Full range PWM (555)

Thread Starter

JMD

Joined Dec 9, 2009
94
Hi everyone

Im trying to make a general purpose PWM circuit. Ive been using the 555 before, so naturally i went with that one again.

Problem is, the standard circuit from the datasheet, calculates the duty cycle by Ra/(Rb + 2*Ra), which means the lowest duty cycle will be 50%.

Anyone know a good schematic, with full range (~0% to ~100%) duty cycle?


Im planning to use a TIP35A to drive whatever i want to control - should be more than enough (got a few in spare anyway..)


Thanks in advance.
 

Razor Concepts

Joined Oct 7, 2008
212
I tried circuit #1 on a breadboard.

5v supply, appears that R2 is 10 ohms (kind of squished together), CR1 and CR2 are 4148 diodes.

The result is that D2 is always on, D1 flickers but it does dim when adjusting the potentiometer.
 

SgtWookie

Joined Jul 17, 2007
22,210
I tried circuit #1 on a breadboard.
Good for you! :)

5v supply
That's mighty low. It'll work much better if you use a higher voltage supply.
appears that R2 is 10 ohms (kind of squished together), CR1 and CR2 are 4148 diodes.
OK.
Replace R1 with a 100k pot.
Replace R2 with a 1k resistor.
Replace C1 with a 10nF (0.01uF) capacitor.
Replace CR1 and CR2 with 1N400x series diodes.
Increase Vcc to 8v-12v.
Add a 24k resistor from pin 5 (CNTL) to ground.
Then see how well it works, or doesn't work.
Keep in mind that if the 555 timer is a transistorized version, the output will approach ground, but won't go higher than about Vcc-1.3v.

If the 555 timer is a CMOS version, it can sink about 10x as much current as it can source. This will throw the circuit out of whack quite a bit. Also, the 24k resistor from CNTL to ground will have to be increased by quite a bit.
 

SgtWookie

Joined Jul 17, 2007
22,210
Bill, have you built and tested the circuit shown in Fig.4.3 ? Does it work as shown?
Cary,
Would you please post a link directly to the circuit?

Not to doubt you , but I've always been under the impression that a 339 does not source current, only sink current. And that is the reason for the pull-up resistor.
You are correct. A quad 339 comparator is not (for all practical intents and purposes) capable of sourcing current, only sinking it.

As the circuit is drawn, what turns on the base of the transistor?
Link needed for further commentary.
 

Wendy

Joined Mar 24, 2008
22,145
He's talking about this image...



Take a good look at the transistor, it is a PNP emitter follower. It really needs a resistor going from base to Vcc (10KΩ would work) to turn the transistor off. When the LM339 goes to ground the transistor will turn on.

I'll correct the drawing. It may work as is, but I wouldn't bet on it. If after I breadboard it it does work I'll go back to the old schematic.

This circuit will work down to 4.5VDC on Vcc. The 555 sets this spec, and it is rated for 4.5VDC to 16VDC (some variations up to 18VDC).

The CMOS may work differently, their 1/3 and 2/3 thresholds are different than a regular 555, which is to say I don't know the exact numbers, but the standard 555 will work as is, the Darlington Transistor output doesn't affect the design at all. It is the thresholds, not the output drivers, that set the triangle wave amplitude. R4, R5, R6 are designed around the 1/3 and 2/3 thresholds.
 
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Razor Concepts

Joined Oct 7, 2008
212
Good for you! :)


That's mighty low. It'll work much better if you use a higher voltage supply.

OK.
Replace R1 with a 100k pot.
Replace R2 with a 1k resistor.
Replace C1 with a 10nF (0.01uF) capacitor.
Replace CR1 and CR2 with 1N400x series diodes.
Increase Vcc to 8v-12v.
Add a 24k resistor from pin 5 (CNTL) to ground.
Then see how well it works, or doesn't work.
Keep in mind that if the 555 timer is a transistorized version, the output will approach ground, but won't go higher than about Vcc-1.3v.

If the 555 timer is a CMOS version, it can sink about 10x as much current as it can source. This will throw the circuit out of whack quite a bit. Also, the 24k resistor from CNTL to ground will have to be increased by quite a bit.
Thanks, the 24k (I used 27k) resistor did the trick.
 

lmartinez

Joined Mar 8, 2009
224
The outcome of the tutorial provided above is dependent on what type of 555 microchip you are utilizing to build the circuits. It is imperative for you to explore!!!
 

Wendy

Joined Mar 24, 2008
22,145
Really, a lot of people have built these, including me. But then, I live in the real world, so I would dispute that statement.

I generally build what I post, especially The 555 Projects, it is part of the fun.

Why do you think it will not work, have you build one?
 

SgtWookie

Joined Jul 17, 2007
22,210
Bill,
Try running one of them from 5v.

It won't work without the addition of a resistor from Control to GND.

If you're running it from 8v, 12v, 15v, 18v - it won't be an issue.

However, at low voltage you WILL need a pull-down resistor to get your circuit to work.

That's just the way it is - in simulation, and in the real world.
 

lmartinez

Joined Mar 8, 2009
224
Really, a lot of people have built these, including me. But then, I live in the real world, so I would dispute that statement.

I generally build what I post, especially The 555 Projects, it is part of the fun.

Why do you think it will not work, have you build one?

I totally agree with you. It is fun to build circuits... And yes indeed, I have built a bunch of them. But it all depends on what manufacturer you buy your 555 microchip from. Texas instruments vs a radio shock brand??? Maybe... I leave it up to your judgment.....
 

SgtWookie

Joined Jul 17, 2007
22,210
I totally agree with you. It is fun to build circuits... And yes indeed, I have built a bunch of them. But it all depends on what manufacturer you buy your 555 microchip from. Texas instruments vs a radio shock brand??? Maybe... I leave it up to your judgment.....
There is no reason to be obtuse and obsequious. Do you expect others to be clairvoyant?

One problem is that as BJT 555's Vcc decreases, they cannot supply enough voltage to trip the high threshold, particularly if the output is under load.

The problem is exacerbated (made worse) if it is a CMOS 555, as these ICs can sink about 10x the current as they can source.

In either case, the solution is to reduce the Control voltage, so that the 555 can still reliably cycle.
 

Wendy

Joined Mar 24, 2008
22,145
OK, I understand what you're saying Wook. With a 5V power supply, 2/3 of 5VDC is 3.33V, and the output can only go to 3.7 (give or take). I missed that one. The charge curve will slow down too. There are ways to beat this using the same number of parts, I'll post it when I get the chance.

BTW, Radio Shack sells nothing but TI chips concerning 555s, it was their parent company, through Tandy. I remember RS at their worst too, but I suspect they have improved (though they are still expensive).
 
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Wendy

Joined Mar 24, 2008
22,145
OK, this will go down to 4.5VDC, and will substitute for the other oscillator. A negative is the square wave isn't anymore.



I found this in the AAC experiments, believe it or not. Substitute a current mirror for the resistor, it becomes linear!
 

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shortbus

Joined Sep 30, 2009
7,756
He's talking about this image...



Take a good look at the transistor, it is a PNP emitter follower. It really needs a resistor going from base to Vcc (10KΩ would work) to turn the transistor off. When the LM339 goes to ground the transistor will turn on.
<snip>
.

What I was concerned about was the "PWM Out" on the transistor base. Any load wired to that would keep the transistor turned on all the time. That would counteract the 339's switching? Right?

I'm just starting to catch on to electronics and how things work, so I didn't mean to disrespect you, just trying to learn.:) You and some of the others here know more about this stuff than I ever will. But I have learned more from you guy's in the short time I've been here than the years trying to learn on my own!

Cary
 

SgtWookie

Joined Jul 17, 2007
22,210
What I was concerned about was the "PWM Out" on the transistor base. Any load wired to that would keep the transistor turned on all the time. That would counteract the 339's switching? Right?
Ahh, I see what you are talking about.

The 339 comparator outputs are open-collector. They cannot source current, they can only sink it.

They have a limited sink current; minimum is 6mA. So, I usually divide the 6mA sink by 2, for 3mA - and then use a pull-up resistor to Vcc that will work out to the 339 sinking 3mA when it's output is low.
 
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