied. Only a few adjustments are needed:
1) The 1st scan will begin with a delay before the scan (longer than remaining scans) as the cap in the 555 circuit charges up to its operating voltage level. I am still working on a solution.
2) The reset time needs to be shortened with a diode and resistor or just a diode for the 555 because now the reset time equals the scan time (it will scan for 1 second then be off for 1 second, then scan again etc.)

Don't worry about the 555's pin 7 ratings, it is normally used to discharge the 555's timing cap which could be huge and is very stong. National Semi has a note about it in their datasheet that says the only problem is if it causes the 555 to get too hot. It will not even get warm by occasionally discharging your puny 1uF cap. Here's what they say:

Note 7: No protaction against excessive pin 7 current is necessary providing the package dissipation rating will not be exceeded.

audioguru said:

Hi Max,
Good, now you have a circuit that will basically scan the LEDs linearly when the power is applied. Only a few adjustments are needed:
1) The 1st scan will begin with a delay before the scan (longer than remaining scans) as the cap in the 555 circuit charges up to its operating voltage level. I am still working on a solution.
2) The reset time needs to be shortened with a diode and resistor or just a diode for the 555 because now the reset time equals the scan time (it will scan for 1 second then be off for 1 second, then scan again etc.)

Don't worry about the 555's pin 7 ratings, it is normally used to discharge the 555's timing cap which could be huge and is very stong. National Semi has a note about it in their datasheet that says the only problem is if it causes the 555 to get too hot. It will not even get warm by occasionally discharging your puny 1uF cap. Here's what they say:

Note 7: No protaction against excessive pin 7 current is necessary providing the package dissipation rating will not be exceeded.

Click to expand...

To shorten reset time where would the resistor need to go and what value?

Maxer said:

audioguru said:

Hi Max,
Good, now you have a circuit that will basically scan the LEDs linearly when the power is applied. Only a few adjustments are needed:
1) The 1st scan will begin with a delay before the scan (longer than remaining scans) as the cap in the 555 circuit charges up to its operating voltage level. I am still working on a solution.
2) The reset time needs to be shortened with a diode and resistor or just a diode for the 555 because now the reset time equals the scan time (it will scan for 1 second then be off for 1 second, then scan again etc.)

Don't worry about the 555's pin 7 ratings, it is normally used to discharge the 555's timing cap which could be huge and is very stong. National Semi has a note about it in their datasheet that says the only problem is if it causes the 555 to get too hot. It will not even get warm by occasionally discharging your puny 1uF cap. Here's what they say:

Note 7: No protaction against excessive pin 7 current is necessary providing the package dissipation rating will not be exceeded.

Click to expand...

To shorten reset time where would the resistor need to go and what value?

Click to expand...

or would this work..

Hi Max,
That's a good idea that you have. There are many ways to connect a 555 but your newest one just might work well. It uses the 1uF cap and current source as the timing parts. The reset pin discharges the cap as fast as it can.
Its problem is that the LM3914's resistor divider between its pins 4 and 6 needs to be set to near the same voltage range as the divider in the 555. You can't simply connect them in parallel because the LM3914's divider resistance is too low. It would need to have its actual resistance measured with an ohmmeter then slightly lower value resistors added from pin 4 to ground and pin 6 to the positive supply. Like my attachment here.

The only problem remaining is the delay on the 1st scan. I am narrowing my search to 555 queries on another web-forum.

audioguru said:

Hi Max,
That's a good idea that you have. There are many ways to connect a 555 but your newest one just might work well. It uses the 1uF cap and current source as the timing parts. The reset pin discharges the cap as fast as it can.
Its problem is that the LM3914's resistor divider between its pins 4 and 6 needs to be set to near the same voltage range as the divider in the 555. You can't simply connect them in parallel because the LM3914's divider resistance is too low. It would need to have its actual resistance measured with an ohmmeter then slightly lower value resistors added from pin 4 to ground and pin 6 to the positive supply. Like my attachment here.

The only problem remaining is the delay on the 1st scan. I am narrowing my search to 555 queries on another web-forum.

Click to expand...

hey AUdio thanks for helping once again..
Does it matter what company i get the LM3914 and 555 timer from? are they all identical?
i am getting all of my stuff from https://www.futurlec.com/cgi-bin/search/search.cgi
i am going to go ahead and order everything i need for 3 circuts.. 1 for practice and 2 for the real deal.. if you think would need anything else besides whats on the schematic please let me know
thank you
max

Hi Max,
Most parts with a dedicated part number are the same from various semiconductor manufacturers, so that companies that make stuff can get "second sources" when their favourite manufacturer runs out of stock. However, some oriental semi manufacturers make an "improved" copy, sometimes with disasterous results! (New Japan Radio's NJM072 copy of a TL072 has "improved" extra wide bandwidth. But it oscillates like crazy and cannot be used in low-gain circuits. They quickly replaced it with their recommended NJM072B).

I would choose National Semi's LM3914, because they invented the thingy and have been good to me. It may cost a little more than ON Semi, Fairchild or others.
I would also order National Semi's LM555.

I haven't yet calculated the resistors in the current source. I will do so soon.

audioguru said:

Hi Max,
That's a good idea that you have. There are many ways to connect a 555 but your newest one just might work well. It uses the 1uF cap and current source as the timing parts. The reset pin discharges the cap as fast as it can.
Its problem is that the LM3914's resistor divider between its pins 4 and 6 needs to be set to near the same voltage range as the divider in the 555. You can't simply connect them in parallel because the LM3914's divider resistance is too low. It would need to have its actual resistance measured with an ohmmeter then slightly lower value resistors added from pin 4 to ground and pin 6 to the positive supply. Like my attachment here.

Click to expand...

i didn't quit understand this statement... there is no resistor in IC555 circuit.... I think you may have changed you original circut because i don't know what resistor you are talking about in LM3914 either.
but will the internal delay of 555 will discharge cap completely? if not we have to add cap and resistor to pin 2 and 6 and then ground? to add a delay of our own.. or will the internal delay of IC555 will work?

Also there are different types of LM555 and LM3914 IC from Semi national here is the list pick 2 so i can pick the correct once. thanks
**broken link removed**
**broken link removed**

Hi Max,
Clarification:
1) Of course the ICs have resistors inside. The LM3914 has a bunch of resistors between pins 4 and 6 that determine what input voltage lights its LEDs, and the 555 has resistors that determine the lowest and highest voltages of its timing cap.
2) Of course the circuit has changed from my original one, now the 1uF cap is the 555's timing cap, which never discharges to ground like it did in my circuit.
3) Inside the LM3914 between pins 4 and 6 are 10 resistors in series, that are used as a voltage divider. This voltage divider determines what input voltage turns on the LEDs. Now that the 1uF cap is used as the 555's timing cap, the cap's voltage will range from 1/3 of the supply voltage for its discharged voltage, to 2/3 of the supply voltage for its highest voltage. The 555's 1/3 and 2/3 of the supply voltage reference points are determined by a voltage divider of 3 resistors inside the 555 that are connected between the supply and ground.
4) We must duplicate the 555's voltage divider with another voltage divider for the LM3914 so that the 1st LED lights when the cap's voltage (input to the LM3914) is 1/3 of the supply voltage (the cap is discharged by the 555) and the 6th LED lights when the cap's voltage is a little less than 2/3 of the supply voltage (cap is at its highest voltage). It would be a lot easier to calculate voltages if the 555's cap voltage simply ranged from ground to a fixed voltage. It would also be a lot easier if you had 10 LEDs instead of only 6.
5) The 1st scan is longer because the cap must charge from zero volts (fully discharged from the circuit being unpowered) to 1/3 of the supply voltage before the timing actually begins. The LEDs will be off during that time.
6) You must measure the LM3914's voltage divider total resistance between pins 4 and 6 because the datasheet says that it could be anywhere from 8K to 17K ohms. We must know exactly what it is.
IC's are lousy at making accurate value resistors, but can make very accurate duplicates and ratios that are used in "normal" LM3914 and 555 circuits.

BTW, the emitter of the 2N3906 current source and its resistors must now connect to the supply voltage, instead of to the LM3914's pin 7. Pin 6 must also be disconnected from pin 7. Pin 4 must be disconnected from ground.

Hi Max,
You need an LM555CN and an LM3914N-1.
They are normal size plastic case DIL ICs for soldering their pins through holes in a circuit board.
The other packages are metal or ceramic for military or outer-space uses, are bare chips or are tiny little things which are usually soldered on top of a circuit board (surface-mount) by robots.