Introduction: Yet Another Simple Pot-controlled 555 PWM Generator

Something that a project that I'm working on has me doing is using a serial to parallel IC (think 74HC595) to control leds. However, rather than drive the leds directly from the pins, I opted for the use of transistors. After testing this out, it became apparent to me that perhaps the leds might be too bright, so I went in search of a simple PWM generator.

Of course, there are a couple of instructables that already feature such a circuit, but I was unable to get them to work correctly for whatever reason. This being the case, I will now present the circuit that I came across and works very well.

Step 1: Yet Another Overview

PWM stands for Pulse Width Modulation, which is a simple way of efficiently supplying varying amounts of power.

For example. Say that you wanted to control the brightness of an led (note: there are many ways to do this, but for the sakes of an example, I'll only note two). The first way would be to put a variable resistor in series with the led. This would alter the amount of current that went through the led, while holding the voltage constant. If you put the variable resistor at 40%, the led would be 40% as bright as it could be.

The second way would be to connect a led in series with a resistor and a power supply that could be turned on and off really quickly. Let's say that you were able to turn on and off the power supply quick enough to the point where 40% of the time it was on, and 60% of the time it was off. This would be reflected by the led being on full brightness, but only for 40% of the time, giving the illusion of being 40% as bright as it could be.

Two different methods, for the same result. What's the difference? About 60% of the energy gets burned off as heat in the resistor in the first circuit, while in the second circuit, almost all of the energy supplied is used.

This is why PWM is useful. It allows a signal to range from completely off to completely on. If a signal is turned on and off quick enough, given a certain ratio, a signal can appear to be that ratio, without suffering from much power loss at all.

Step 2: What Will Be Needed

The schematic that this circuit is based on is so general, that instead of giving hard values for each component, the author gave relations that would allow for any combination of parts.

However, for the sake of getting it right the first time, I'll list the values that did work.

  * 1 - Generic 555 IC (NE555 was used)
  * 2 - 1K Resistors (R1, R3)
  * 1 - 100K Linear Potentiometer ( R2 = 100 * R1 )
  * 1 - 1n4004 Diode (Pretty much any diode will do)
  * 1 - .1uF Ceramic Capacitor (Unsure about the relation of the value of this to the resistors)
  * Breadboard

Step 3: Build It!

There's nothing much else to this circuit except for just going for it and building it.

Know of the proverb "Measure twice, cut once"? The same thing applies. Build the circuit, check the circuit, then apply the power.

Using the circuit is simple! Your PWM signal will be coming from pin 3. From there, make the standard led circuit, except route the voltage to pin 3. Play with the pot and enjoy!

Step 4: Credits / Sources

Article discussing the schematic:



dennisg8 (author)2016-11-14

Got this working, with asspumtion they print north to south, with assumption pin 7 comes in at same point as r1 off +.

Too many wires. But I can't see a breadboard working with a 555 as pin spacing is 1.25m, while bread board is 2.5mm.

Either I am wrong with bb spec, or my 555 is wrong brand.

JohnL79 (author)dennisg82016-12-15

Please slow down and try to absorb things step by step. Always look through the datasheet for any part you're using, it will list what marking is used to indicate pin 1. In the case of 555s, they are usually interchangeable as far as pinout but may have small differences in details of spec and performance. Keep in mind that some parts come in different packages, they're not all made to work with your breadboard. Almost all breadboards are 0.1"/2.54mm spacing, and most all *DIP* (dual inline package) chips are 0.1"/2.54mm spacing as well and should plug right into a breadboard. It sounds like you got an SOIC package which has 0.050"/1.27mm spacing and is for surface mounting, not meant for breadboard without a DIP adapter. Using pots can also be searched and studied, they use certain conventions that you just have to learn first. The "CW" with arrow indicates which direction you turn the pot to move the wiper (pin 2) from one voltage to the other, thus sweeping its output. The pot forms a voltage divider from pin 1 to pin 3 voltage, with the divided value being output on pin 2.

dennisg8 (author)2016-11-13

My 555 that arrived looks nothing like the wikipedia, where there is a dot or two in the north side. How are we supposed to figure out the north south orientation of the 555 without burning out the chip, if it doesn't look like the wikipedia picture? There is random number on it too small to read. I could assume they would print north to south. Or I could assume they would print south to north.

dennisg8 (author)2016-11-13

Of course, I will go with the detailed diagram over rough. Trial and error and google on the diode/cap. However, I don't want to trial and error pot hookup. This pot hookup needs explanation of what is actual hookup, described by the three arrows pointing into r2.

dennisg8 (author)2016-11-13

The rough and detailed schematic don't jive. On the detailed pin 5 is not hooked up. On the rough schematic, it is hooked to 2 and 6 pin.

doctaven (author)2016-10-16

=== An updated Link for your Credits ===

dennisg8 (author)doctaven2016-11-13

Also, on rough schematic, what does 2 cw refer to?

dennisg8 (author)2016-11-13

I haven't yet built, but studied with microscope the laid out schematic.

Firstly, I do not yet understand the pot hookup. Most pots have three pins. I assume only two to be tapped, not three. But assuming as nothing mentioned in diagrams. Now assuming only two pins the Vs through R1 and pin 7 go into center pin of the pot, while third pin of the pot goes to r3. ?????????????????

Also, I find the schematic a mere jumping off point for a physical diagram, which takes another half hour to draw out. On an actual diagram, I see these pins shorted together: 8:4 , 5:2 and 7 to 6 via Diode. If I am right, 7 and 8 are shorted through r1. And 7 and 5 are shorted/connected through pot and r3. And 5 and 1 through the capacitor.

Not having built circuits recently, I am also going cold on diode and cap orientation. Hope it is obvious, when building.

dennisg8 (author)dennisg82016-11-13

Blowing up the pot hookup on the jpg of the breadboard, I am even more confused, since they all look connected together.

chinmayj made it! (author)2015-10-31

works fine

Sir Dean (author)2013-02-13

Any diode will work, schottky just has a lower voltage drop but that doesn't affect things in a meaningful way here.
When R2 is at the top, the Capacitor is charged via R1 and the PWM is High. So
roughly speaking, the smallest PWM High time is R1 (1K) * C1 (0.1u) = 0.1ms, and the Max Low PWM time is then (R2 (100K) + R3 (1K) ) * C1 (0.1u) = 10ms. Therefore the max On duty cycle is 99%, and the min On duty cycle is 1%. One can play with these by varying the values of Rs and C1.

StevenM8 (author)Sir Dean2015-05-17

So when you say "any diode", do you really mean ANY? I really don't have to worry about any of the specs? Could you confirm this one would be fine:


Sir Dean (author)StevenM82015-05-17

Yes 1N4148X will work.

Good luck!

stefco (author)2014-07-23

i believe the cap value is regulating the pwm frequency , i have 0 experience with pwm this is my first project , but i believe lower value caps = higher frequency and vice versa. You don't want too low because it will either be audible ( i've heard motors produce a sound from the controller frequency if it's within hearing range) and you don't want too high because well... i have no idea (maybe it will heat up ? )

stuuf (author)stefco2015-04-29

The capacitance controls the PWM frequency by changing the time it
takes to charge to a given voltage through a current determined by the
resistor value.

You're right, increasing the frequency will create more heat. The reason is switching losses in the transistor: it spends most of its time in either the cutoff (~infinite resistance, ~0 current, ~0 power dissipation) or saturation (~0 resistance, ~0 voltage drop, ~0 power) regions, but each transition moves through the linear region where current is flowing through an intermediate resistance for a short amount of time. Each transition dissipates some energy, and the more transitions per second (higher frequency), more power is dissipated overall. (This is why SMPS designers use lower PWM frequencies to increase efficiency... but that also slows response to large load current steps, so for things like CPU power supplies they also have to use other tricks like synchronous multiphase topologies to get the right performance)

Xavierxf (author)2013-11-15

What is pin 5 connected to? Is it +5 or GND?

stuuf (author)Xavierxf2015-04-29

I generally bypass it to ground with a small cap. (Internally it's connected to the other side of the threshold comparator so you can override the 2/3Vcc with a different control voltage if necessary.)

desmondtheredx (author)Xavierxf2014-04-27

just leave it unconnected

OhYeahThatGuy81 (author)2014-11-06

Awesome, this circuits works brilliantly.

I also had a similar issue like you - other circuits on Instructables did not work for me (maybe I didn't wire them correctly?), but this one works great! Thanks!

desmondtheredx (author)2014-04-10

what is the frequency range of this PWM generator.

BTW AMAZING instructable


You can play with various values of R and C in this calculator:

The calculator does not incorporate the diodes that allow this circuit to go below 50% duty cycle, so I don't know how that might affect the frequency. To simulate the effect of turning the pot all the way one way and then all the way the other way, I set R1 to 1 kOhm and R2 to 101 kOhm, and vice versa. I got a frequency of 70 and 140 Hz, so it may be the case that the frequency changes depending on which way the pot is turned.


interesting. I also played around with the capacitor as well, and it also affects the frequency directly. So i was thinking of trying to put multiple capacitors in series and parallel with switches to modify the capacitance. By having them in series allows you to increase the frequency and having it in parallel gives a lower frequency.


I built a version of this circuit in the Java Circuit Simulator App. You should be able to view it using this link. You can use the slider on the right to move the pot around and see the resulting frequency. The graph at the bottom should show frequency in the upper left, as long as the frequency has been stable for long enough.


Sorry. It appears that the link isn't working right. Try opening the applet at and then clicking file/import and pasting in this text:

$ 1 5.0E-6 19.056626845863 50 5.0 50

165 464 224 480 192 2 12.0

R 528 128 528 80 0 0 40.0 12.0 0.0 0.0 0.5

w 592 256 592 128 0

w 592 128 528 128 0

r 592 288 688 288 0 100.0

g 688 288 736 288 0

w 464 352 384 352 0

w 464 320 384 320 0

w 384 320 384 352 0

w 464 256 384 256 0

z 384 256 384 320 1 0.805904783 5.6

174 336 224 384 288 0 100000.0 0.9950000000000001 Duty Cycle Pot

w 528 128 528 192 0

r 336 224 336 128 0 1000.0

w 336 128 528 128 0

r 336 288 336 352 0 1000.0

w 384 352 336 352 0

c 336 352 336 416 0 1.0E-7 6.801312165289001

g 336 416 336 448 0

o 4 64 0 46 20.0 0.1 0 -1

freenergyfuture (author)2014-01-07

I understand all electrical parts/components have a mathematical equation to them.
What I am Currently in need of, is the math relating to pulse frequency and high voltage.
IE; as High frequency as possible and as high voltage as possible.
Resistors, Capacitors, Transistors, and whatever else is required to Increase DC voltage and frequency of number of pulses per second.
What I desire it to have a circuit, on the Cheap, that I can Easily adjust the frequency and voltage output.
Type of signal is currently not important, although, I'll eventually be using Scalar waves frequency which is more Potent than Radio waves.
Cold Current Generation.
Can you Help me out?

gwood6 (author)2013-04-04

Hi - Thanks for this! It looks nice and simple and very like a simple 555 pwm that I constructed recently and actually worked so thanks and well done : )
May I ask what you would do with pin 5 on the diagram..????
Would it go to ground or connect with a .01uf cap on its way to ground ??
Thats what my present simple 555 does..!

LesB (author)2013-02-07

Thanks for the 'ible. I'm gonna use it tomorrow. Looks like a good simple circuit.

The diode symbol in your schematic (not the hand-drawn one) is the symbol for a schottky diode. In your parts list the diode specified is a 1N4004. The 4004 is a general junk diode and not a Schottky. A regular diode symbol has just a simple straight bar without the switchbacks. Although a Schottky would probably work fine.

Another PWM generator on this site did only 10% to 90%. From the looks of the scope pictures it looks like yours does more like 1% to 99%.

rimar2000 (author)2010-07-11

Good work! Now, a question: can I use this PWM to vary the velocity of a desk fan? And a drill? In other words: how many potency can this circuit manage?

MrCruz (author)rimar20002010-07-11

Directly? No. If you were to directly connect the output pin to the device, the most current you'd get would be something around 30mA, I believe. However, that's still not to say that it couldn't be done with this circuit. What you could instead do is use the output of this circuit and use it to control the gate pin on an NPN transistor, similar to what I did on the very last image. You'd have to make sure that the transistor could take the kind of currents that you need (drills and fans tend to use a lot of current, around the range of Amps)

edembowski (author)MrCruz2012-12-19

Can you show that part of the circuit?

rimar2000 (author)MrCruz2010-07-11

Thanks very much for your response. Electronic, as you can see, isn't one of my skills...

coolstuff14 (author)2011-02-25

Did you know you can play tetris on that oscilloscope?

rcisneros (author)2010-10-15

Thanks for the post.
I know it's my own ignorance, but there is not info here for me. The levels of the readers electronics knowledge varies greatly.
Like for me, I don't see how this even works since I can't find where to attach the power, where it comes out. I have to figure out ground, might be the (-). I know it's me, but help a guy out and label those things.
Keep in mind that the people that can read this cold, probably know how to do it already.


MrCruz (author)rcisneros2010-10-19

Thanks for the reminder. It's true that I do often overlook this.

The 555 IC can take a voltage of between 5-15v, which means that you could power it with a simple 9v battery. In schematics, Vs stands for Voltage Source, which is where you would plug the positive (+) terminal of the voltage source, while the triangle (in the first picture in this step) is where you would connect the negative (-) terminal.

Schematics that have multiple ground symbols in actuality only have one. When creating the circuit, all of those ground symbols will connect together and meet at the battery's negative terminal.

The output voltage, where the signal is generated, comes from pin 3 on the IC.

Remember, all grounds meet together, unless otherwise specified.

omnibot (author)2010-07-11

Nice one.

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