Hi,

I have a circuit that drives 5 leds in series. The Leds are turned on and off via a P-channel Mosfet with Low RDS On.
The gate is connected to a microcontroller. The Led series have a current limiting resistor as normal.

My question is: if I drive the mosfet gate with a PWM signal (1Khz for instance) with a maximum duty cicle of about 95% (so the Leds never get fully on all the time) it will be possible not use the current limiting resistor?

I'm aware of the avalanche effect but id the led only stays on for a limited time with will be possible without destroy the LEDs in a short time?

Thanks

Joaquim

 

No you cannot because a slight variation in the voltage can cause the LED current to exceed the maximum instantaneous current rating of the LED and burn it. Better to use a current limiting resistor or a MOSFET with a saturation current of less or equal to the operating LED current.

 

I agree with mik3. Some means of limiting the currrent to each LED should alway be present. You can use a series resistor or you can use a constant current driver that is set to the proper drive current.

hgmjr

 

Thanks

And if I use Bipolar transistors with a base current that put the transistors in the active area (just below the saturation point) ? That will limit colector current right?

Joaquim

 

That will limit the current but not reliably since the beta of the transistor and thus the collector current will change with the temperature. Beta will also vary from device to device.

You could consider using a current-mirror but I sense that your desire is to reduce the number of components used rather than increase the component count as the current mirror circuit would do.

A current limiting resistor in series with the LED is the simpliest means available.

hgmjr

 

Yes.. IC = Beta * IB.

Some devices can have huge diferences on gain.

You are right. I will keep the design as is.

Thanks all for the anwsers.

Joaquim

 

每个LED一般用3V电压（drive voltage), 5 leds in series，i.e , 3*5=15V。而单片机的工作电压一般为3.3V~5V，你又用P-channel MOSFET, 那么，要让它正常工作就得用到负压。
那么怎么设计电路呢？
1、不要用P-channel MOSFET，用N-channel，因为这样便宜一点.
2、用并联，别用串联。如果要让串联时正常工作，就必须加入升压电路（boost)，这样一来，电路成本就高了。
3、三极管(or mosfet)工作在开关状态，所以必须要限流电阻(limiting resistor)

 

每个LED一般用3V电压（drive voltage), 5 leds in series，i.e , 3*5=15V。而单片机的工作电压一般为3.3V~5V，你又用P-channel MOSFET, 那么，要让它正常工作就得用到负压。
那么怎么设计电路呢？
1、不要用P-channel MOSFET，用N-channel，因为这样便宜一点.
2、用并联，别用串联。如果要让串联时正常工作，就必须加入升压电路（boost)，这样一来，电路成本就高了。
3、三极管(or mosfet)工作在开关状态，所以必须要限流电阻(limiting resistor)

Hey,
This is an international forum so you have to talk in english.

 

Each LED in general use 3V voltage (drive voltage), 5 leds in series, ie, 3 * 5 = 15V. MCU operating voltage and is normally 3.3V ~ 5V, you have to use P-channel MOSFET, then, should it have to use negative pressure to work.
So how does the design of the circuit?
1, do not use P-channel MOSFET, using N-channel, because it cheaper.
2, using parallel, do not use in series. If you want to work when the series, it is necessary to join the step-up circuit (boost), this way, the circuit on the high cost.
3, transistor (or mosfet) state in the switch, it must be current-limiting resistor (limiting resistor)

Here is a translation courtesy of Google Translate.

hgmjr

 

My mistake. I use N-Channel not P-channel.

The anode are connected to a 12V supply, the leds are in series (with the liminting resistor) and the cathode is switched to the ground via the mosfet.

A simple 78L05 takes the 12V down to power up the Micro.

Sorry for the mistake.

 

Name-brand LEDs have a maximum allowed instantaneous peak current at a certain pulse width spec'd. Their max forward voltage is listed at 20mA but not at the very high current pulse.

My red LEDs have a max allowed continuous current of 40mA and a max peak current of 200mA (at f= 1kHz, duty factor= 1/10).
Their forward voltage at 20mA is 1.5V to 2.4V with a typical voltage of 1.7V. A graph shows the typical forward voltage at 2.8V at 200mA but the temperature is not listed.