I purchased this LED bulb:
https://www.liquidleds.com.au/colle...und-dimmable-led-light-bulb-e14-in-warm-white
I'm trying to work out what the best method to dim it would be.
I was thinking a 555 timing chip with a potentiometer to dim it by PWM? What I'm trying to work out is, will I need a series resistor to limit current flow to this bulb, or will it likely be safe if I just hook it up to 12 volts? The company I bought it off don't really know about circuit design but said if you hook it up to a battery it will work, so that leads me to believe it should be fine. No idea how much current I would need either, I'm guessing 0.3A should be fine since it's 4W?
Thanks for any help in advance

 

Here is a circuit found on line.
There are many to chose from. If you run it on 12V, no resistor is needed as it is advertised as a 12V lamp.

 

View attachment 304728
Here is a circuit found on line.
There are many to chose from. If you run it on 12V, no resistor is needed as it is advertised as a 12V lamp.

Excellent thank you so much for your reply! I have also seen that circuit so I think that’s the one!

 

One thing I would add is a 1000uF electrolytic capacitor across the supply. I just noticed there is no supply bypassing. A good filtered supply is important in any electronic design. Of course, the power supply you use may have that included but more does not hurt.

 

One thing I would add is a 1000uF electrolytic capacitor across the supply. I just noticed there is no supply bypassing. A good filtered supply is important in any electronic design. Of course, the power supply you use may have that included but more does not hurt.

I am powering the bulb with a battery and boosting it to 12V, so do you think the capacitor would be a good idea? The battery charge is being controlled by a tp4056.

 

Adding a filter capacitor across an unfiltered DC supply will make the bulb brighter because the average voltage will be higher.
The PWM circuit will allow dimming, and if the FET, Q1 is adequately saturated heating should not be an issue. Ifthe FET runs hot, then perhaps it will need more gate drive.

 

I purchased this LED bulb:
https://www.liquidleds.com.au/colle...und-dimmable-led-light-bulb-e14-in-warm-white
I'm trying to work out what the best method to dim it would be.
I was thinking a 555 timing chip with a potentiometer to dim it by PWM? What I'm trying to work out is, will I need a series resistor to limit current flow to this bulb, or will it likely be safe if I just hook it up to 12 volts? The company I bought it off don't really know about circuit design but said if you hook it up to a battery it will work, so that leads me to believe it should be fine. No idea how much current I would need either, I'm guessing 0.3A should be fine since it's 4W?
Thanks for any help in advance

Hi,

The only advantage PWM has is that you can *dim* the LED without using a big rheostat (a really big potentiometer).
There is no efficiency gain for example, and that is because even a small series resistance looks large when used with PWM. This might be hard to understand at first but it's true and we can talk about that more if you like.

If you want better efficiency you have to use a switching converter like a Buck Converter. This are widely available now for low cost and you don't have to know much about how they work because you get the whole PC board. You just connect the input and output and turn the potentiometer to the setting you want.
You may have to opt for a current regulated model though rather than a purely voltage regulated model.

 

A TP4056 charges a single lithium cell battery at 1A max. If the battery charge is low at 3.2V and you are boosting it to 12V then the charging current will be zero while the TP4056 and the booster are dimly lighting the 4W LED.

Charge the battery cell before turning on the LED. The battery voltage will drop as it powers the LED then the dimming will be automatic.

 

will it likely be safe if I just hook it up to 12 volts?

Yes, it states the bulb is designed to operate from 12VDC.

 

Depending on the characteristic of the lamp, logarithmic dimming may be required for the required visual effect.

 

the bulbs i've bought with filaments like that didn't last.

 

Why not use this chip: MIC3203


ADDED:

What would I need coming into pins 5 and 6? Would my bulb be connected where the string of leds is on this diagram?

Pin 5 Vin/GND = ON/OFF.
Pin 6 put here your PWM signal for dimming.
Yes you can connect your bulb instead string of LEDs.
Please, read PDF file.

I think the 555 circuit is a lot simpler and cheaper. Try that first and see if it does what you want.

@organic.parfait, @dendad
Sorry, my bad. This bulb contains ballast inside, so MIC3203 is not needed.

 

Why not use this chip: MIC3203

View attachment 304775View attachment 304776

What would I need coming into pins 5 and 6? Would my bulb be connected where the string of leds is on this diagram?

 

I think the 555 circuit is a lot simpler and cheaper. Try that first and see if it does what you want.

 

I think the 555 circuit is a lot simpler and cheaper. Try that first and see if it does what you want.

Alright, thanks, appreciate your responses a lot

 

Hi,

The only advantage PWM has is that you can *dim* the LED without using a big rheostat (a really big potentiometer).
There is no efficiency gain for example, and that is because even a small series resistance looks large when used with PWM. This might be hard to understand at first but it's true and we can talk about that more if you like.

If you want better efficiency you have to use a switching converter like a Buck Converter. This are widely available now for low cost and you don't have to know much about how they work because you get the whole PC board. You just connect the input and output and turn the potentiometer to the setting you want.
You may have to opt for a current regulated model though rather than a purely voltage regulated model.

So far it is not "difficult to understand". It is impossible to understand. Of course the efficiency of the dimmer is greater than the efficiency of a big rheostat. A simple set of I / V = P measurements over the operating range will show that. The heat produced by the rheostat vs no heat produced by the PWM approach declares the inefficiency of the rheostat.

 

So far it is not "difficult to understand". It is impossible to understand. Of course the efficiency of the dimmer is greater than the efficiency of a big rheostat. A simple set of I / V = P measurements over the operating range will show that. The heat produced by the rheostat vs no heat produced by the PWM approach declares the inefficiency of the rheostat.

An additional advantage of PWM dimming, at least for some LEDs, is no color change. With some LEDs the color of the light changes as the voltage and current change. But with PWM that is avoided. And the not producing of heat is a benefit as well. Less heat in both the controller and the LED.

 

So far it is not "difficult to understand". It is impossible to understand. Of course the efficiency of the dimmer is greater than the efficiency of a big rheostat. A simple set of I / V = P measurements over the operating range will show that. The heat produced by the rheostat vs no heat produced by the PWM approach declares the inefficiency of the rheostat.

Hi,

Yes, that's not accurate at all and that is one reason why I stated that it may be hard to understand at first.
I have proved this in forums many times. The simplest explanation is that there is always some resistance in the circuit and the PWM switching plays into that. The expression is fairly simple. The main idea is that the PWM switch makes the physical resistance absorb all the energy that is not being used to power the LED. A comparison between a rheostat and a PWM circuit shows how this works.

A rheostat of course changes the physical series resistance Rs, and resistance suffers from R*i^2 losses.
With PWM, there is a small, constant, physical series resistance Rsc, but how can a small series resistance drop as much voltage and reduce the current? It can't. The PWM makes the equivalent resistance Req look larger by a factor:
Req=Rsc/D

or in terms of the original Rs:
Rs=Rsc/D

and so at all times Rs=Req. That means that Rs*i^2 equals the same as Req*i ^2 of course after averaging.

Another explanation comes from analyzing the energy conversion of a linear circuit versus a non-linear switching circuit. The non-linear switching circuit acts the same as the linear circuit in the averaging solution, unless it contains at least one energy storage element.
An energy storage element allows a true energy conversion, and that is the only way to get higher efficiency in the most general case.

An extreme example would be Rs=10 Ohms, and Rsc=5 Ohms with a PWM switch at 50 percent duty cycle (D=0.5).
If Rs causes a dimming to 50 percent brightness, then we must make Req 10 Ohms and to do that we need a duty cycle of 50 percent, or D=0.5, and with that D=0.5 we see that Req comes out to:
Req=5/0.5=10 Ohms.
Now the question arises, how does it accomplish this.
The answer is that it allows a higher peak current in exchange for a lower average current. If with Rs=10 Ohms we get say 1 amp average and that means 1 amp peak also, then with Rsc=5 Ohms and a 50 percent duty cycle we still get 1 amp average but now we have 2 amps peak current (2 amps at 50 percent duty cycle averages to 1 amp). Now the instantaneous power in Rsc is 5*2^2=20 watts, but the average is just 10 watts. The instantaneous power in Rs is 10 watts and the average is 10 watts, so what changed.
The only thing that changed was the fact that we used PWM with the smaller value series resistance, as they both dissipate an average power of 10 watts.

It actually gets a little worse for PWM though, because LEDs tend to have their efficacy drop with higher currents, and that includes peak currents. Since the peak current in the PWM case is twice as high as the non-PWM case, we can expect to see a drop in brightness using PWM. It may not be significant enough to warrant not using PWM though.

You can also prove this with a simulation. You have to calculate the average power in Rs for the non-PWM case, and calculate the average power in Rsc for the PWM case. When you have the correct duty cycle the average current in both circuits will be the same, then you can compare the two average power calculations.

Now if you use a buck for example, which is a PWM circuit with an energy storage element like an inductor, you would see true power conversion. This means that when you dim the LED, you use much less power.

I think the confusion arises because the PWM in the non-energy storage version causes a reduced current just like in a circuit like a Buck, and that makes it seem more efficient. However, a reduced current does not equate to higher efficiency by itself, it's about power ratios.

There is one caveat to all this. That is when the output voltage is very very close to the input voltage. This is a somewhat rare case though.
Because of the efficiencies of most bucks, the PWM could come out slightly higher for this rare case. It's only important though when you intend to run the LED at nearly full brightness and you happen to have an input voltage that is very close to the output voltage. That makes it less important than the more usual case where we want to lower the current over a decent range so we get a variation in brightness.

There actually is another caveat to this though, and that is the human eye physiological response. There may be pulse timings that make a light seem brighter to a human eye and thus make the setup seem to have a higher efficacy than a pure physics calculation would yield.

The other thing to remember is that PWM does in fact cause a decrease in current, so it does function as a brightness control without having to use a large rheostat.

Lastly, you have to be sure to read the data sheet of the LED carefully because some LEDs are not made to be able to tolerate pulsed currents. This is usually stated clearly in the data sheet.

 

PWM, (Pulse Width Modulation) is simple if you ignore the resistors, real or imagined. If full on is 100% average, then 50% on time averages to half of that. And the same thing can also be proved through integral calculus. But why bother.
And if adjustable intensity is not a requirement then just a simple series resistor will dim the LED quite well, although it is not so efficient.

 

PWM, (Pulse Width Modulation) is simple if you ignore the resistors, real or imagined. If full on is 100% average, then 50% on time averages to half of that. And the same thing can also be proved through integral calculus. But why bother.
And if adjustable intensity is not a requirement then just a simple series resistor will dim the LED quite well, although it is not so efficient.

It might benefit you to spend more time on the bench.