Beginner question - please help

So I have been playing with this string of 25 lights for hours, trying to figure out how to light all of them up with one 9v battery but I dont understand how to. I have a square wooden frame and I want the lights to go around the frame.

I'm using a 1/4 watt resistor for every 4 led lights. (they are all white I just colored some of the leds red)



Every light is a 5mm white

5mm (T-1 3/4) LEDs and some 1/4-watt resistors for making connections.
In water clear casing/lens. Emits focused, ultra bright white light.
Great for electronic and electrical experiments
Forward Voltage: 3.0-3.2V

I understand the FV is 3v per led but how can I get all lights to light up? DO I have to buy different lights?

Thank you

@SunnySkyguy Thank you for the very detailed explanation.

However, this is what I am trying to achieve in my project. I'm unfamiliar with wiring LEDS in parallel I only know series. Based on the details you gave me I dont think the below schematic is possible unless I get a much higher voltage battery. Am I correct?

It looks like you connected four white LEDs in series. Then you need a battery voltage that is 3V + 3V + 3V + 3V + maybe 3V for the resistor= 15V. Obviously a 9V battery will not work.

You never said what is the resistance of the resistor nor how you calculated its required resistance.

Since a little 9V battery drops to 6V or 7V during its life then why not connect two LEDs in series and in series with a resistor?
When the battery is new then its voltage is about 9.6V then two LEDs need 6V and the resistor will have the remaining voltage which is 9.6V - 6V= 3.6V. LEDs are rated at 20mA which is nice and bright so the resistance should be 3.6V/20mA= 180 ohms. Its power rating should be more than 3.6V x 20mA= 0.072W so a 1/4W resistor is fine.
Then you will have 12 sets of two LEDs, each set with its own resistor.

If you want one more LED then its resistor resistance should be (9.6V - 3V)/20mA= 380 ohms which is not a standard value. Use 390 ohms.

The current in the poor little 9V battery will be (12 x 20mA) + 20mA= 260mA. The datasheet for a 9V alkaline battery shows that with a current of 260mA its voltage will be so low that the dual LEDs won't light anymore after about 2 hours. The LEDs will be slowly dimming during the 2 hours.
The single LED will dim less and will still be pretty bright when the dual LEDs do not light anymore.

No Bob,
24 series means 24x3V=72V

You need two wires V+ & V- to run around the strings to supply the 3 or 4 in series strings on this bus tapped out to each string +\- .

The pair feeds power to all circuits in parallel!!!


The closer you match the supply to the load string voltage ,the more efficient it becomes with a low resistor value, as I designed for you.

Your sketch of a picture frame wrongly shows the 9V battery directly connected to one LED. DO NOT DOO DAT! It will cause the LED to instantly burn out.

Look at the datasheet of a 9V alkaline battery. When it is loaded it shows its voltage dropping fairly quickly to about 7.5V then slower to about 7V then quickly again to 5V. Then you cannot use three LEDs in series unless you use a 12V AC/DC power supply. Use two LEDs in series and in series with a 180 ohm resistor and power 12 of these circuits with a 9V battery.