First post here, and I'm new to making circuits of any kind. Attached is my proposed circuit I have 9v to a spst swx then a 50 Ω resistor, 5 -2v 20mA LEDs another 50Ω resistor to ground. I don't believe the resistors are helping me because even without them I have a 10V drop across all 5 LEDs but only 9V avail. In real life the LEDs will light up just not at 100% which is fine but my question is how hard of a hit does the battery take when the load is beyond its capability? Should I take out the resistors? Size is an issue so adding a larger power supply is not an option unless a 12V A23 battery will do the job but from what I've read they don't seem to have much capacity. Any help would be appreciated.

 

Firstly, why are you using two resistors, one before and one after the five LEDs?
The current doesn't care where you put the resistors. The current still has to go through both resistors.
One 100Ω resistor is the same as two 50Ω resistors placed in series with the LEDs.

Connect the circuit with the two 50Ω resistors. Measure the voltage across one of the resistors. Observe the brightness of the LEDs.

Remove one of the resistors and repeat the above measurement and observation.

Remove both resistors and observer the brightness of the LEDs.

Welcome to AAC!

 

they may not turn on.
look at their datasheet for voltage/current graph.
i would use 4 LEDs per string.
or if you must use 5 LEDs, divide them in strings of 3 and 2 or 4 and 1 with appropriate resistors.

 

Firstly, why are you using two resistors, one before and one after the five LEDs?
The current doesn't care where you put the resistors. The current still has to go through both resistors.
One 100Ω resistor is the same as two 50Ω resistors placed in series with the LEDs.

Connect the circuit with the two 50Ω resistors. Measure the voltage across one of the resistors. Observe the brightness of the LEDs.

Remove one of the resistors and repeat the above measurement and observation.

Remove both resistors and observer the brightness of the LEDs.

Welcome to AAC!

Thank you MrChips I used the resistors on each end of the LEDs for symmetry because in my mockup they are visible.
I'll try to systematically remove the resistors and see how it goes.

they may not turn on.
look at their datasheet for voltage/current graph.
i would use 4 LEDs per string.
or if you must use 5 LEDs, divide them in strings of 3 and 2 or 4 and 1 with appropriate resistors.

Unfortunately, I can't divide them up, all 5 have to be in series, see attached photo clarification.

 

I have put you picture on display to make the post more clear.

My prediction: The LEDs will be dimmer. but the resistors are needed to prevent burn out. With small batteries resister are probably not even needed,

LEDs, 555s, Flashers, and Light Chasers

 

Just put 2 9V batteries in series. You will have plenty of voltage.

 

I have put you picture on display to make the post more clear.

My prediction: The LEDs will be dimmer. but the resistors are needed to prevent burn out. With small batteries resister are probably not even needed,

LEDs, 555s, Flashers, and Light Chasers

Cool, thanks for your help, I actually want the LEDs to be less than 100% bright so the thing isn't annoying. I am curious if the resistors help "protect" the LEDs even if they are under powered?

Just put 2 9V batteries in series. You will have plenty of voltage.

The enclosure I plan to make, that will have the swx and 9v holder, will not have enough room for two 9Vs, but I appreciate the suggestion.

 

Yes the resistors will help. Batteries come in many sizes &voltages. Lets see 5 x 2,.5=12.5 Vf That should be your minimum power supply voltage.

 

Of course a joule thief will boost the voltage nicely they would be bright then.

 

I try to simulate but it didn't work with the red LED even I connected them directly to the 9V battery.

Here's a quote from techlib.com

A circuit that draws 10 ma powered by a 9 volt rectangular battery will operate about 50 hours: 500 mAh /10 mA = 50 hours The cell voltage of alkaline cells steadily drops with usage from 1.54 volts to about 1 volt when discharged. The voltage is near 1.25 volts at the 50% discharge point. Alkaline cells exhibit a slightly increased capacity when warmed and the capacity drops significantly at temperatures below freezing.

If it works and you limit the current to <10mA, theoretically it can work for 50 hours. But I doubt it can last for even 10 hours, as the voltage would drop to a point the LED would not conduct anymore. The new super bright LEDs use very little current and it should be able to stand longer.

Allen

 

Nobody makes a 2V LED, LEDs have a range of voltage even if they have the same part number and you get whatever voltage they have. Some of your red LEDs might be 1.8V and burn out soon without a current-limiting resistor and other LEDs might be 2.2V and not light with a 9V supply (5 x 2.2V= 11V). You are simply lucky that your LEDs do not burn out or light up without a resistor and with only 9V. But the battery is 9V for only a short time and you will probably see the LEDs dim as the battery voltage runs down.

You need a battery with a voltage high enough that the LEDs still produce light when the battery is replaced and a resistor so that the LEDs do not burn out.

 

Of course a joule thief will boost the voltage nicely they would be bright then.

A joule thief sounds very interesting, it would be cool if I could make one to fit in the space provided and jump it up a couple of volts.

I try to simulate but it didn't work with the red LED even I connected them directly to the 9V battery.
If it works and you limit the current to <10mA, theoretically it can work for 50 hours. But I doubt it can last for even 10 hours, as the voltage would drop to a point the LED would not conduct anymore. The new super bright LEDs use very little current and it should be able to stand longer.
Allen

When I ran the circuit on the Everycircuit app it didn't work either but it does work in real life.... go figure.

Nobody makes a 2V LED, LEDs have a range of voltage even if they have the same part number and you get whatever voltage they have. Some of your red LEDs might be 1.8V and burn out soon without a current-limiting resistor and other LEDs might be 2.2V and not light with a 9V supply (5 x 2.2V= 11V). You are simply lucky that your LEDs do not burn out or light up without a resistor and with only 9V. But the battery is 9V for only a short time and you will probably see the LEDs dim as the battery voltage runs down.

You need a battery with a voltage high enough that the LEDs still produce light when the battery is replaced and a resistor so that the LEDs do not burn out.

I'd rather be changing batteries than LEDs so hopefully, the battery will drain at an acceptable rate (perhaps faster than ideal, but still acceptable) and the LED's will be safe.

 

May be you would like this circuit?
Calculate resistors for current 10 mA, or even 5 mA. Brightness of luminodiodes will be enough.

 

According to someone who is knowledgeable, I think, for 5 LEDs in a string, and 2V per LED, the supply voltage should not be less than 12.5V. This is the minimum to insure that current drawn is predictable and the lighting is stable. That the intensity of the LEDs is stable is probably not that important for your application.

If 10 mA is the current through the string, that means that the resistance of the current limiting resistor should be (12.5V- 10V)/ 10 mA equals 250 Ohms.

Where the voltage supply is a battery, the voltage of the battery fully charged or new off the shelf would preferably be higher than 12.5V as the voltage of all batteries droop to some extent corresponding to the total amount of time that the circuit is operated.

 

You may find this helpful...

http://www.quickar.com/bestledcalc.php?session=

 

"Nobody makes a 2V LED" Well, check the voltage of high brightness red and green LEDs, some indeed are 2V.
But to the TS problem: as explained you must limit the current in your LEDs or they will burn out. So if one of your LEDs does not light on a 9V battery it probably has. On the other hand typical red LEDs have voltages of 1.6-1.8V, so you will need at least 8V and maybe 10V to make them light, if they are in series. Most LEDs will light at quite low currents, so you don't need to run them at 10mA if you only need to run them at 5mA and let the battery run down as the current will drop a lot as the voltage falls, and so a battery could light them for quite a while.
You might want to use a 12v battery instead of 9V and then a 1k resistor will limit the current to 2 to 4mA (470 ohms if you want more).
From the resistor colours in your picture I think they are 100 ohms, not 50.
I've run 4 LEDs from a single NIMH cell using a very simple converter before now. If you want to run them from 9V a very simple converter would control the power and flash them on and off quickly enough not to notice.

 

"Nobody makes a 2V LED" Well, check the voltage of high brightness red and green LEDs, some indeed are 2V.

Some people wrongly believe that a red or old green LED rated at a typical 2V is a 2V light bulb made with an accurate length and thickness of tungsten wire so all 2V Leds should be the same and work fine if connected to 2V like a 2V light bulb does.
But an LED is not accurately 2V, it has a range of voltages that could be 1.8V to 2.2V and the actual voltage cannot be bought, you get whatever is there, unmarked. In addition to the range of voltages is that an LED is a diode and if it is fed a voltage a little higher than its actual forward voltage then the current can be massive and it will soon burn out. If the voltage is a little less than its actual forward voltage then its current will be very low or nothing and it will be dim or not produce any light. That is why the supply voltage must be higher than the LED voltage so that the current-limiting resistor has enough voltage across it to reduce the wide range of current.

 

Thank you all for the input. When I put the resistor in I might use a single 250 Ω or something close if I have one.

I'd love to use a 12V battery, I just don't know that they make one that will fit in the area provided, as the box that the battery will be stored in is only slightly longer than the battery itself, and maybe 3x as wide, but I also have to get my stripboard in there too so I don't have tons of room. If someone knows of a 12v that isn't much larger than a 9v please point me in that direction.

 

Is there an off the shelf converter that will take me from 9 to 12v?

 

Is there an off the shelf converter that will take me from 9 to 12v?

It is called a boost converter. They are available on Ebay, But when voltage goes up current goes down, not to mention the current consumed by the boost convertor. Your little 9v battery won't last long.

As we say you can't get blood out of a stone.