2025-09-11 6:25 AM
Hi Sir,
We have connected the REXT resistor with a value of 241 Ohms, which sets the current to 80 mA. I would like to confirm whether it is possible to use an external current control circuit to dim the LED.
For example, in the case of STP08CP05TTR, with VDD = 3.3V REXT = 241 Ohms, can I control the LED current from 20 mA to 80 mA using an external current control circuit to adjust the brightness?
In this setup, I plan to apply a constant 5V as the LED supply voltage.
Additionally, I would like to clarify the following:
Kindly confirm if this approach is suitable.
2025-09-11 6:35 AM
Welcome @Anjaneyulu, to the community!
It is generally NOT a good idea to adjust the brightness of LEDs via the current. The brightness not only depends non-linearly on the current, but also on parameters such as temperature and the manufacturing technology of the LEDs. In the case of "white" LEDs with a yellow-looking layer in front of the LED chip, there is also a colour location shift of the white point.
For these reasons, the variable brightness of LEDs is not controlled by current, but by PWM.
Hope that answers your questions?
Regards
/Peter
2025-09-11 6:38 AM - edited 2025-09-11 8:06 AM
I am using an IR LED with a 940nm wavelength and need to control its brightness. Can I use external current control with the STP08CP05TTR IC for this purpose?
2025-09-12 2:02 AM
You are welcome to try controlling the intensity of the IR LEDs using a variable REXT. Since you will probably want to control several LEDs, you should note that the characteristic current-brightness curves may differ, which means that the individual LEDs may have different brightness levels.
To answer your original questions:
However, I would also like to point out that the output channels of the STP08CP05T regulate the LED current linearly, i.e. they have to dissipate the difference between the voltage at the LED anode and the forward voltage of the LED (usually 1.5-2V for IR LEDs), which means that the set LED currents results in a correspondingly high power loss in the chip that has to be dissipated.
Regards
/Peter
2025-09-12 2:46 AM - edited 2025-09-12 3:10 AM
3. The LED supply voltage at the anode (VLED) is 5V, while the IC STP08CP05TTR is powered by a 3.3V VCC supply. I need to continuously adjust the IR LED current between 20mA and 80mA to control its intensity.
If I use a constant REXT = 241 Ω, then:
The forward voltage of the LED (VF) is 1.5V.
So the voltage across the LED circuit is:
5V−1.5V=3.5V
At maximum current of 80mA, the power dissipation is:
Pd=3.5V×80mA=280mW
Question:
If I reduce the current to 20mA using an external current control/driving circuit, will there be a voltage drop at the 5V (VLED) supply?
According to the functional block diagram in the datasheet, the sources of the MOSFETs are not connected to any current sensing circuitry.
Is the “I-REG” block used for generating a PWM signal, or is it for other types of control based on feedback from the current?
If it is PWM based, how can I calculate the PWM frequency? Specifically, I want to understand how the turn on time is determined and when it occurs if the control is indeed PWM-based. Because at the same particular turn on time only, i will have to take the reading from receiver.
So, how it is maintaining the constant current?
2025-09-12 4:14 AM
As already mentioned, the current per output channel is regulated linearly. Figure 2 from the data sheet only shows the output stages schematically; in reality, each of the channels is a constant current source, not PWM based.
You have calculated the power of 280mW for only one channel, so with 8 LEDs you have to multiply this value by 8. If you want to reduce the current to 20mA, you do not need an external circuit (which would not make sense anyway, because then two regulators would be working against each other), but only need to dimension the REXT accordingly, e.g. with a controllable resistor, which you can find from other manufacturers, e.g. with I2C.
However, to minimise power loss at the STP08CP05, you should consider setting the LED supply to 3.3V as well, because then 1.7V less voltage has to be burned.
Regards
/Peter
2025-09-12 4:18 AM
I forgot to mention how the linear power sources are implemented: without going into too much detail, I would just like to mention the term current mirror.
Regards
/Peter