Count down clock

[jldavis1969]

I want to remove the extra lights on a count down clock that I have.

Problem is when I cut the wire on the outside the clock quit working other than one of the digits turning to an "F"

There are 20 Red LED's on the outside with a resistor (100 ohms)checked via color coding on resistor (Brown Black Black Gold).

(Edit)The power supply is 8.0 VAC 650mA

Question is what size resistor would I need to replace the whole string with to make it function?

Thanks
Jeff Davis

[jldavis1969]

UPDATE:

after pulling on the string of lights I found that there are a series for four (4) Red LED's and another resistor (Yellow Brown Black Gold).

Totalling 5 sets of 4 Red LED's and the initial resistor with Brown Black Black Gold on it

[jldavis1969]

UPDATE 2:

Positive connection, 4 Red LED's in series, at last LED there is a resistor (Yellow Brown Black Gold).

There are 5 sets of the above.

[jldavis1969]

UPDATE 3:

Yellow Brown Black resistor = 410 ohms 5% tolarance

So one strand of 4 led's would look like the following

+ LED wire LED wire LED wire LED resistor (Y,BR, BL) wire (BR, BL, BL resistor)

________________________________________________________

What is the ohm rating of this circuit considering the power at the connection is 9.0 VDC

[jldavis1969]

UPDATE:

+ Wire LED- Wire- LED- Wire- LED- Wire- LED- Wire- LED
| | | | |
Wire Wire Wire Wire Wire
| | | | |
LED LED LED LED LED
| | | | |
Wire Wire Wire Wire Wire
| | | | |
LED LED LED LED LED
| | | | |
Wire Wire Wire Wire Wire
| | | | |
LED LED LED LED LED
| | | | |
R2 R2 R2 R2 R2
| | | | |
- R1 - Wire ----- Wire ----- Wire ----- Wire ----- Wire

R1 = Brown Black Black Resistor (100 ohms ?)
R2 = Yellow Brown Black Resistor (410 Ohms ?)

+ = 9.0 VDC

Hope someone can understand the diagram above.

[Steven]

jldavis1969 wrote:

R1 = Brown Black Black Resistor (100 ohms ?)
R2 = Yellow Brown Black Resistor (410 Ohms ?)

Brown Black Black is 10 ohms.
Yellow Brown Block is 41 ohms.

The 3rd band in a 4-band resistor indicates the multiplier, or how many zeros to add. Black is 0 (no zeros, or x 1).

It would be easier to understand the circuit if you posted a photo.

However, it sounds like you have several strings that each consists of 4 red LEDs in series with a 41Ω resistor. 4 red LEDs drops nearly 9v, which explains why the resistor is so small. With these tight tolerances, it's difficult to tell the exact current flowing through the LEDs, so if you can't measure it, let's assume 25mA, which is typical. Thus 2.2v (typical red LED) divided by 25mA gives 88Ω. Thus, you could short the unwanted LED and replace the resistor with a 120Ω.

Is this what you want to do?

[jldavis1969]

Steven wrote:

jldavis1969 wrote:

R1 = Brown Black Black Resistor (100 ohms ?)
R2 = Yellow Brown Black Resistor (410 Ohms ?)

Brown Black Black is 10 ohms.
Yellow Brown Block is 41 ohms.

The 3rd band in a 4-band resistor indicates the multiplier, or how many zeros to add.  Black is 0 (no zeros, or x 1).

It would be easier to understand the circuit if you posted a photo.

However, it sounds like you have several strings that each consists of 4 red LEDs in series with a 41Ω resistor.  4 red LEDs drops nearly 9v, which explains why the resistor is so small.  With these tight tolerances, it's difficult to tell the exact current flowing through the LEDs, so if you can't measure it, let's assume 25mA, which is typical.  Thus 2.2v (typical red LED) divided by 25mA gives 88Ω.  Thus, you could short the unwanted LED and replace the resistor with a 120Ω.

Is this what you want to do?

Yes replacing the whole string is what I want to do.

There are 5 parellel strings consisting of 4 Red LEDs and resistor in series. plus the 10 ohm resistor.

[jldavis1969]

Will the powers that be the moderators here on this forum Kindly Close this thread for me. I am done with this project.

Thank you

[Don]

Closed at the request of thread creator.