Basic Led Circuit not lighting

[wizman]

Howdy all,

I've made a basic LED circuit based off of this schematic:

And in physical form as:

I have tested every single diode for function and polarity, as well as the wires and components for continuity.
Prior to soldering the circuit I tested it on a breadboard, and of course it worked. Now when I apply 9V none of
the LED's light and I'm wondering what I could be missing. Any thoughts would be much appreciated.

Thanks, Wizman

 

[std_match]

If you don't have a multimeter, you should buy one now. Even the cheapest one is enough for this troubleshooting.
It is better for your learning to measure voltages and try to logically find explanations, than to measure resistance to verify that the connections are correct.
As an example, say that one of the LED's has an internal open circuit. You can quickly find this with voltage measurements, but resistance meaurements of the wires/connections will not help.

 

[wizman]

I used a multimeter on the diode test setting to test each LED and they all lit up and showed resistance. How would I test the circuit through voltage measurements?

 

[std_match]

Connect the multimeter "-" to the negative battery terminal. Measure 9V on the battery positive terminal. Then move the multimeter "+" one "step" at a time to the diodes, the resistor etc. When you see a sudden jump to 0V, you have just passed the problem.
On one side of the problem there should be about 9V, on the other side about 0V.

 

[wizman]

well that might be my issue sorted right there! I measured voltage of the battery it only read 4.6V. Is it normal for cheap batteries to have unreliable voltages, or does the voltage output decrease as the battery is used?

edit: the battery was fine. Weirdly I measured it again (read 9V) and plugged in and i have light? Not sure what is going on

 

[dr pepper]

Quickly short each led legs with a screwdriver, if doing this on all has no effect you might have them all in the wrong way round, quickly try reversing the battery.

 

[wizman]

I sorted out the last circuit but the LED's weren't bright enough. I bought a set of LEDs, however they are incredibly dull despite following this calculator: https://www.amplifiedparts.com/tech-articles/led-parallel-series-calculator . Here is an example of some circuits. The Circuit 6 LED's used 150R resistors, however i gradually reduced this resistance to 0 without any effect. The single LED circuit uses a 330R resistor however the results were actually worse! The light is barely noticeable and they are supposed to be 3000mcd. Am i missing something obvious?

 

[KlausST]

Hi,

or does the voltage output decrease as the battery is used?

For sure the voltage goes down on old/used/drained out batteries.

Replace the battery.

The voltage of a new battery shoud be obove 9V. Maybe 9.2V, maybe 9.5V.
during operation the voltage goes down slowly until they are almost drained out.
Then below 8.5V the battery is almost dead and the voltage will break down much faster.

But you have a multimeter. Just use it and measure voltages.
If you calcultate your resistors for 9V .... you shouldn´t be surprised if the LEDs have low brightness on (real) 8V or below.

and.. measure the battery voltage when the LEDs are connected.

Klaus

 

[wizman]

Hi,

For sure the voltage goes down on old/used/drained out batteries.

Replace the battery.

The voltage of a new battery shoud be obove 9V. Maybe 9.2V, maybe 9.5V.
during operation the voltage goes down slowly until they are almost drained out.
Then below 8.5V the battery is almost dead and the voltage will break down much faster.

But you have a multimeter. Just use it and measure voltages.
If you calcultate your resistors for 9V .... you shouldn´t be surprised if the LEDs have low brightness on (real) 8V or below.

Klaus

You are absolutely right the voltage is around 8.2V. Is the reason that LED lights so dimly due to the lack of 'force' so to speak that the voltage pushes the current through the circuit?

 

[stenzer]

Hi,

if the battery voltage decreases, the voltage dropping at your resistor also drops wich determines the current through your LEDs by

\[I_{f,0} = \frac{V_{R,0}}{R_{0}}\],

where \[V_{R,0}\] is the voltage drop at \[R_{0}\], which is given by

\[V_{R,0} = V_{Battery} - 3 \cdot V_{f}\].

This leads to an reduction of the LED the brightness.

BR

 

[KlausST]

Hi,

absolutely.

The LED voltage is very critical. Dropping the LED voltage (mind: not the battery voltga) by 0.2V may make the brightness go down to almost zero.
Increasing the LED voltage by 0.2V may make the LED to be killed.
LED brightness is very non linear to the applied voltage. But it is rather linear to the applied current.

Thus LED´s need to be "current driven". In simplest case a series resistor is used to keep the LED current in the desired range.

Klaus

 

[stenzer]

Hi,

LED brightness is very non linear to the applied voltage.

have a look at the \[V_f\] vs. \[I_f\] curve [1] of a LED to see the influence of the forward voltage.

[1] https://lednique.com/current-voltage-relationships/iv-curves/

BR

 

[std_match]

The correct solution in this case isn't to replace the battery (but it may be necessary anyway).
6 of your LEDs in series is too much for a 9V battery.
A circuit designed for a 9V battery should work down to 7.2V or lower (1.2V per internal cell):

https://www.silabs.com/documents/public/application-notes/AN834.pdf

You need a certain voltage drop over the series resistor to keep the current reasonably constant when the battery voltage decreases from 9V to 7.2V
A more advanced solution is to replace the resistor with a constant current regulator.

 

[KlausST]

Hi,

6 of your LEDs in series is too much for a 9V battery.

I don´t see 6 in series...I just see 3 in series
But we don´t have any technical information about the LEDs... Without a datasheet it´s just guessing...

Klaus