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PopularAltering a LED Flasher Circuit? - Will it support MANY LEDs???

Discussion in 'LEDs and Optoelectronics' started by auslander, Jul 18, 2013.

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  1. auslander


    Jul 18, 2013
    Hello All,

    I have been trolling the forums for about a week or so and decided on a good first post. Im still new to building circuits, but have a little experience when it comes to rigging up LEDs. I finally decided to get around to making a flashing circuit. I built one, but I think I am seeing a problem that will arise as I make changes. Let me explain:

    I have built this circuit...I have seen variations of this circuit around. some have different resister values and another curcuit that I have as a kit I bought also includes two potentiometers. Anywho, this is the specific circuit using a single 9 v battery: is the very first circuit you see on the page)

    It works great for the 2 LEDs it shows you to use. My situation is that I am using much more than just two...In fact, on a project I am wanting to work on, I will probably be operating more like 200 LEDs off of the circuit..
    (using may different LEDs in series to equal 9 volts)

    Would someone be able to help me understand what I may need to alter with this particular circuit in order to be able to drive so many LEDs?

    If any additional info is needed, feel free to ask and I will answer to the best of my abilities. Im still a n00b when it comes to circuitry. :D
  2. GreenGiant


    Feb 9, 2012
    The problem you will run into with this is the 2 LED's in the circuit are drawing somewhere around 12mA each, accounting for some loss and the power consumed by the rest of the circuit you can reasonably say that 30mA is constantly being drawn from it.

    Increasing the number of LED's (putting them in parallel) is going to directly increase the current, 11mA per LED or so depending on what resistors you pair with them, after a certain point the battery will start dying in a matter of seconds (200 LED's @ 11mA, accounting for only half being on at a time is 1.1A give or take, thats more than most 9V batteries can output.

    You would want to use a power supply that is capable of AT LEAST 1.5A, you can use a 5V supply as long as the LED's will run at 5V (the most common ones will) you can put LED's in series (I would say 2 or 3, you can try 4 or 5 but then youll start getting dimmer or have no voltage at the end) now depending on how many you have you need to recalculate the resistor value so you end up with 11-20mA accounting for the voltage drop acrss the LED.
  3. duke37


    Jan 9, 2011
    That circuit is a bit dubious, at 9V it could bias the transistors negatively more than they are can deal with. A diode in the emitter lead will stop reverse current.

    You could then increase the voltage supply to 12V which is nice to drive some fets. Just connect the fet gate to the collector of your oscillator.
    If you use 180 leds, put three in a string with a current limiting resistor, that means 60 strings each taking 20mA. Total current 60 * 20m = 1.2A. This is well within the capabilty of a power fet to switch They can do several amps if provided with a heat sink and are rapidly switched.

    Logic level fets do not need such high driving voltages but if you reduce the supply to 9V, then you will only be able to put two fets in each string, you now have 90 strings, total 1.8A.
  4. auslander


    Jul 18, 2013
    Hmmm, Would you guys suggest possibly using a different circuit?

    Increasing the power supply to up the amperage would not be a problem in most cases, but the project I am currently working will have to run on battery. Also, I forgot to mention earlier was that I will be running different colors, which will def have higher amps. well, I guess to just fill you guys in on the scenario, I am trying to deck out one of my friends dance costumes with LEDs (hense the need for portability)

    I will have to do more reading on these fets you speak of (Im assuming you are meaning field-effect transistors).

    This spawns another question (a noob question at that). is the source power given to the circuit passed along to the LEDs? in this case, 9v is used. does the circuit still pass those 9 volts to the LED? I have thought this the whole time, but something made me feel i should ask in case somehow the circuit is taking away from this. (if any of that made sense)

    ==Thanks for the feedback!==
  5. duke37


    Jan 9, 2011
    Your oscillator circuit does not give a really sharp tun on/off signal and the better circuit in your reference uses a 555 chip. This can provide enough current to drive a few leds directly.

    If you want to use a lot of leds, then a separate switch is required which will take the current. A fet is the best for this since it takes very little power to drive it and will turn on with a very low resistance, giving minimal power loss.

    An even simpler oscillator can be made using a CMOS 4093, this just needs a resistor and capacitor. The disadvantage is that it can provide only very limited output current but this is no problem if you are driving a fet as a switch.

    I do not see why different colours should need higher current. The total current will depend on the number of leds that you have.

    Your last paragraph is a little unclear. The oscillator that you preferred drives the leds directly but can only do a few. To drive more, the oscillator and switch should be separated, then the sky is the limit.
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    Flashing multiple LEDs

    This diagram shows a general way to flash a large number of LEDs in unison.


    On the left is the power source. It is shown as a battery but would normally be a power supply, since the LEDs can draw quite a lot of current.

    The ideal voltage is 12V. Do not exceed 15V because U1 (the 555 timer) and Q1 (the MOSFET) may be damaged.

    The ON/OFF switch, SW1, is optional.


    U1 generates a square wave at its output (pin 3). The frequency is determined by CT (shown as a 33 uF electrolytic capacitor) and RT (shown as a 100 kilohm preset potentiometer or trimpot).

    The LEDs flash ON and OFF for equal durations, i.e. the duty cycle of the oscillator is 50%. The LED ON time (which is equal to the LED OFF time) can be calculated as approximately T = 0.7 RT CT, as shown on the schematic.

    T is the ON time (which is equal to the OFF time), in seconds;
    RT is the RT resistance in ohms (as shown, RT is adjustable from 0 ohms to 100,000 ohms);
    CT is the CT capacitance in farads (33 uF is 0.000033 farads).

    So with the value given, the ON time is adjustable up to approximately 0.7 * 100,000 * 0.000033 which is about 2.3 seconds. So with RT fully anticlockwise, the LEDs will flash ON for 2.3 seconds, and OFF for 2.3 seconds.

    You can change the values of CT and RT, and replace RT with a fixed resistor if you like, to get the flash rate you want.


    Q1 is driven from the oscillator and controls the LEDs. I have suggested the NTD4906N, which is a small, cheap and high-performance device available from Digikey ( but many other types are suitable, including the MTP3055, and most likely any MOSFET with 3055 as the numeric part of its part number.

    The important selection parameters for Q1 are:

    1-1. It must be an N-channel MOSFET, not a JFET

    1-2. Its Vds voltage specification (maximum allowable drain-source voltage) must be at least 20V. The NTD4906N is rated for 30V and the MTP3055 for 60V.

    1-3. Its ON-resistance (the Rds(on) parameter) should be as low as possible. This parameter is specified in ohms, or milliohms (one milliohm is 0.001 ohms). Don't use a device with Rds(on) greater than 0.1 ohms (100 milliohms). Under 50 milliohms is good.

    This parameter directly affects the amount of voltage lost in the MOSFET, and the amount of power it dissipates, while the LEDs are ON. Keeping Rds(on) under 50 milliohms means that a total LED current of up to 2.4A can be driven with less than 1% voltage loss in the MOSFET, and very little heating (no need for a heatsink on the MOSFET).

    1-4. It must be in a suitable package. If you are a hobbyist, you may prefer a through-hole package (the type with wire leads that can be pushed through holes in a piece of stripboard). Of these, the TO-220 package (and its variants) are convenient, because the pins are spaced at 0.1 inch pitch, to suit stripboard. The NTD4906N I recommended is smaller than TO-220 so you'll need to spread the leads a bit.

    Most MOSFETs nowadays are in SMT (surface mount technology) packages, which mount directly onto a copper surface. This is quite workable with stripboard, and gives a more robust result than a through-hole package, which can become bent over, and can short onto other components.

    Other MOSFET parameters are not really important in this application.


    LEDs are driven in series strings, each string with a current limiting resistor. The strings are connected in parallel. When Q1 turns ON, the full supply voltage is applied across all of the strings.

    The number of LEDs in each string, and the value of the series resistor, can be calculated as follows.

    2-1. The forward voltages of the LEDs in a series string add together. The resistor "takes up the slack" between the sum of the LED voltages and the total applied voltage.

    The current in all parts of a series string is the same. This current is controlled by the resistor - specifically, by its resistance and the voltage across it - and can be calculated using Ohm's Law: I = V / R, where I is the LED current in amps, V is the voltage across the resistor, and R is the resistor value, in ohms.

    Ohm's Law can be rearranged to R = V / I, where V is the voltage across the resistor, I is the desired string current in amps, and R is the resistor value in ohms.


    Say we want an LED current of 20 mA (0.02 amps), with four red LEDs in the string. Each LED has a forward voltage of 2.1V at 20 mA (from the data sheet).

    The total voltage across the LEDs is 4 * 2.1V = 8.4V. If the supply voltage is 12V, that leaves 3.6V across the resistor.

    R = V / I where V=3.6 and I=0.02 so R is 180 ohms.

    2-3. The forward voltage of an LED depends mainly on what colour it is. It is also affected by the current that you pass through the LED. It also varies from one device to another within a batch, and between batches, and between manufacturers.

    Manufacturers usually give a "typical" voltage at one or more operating currents. You should base your calculations on this typical value, but be aware that with the simple driving method shown here, variations in LED forward voltage WILL affect the string current.

    If you're running LEDs at high currents (e.g. over 50 mA), you may want to consider using a current source instead of a simple current limiting resistor, to get more accurate control of the LED current.

    2-4. As a rule of thumb, the voltage across the resistor should be at least 15~20% of the total voltage. For a 12V supply, this means you should have 1.8~2.4V across the resistor. So the total voltage available for the LEDs is 9.6~10.2V.

    Say your LED forward voltage is 2.9V. Although 4 * 2.9V is 11.6V, which is less than the 12V supply voltage, that would leave only 0.4V across the resistor. That is only 3.3% of the total voltage.

    In this case, you wouldn't be able to have four LEDs per string. You would use three LEDs per string. Total LED voltage would be 8.7V and there would be 3.3V dropped across the resistor.

    The reason for this rule of thumb is that LED forward voltages are not tightly controlled, so you want a reasonable amount of "slack" voltage across the resistor. In the example above, with only 0.4V across the resistor, a small change in LED forward voltage from 2.9V to 2.8V would cause the voltage across the resistor to double, from 0.4V to 0.8V. Since this voltage determines the string current, the current would double too, and this could damage the LEDs.

    2-5. The total current drawn from the power source when the LEDs are ON is the sum of the currents of all the strings. If each string operates at 20 mA and you have 50 strings, there will be 1A load on the battery while the LEDs are ON. Load on the power source while the LEDs are OFF will be about 10 mA.

    2-6. You can mix and match colours within each string. As always, all LEDs in the series string will operate at the same current. This may make the layout more convenient, or it may enable a better combination of voltages.

    For example, say you're using a lot of white LEDs with forward voltages of 3.4V. Three of these adds up to 10.2V, which is pretty close to the 12V total voltage. You would probably use two white LEDs per string, so each string would have 6.8V across the LEDs and 5.2V across the resistor. This means more strings are needed, which wastes power. But you could add a red LED (forward voltage about 2V) into each string.

    If you want different colours to operate at different currents, you can put them in separate strings. An alternative is to connect a resistor across the LEDs that you want to run at a lower current; the resistor will carry some of the current, leaving less current flowing through the LED. The resistance can be calculated using R = V / I where V is the voltage across the LED and I is the current that you want to divert through the resistor, in amps.


    Several part numbers are listed for U1. The LM555 and NE555 types are the original "555 timer"; TLC555 and TS555 are enhanced modern equivalents. Any type is suitable for this application.

    CD and CC are decoupling capacitors. CD should be connected close to pins 1 and 8 of U1. RG limits the current into and out of Q1's gate at switching times (MOSFET gates have significant capacitance.)

    MOSFETs are static-sensitive devices. You should keep the pins connected together using a piece of conductive foam, a paper clip, or some thin wire, until it has been installed in the board. (Install it last.)

    The circuit can be constructed on stripboard. Layout is not critical.

    Attached Files:

    Last edited: Jul 20, 2013
  7. auslander


    Jul 18, 2013
    Thanks, Duke and Kris. I will do some reading on those items and get back to ya if I have any questions. Duke, about using different Colors, some of my leds (like my whites) use around 20 mA a piece
  8. duke37


    Jan 9, 2011
    A CMOS 4093 contains four gates, one can be used as an oscillator with similar values of R and C as the 555 circuit. A spare gate can be used as in inverter so that one set of leds can be turned on when another set goes off (Two fets required).

    Alternatively you could have four oscillators, running at slightly different frequencies.
    Unused gates will need their inputs connected to something or the chip may go bananas.

    It seems as if your leds all use 20 mA. Different coloured leds take different voltages so the current limiting resistance will need to be calculated to suit with different string compositions as Kris says.
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