Hello,

I have a CMOS (4029) output driving a transistor base/gate to switch a LED.
a capacitor is connected to the transistor base, to soften the power-on and power-off.

the result is a led fading in and fading out according to the the 4029 output:

this circuit works great with 20mA LEDs.
I use 400mA LEDs (with current limiting resistors like in the schematic)

the problem is:

when i use the 2N3904 (which is 0.2A), the LED draws only 0.06A (although when i measure the current across the transistor it gave me even 1A)

someone told me that it's a BJT transistor that current output is relative to input current, and i should use MOSFETs.

i got the IRF510 (100v 5A)
i just replaced the 2N3904 with the MOSFET,
then the LED draws about 0.19A

when i connect the led to +12v it draws full 400mA it should,
but at the MOSFET emitter it draws only 0.19A as i wrote...
here also, measuring the MOSFET's output gave me over 1A of current.

then i tried driving the MOSFET with the 2N3904, but once it gave me the same results as with the MOSFET alone,
and then the LED like stayed always on dim, and with higher pulses (from the cmos)

now i cant get back to use only the mosfets, the leds are steady on dim -
I probably fried the MOSFETs ?

also tried TWO 2N3904 in parallel, they gave 0.09A instead of 0.06A with one.

what's next ?
what transistor type should i use ?
what's the explanation to that issue ?

thanks !!

 

Hello,

The problem is probably the gain of the transistor.
The base resistor is 10 K this will give a to low base current to drive the 400 mA.
You could try a darlington transistor like the TIP 120.

Greetings,
Bertus

 

OK, thanks,

but the 10K resistor allows slow charge and discharge of the capacitor by the CMOS 4029 (making the led fading)
if i lower the resistor value to 1K, the LED will fade much shorter - which i dont want.

you mean i should keep the 10K resistor, and that a TIP120 will do better ?
i just have to replace the 2n3904 with it ? or it has to be driven somehow ?

thanks

 

Hello,

The gain of the TIP is higher, but also the voltage drop will be higher.
You can try the TIP and measure the voltage drop when the 10 K has high input.
Then you can recalculate the resistors for leds.

Greetings,
Bertus

 

I'm driving power LEDs,
so if there's a voltage drop from 12v it doesnt matter,
even if i end up with 6v, that would be fine, but i want to get the full current possible.

how do you explain that:
when i check with the led connected to the mosfet or 2n3904 it draws too small current,
but when i check the current just without the led (like short circuit between 0v and 12v through the mosfet, it gives me over 1A.

that means that the mosfet/2n3904 gate is "ON" enough and allows over 1A,
but why doesn't the led draw its calculated current.

when i think of it -
the led resistor is calculated to let 400mA at 12v,
but i didnt measure the voltage at the MOSFET output.
maybe that drop makes the lose of 200mA out of the 400mA ??
according to reverse calculation, to have 200mA at the same LED voltage, that means the source voltage is 7v.

does it makes sense that the MOSFET i used (IRF510) dropped from 12v to 7v ?

if it is the case,
can i re-calculate the led resistor on 7v source basis ? then it will give me the full current at 7v ?

thanks !

 

Hello,

When driving an emittor follower, the output voltage will always be lower as the supply voltage,
You have the voltage drop on the base resistor (I base X resistor value) and the base - emittor junction (about 0.7 volts).

An other way is using PWM to have the led dimmed.
There are voltage controlled PWM circuits around.

Greetings,
Bertus

 

thanks.

the PWM looks very interesting !
this is the best i found:

this PWM is controlled by P1 dimmer.
supposing i still use a 4029 cmos to drive a capacitor,
how can i convert the PWM driver to be controlled by the voltage of the capacitor (driven by the cmos) ?

Thanks !

 

Hello,

Take a look at fig 4.3 of this post:
http://forum.allaboutcircuits.com/showpost.php?p=117641&postcount=6

There the 555 is used as oscillator and an opamp as comparator to create the PWM.

Greetings,
Bertus

 

as i see there the op amp uses to make the square wave of the 555 more spiky/triangle.
why is it better than square waves ?
i thought PWM idea was just ON\OFF, so why do we need to soften the edges of sq waves ?

and 2nd question:
the R5 pot will set the duty sycle\led intensity ?
it seems like it will set the threshold, no ?

i want to set the duty cycle, the led intensity, by a voltage...

thanks again

 

Hello,

By design the 555 creates a triangle signal between 1/3 and 2/3 of the powersupply voltage.
The comparator has a reference voltage between these levels.
On the other leg of the comparator there is the triangle voltage of the 555.
On the output the PWM signal will vary with the position of the potmeter.
Of course you can give a voltage externaly to control the PWM, on stead of the potmeter.
The voltage must be between 1/3 and 2/3 of the powersupply-voltage.

With a PWM signal the on / off ratio determines the brightness of the led.

Greetings,
Bertus

 

OK,

so if we look here:

R5 determinate the duty cycle ? (PWM brightness)
R1 only determinate the frequency ? (which should be above 33Hz)
did i get it right ?

in other words, the + of the comparator (between 1/3 and 2/3 Vcc) will set the led brightness ?

what is the difference between
using this 555 with op amp
and
use only a 555 like here -
can i drive the pwm with a voltage without an op amp ?

 

Hello,

With the 555 only version you can not control it with an external voltage.

Greetings,
Bertus

 

thanks.

the PWM looks very interesting !
this is the best i found:

this PWM is controlled by P1 dimmer.
supposing i still use a 4029 cmos to drive a capacitor,
how can i convert the PWM driver to be controlled by the voltage of the capacitor (driven by the cmos) ?

Thanks !

This design shows pin 7 connected, which works, but not the best for this kind of PWM.

You can go as low as 120Ω for R2, but you'll need a 2W resistor. Try a 470Ω ½W, or two 1KΩ ¼W resistors in parallel.

I think circut 1 is better.