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Question about PWM LED dimmer circuit

Hi all.

I built (on a breadboard) the following circuit for my first electronics project:

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Note that the parallel resistor array (R4-R8) is there simply because I ran out of 1K resistors. :)

It works, generally; rotating the pot controls the brightness. However, I note that the LED never actually goes out, it just gets fairly dim. When I test the duty cycle across the LED with my DMM at the minimum, I get 1.8% (99% at maximum).

Can anyone suggest a change that would allow me the full range of brightness from off to bright?

Also, removing one of the resistors from the parallel array (thereby raising the resistance of the array as a whole) brightens the LED. Can anyone explain why? (I'm clearly just getting started with all this.)

Thanks for helping a newbie.

/afb

Reply to
adam.block
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** See:

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The duty cycle control arrangement shown here gives 0 to 100% .

** Your schematic has a least one error ( pins 2 and 6 must be linked).

Is there really NO resistor in series with the LED ??

Time to add one.

Is the DC supply a 9 volt radio battery ?

........ Phil

Reply to
Phil Allison

Phil, thanks for your input. You are correct. Pins 2 and 6 are linked, but I left that off the schematic.

I added a 3.3K resistor in series with the LED, but got the same effect.

The DC supply is a wall wart.

I will review the URL you posted.

Thanks again!

/afb

Phil Allis> >

Reply to
adam.block

** That was the missing fact.

The voltage from the ( I assume unregulated and so NOT actually 9 volt ) wart will drop under load.

Increasing the drive resistance to the TIP 3055 reduces the total current load on the wart - ergo the voltage goes higher and the LED becomes ( very slightly ) brighter.

...... Phil

Reply to
Phil Allison

I second what Phil said. You need a resistor in series with the LED to limit the current. Something like 470 ohms. Some other points with the existing circuit: you don't need a power transistor like the TIP 3055 for the LED - any small NPN will do. You can stick with the 3055 if you want, but in any event, you don't need ~1K base resistance. A single 4.7K will work. You can add a large value resistor from base to ground - 22K, 33K, 47K something in that neighborhood. I'm guessing at what you might have in your parts pin. Don't forget the 2-6 connection Phil mentioned.

The 555 can drive an LED directly, so you could eliminate R4-8 and Q1 entirely, if you want. Just connect pin 3 to the top of the LED in your diagram, and remove the existing connection that goes from the top of the LED to D3. You don't need R3. You still need a resistor in series with the LED.

Ed

Reply to
ehsjr

Thanks Phil and Ed for your feedback.

/afb

Reply to
adam.block

Not to be rude, but where did you get this circuit design? It's the best example I've seen in a long time of how to do a really bad engineering design - overpowered, and overly complex.

A circuit for this could have been done with just a pot, 2 resistors, and one small transistor.

Luhan

Reply to
Luhan

"Luhan"

** A false disclaimer as usual.

** Try reading the original post properly - d*****ad.

** Completely beside the point of learning about EFFICIENT electronic techniques like PWM.

....... Phil

Reply to
Phil Allison

Try changing your transistor drive like so: View in a fixed-width font such as Courier.

. . . . +9V . | . | . --- . \\ / ~~ . --- . | . | . [330] . | . | . 1n |/ . 555 pin3 >--+--[470]---||---+------| . | | |>

. | | | . '----[4.7k]-----+ | . | | . [10k] | . | | . '----+---' . | . --- . /// .

Reply to
Fred Bloggs

What's that upper r-c for?

Why not

+9 | | c 555----------------b e | | r | led | | gnd

and save a lot of parts and power?

John

Reply to
John Larkin

I'm just guessing here - I'm only a tech, after all - but could it be to pull charge out of the base to make sure the transistor turns off cleanly?

Thanks, Rich

Reply to
Rich Grise

It's just an led... who cares?

John

Reply to
John Larkin

Or:

555 (p3) ---+ | .-. | | | | '-' | | led | | === gnd

James Arthur

Reply to
dagmargoodboat

Ahh, I see Ed (ehsjr) beat me to this suggestion... but I get credit for 1rst ASCII version !

Cheers, James Arthur

Reply to
dagmargoodboat

Or

Or

+9v | | r | | pot | | led | | | gnd

John

Reply to
John Larkin

The original question was, "Howcome I can't turn it down to 0%?". Either the transistor is leaking, or he's really not getting a 0% duty cycle.

If it's the latter, of course, he'll have to simply fix the PWM.

Cheers! Rich

Reply to
Rich Grise

Or: John Lark>

Or:

+9v | o \\ \\ SW1 \\ o | R1 | | led | | gnd

Cycle SW1 quickly for PWM.

James

Reply to
dagmargoodboat

The OP was not about alternative circuit schemes, it was about his particular circuit. And nobody uses a 3055 in that kind of follower configuration with Vce,on~2V.

Reply to
Fred Bloggs

A '3055 is a strange choice here, but Vce_on will be just like any other NPN at modest currents... better than most, in fact. And I realize you don't like ideas.

But what was that upper series r-c all about?

John

Reply to
John Larkin

If you want the LED to go completely off then you can add a second 555 that blocks output pulses going to the drive transistor when those pulses have duration less than the time constant of RqCq like so:

View in a fixed-width font such as Courier.

. . . 9V . | . [10K] . | +9V . .-----+ | . | | --- . | | \\ / ~~ . | |/ --- . .-------| 2N3904 | . | | |> | . | | | [330] . [10K] | --- | . | | /// | . | | +---------+ 1n |/ . 555 pin3 >-+---------|RST OUT |-+-[470]-||--+------| . | | | | | |>

. Rq [10K] | | | | | . | | | '---[4.7k]--+ | . +-----|TRIG | | | . |15n | | 10k] | . Cq === | | | | . | | | '----+---' . +-----|THRESH | | . | +---------+ --- . -+- 2ND 555 /// . /// . .

Reply to
Fred Bloggs

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