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| Questions about infrared LEDs for remotes |
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| Peabody:
My completed remote works, kinda, but the IR isn't getting through as well as it should, and some commands are missed at the receiver. And when I look at the transmitting remote in my camera's live view, I barely see anything, whereas the original Roku remote LED is plainly visible. So I'm not putting out enough IR. Attached is the circuit I'm using, which includes a Darlington. The LED itself is a mystery because I took it from an old remote, which worked fine in its day. The 22Ω resistor may still be way too high, but I have a question about the use of a Darlington here. It seems to me that while current needed from the I/O port to drive the LED is smaller with a Darlington than with a regular NPN, the total current actually flowing through the LED is less with the Darlingto than with the NPN, provided the NPN's gain is enough to fully saturate it from base current provided by the port. The reason is that the Darlington has a saturation voltage of one junction drop - call it 0.6V, and the NPN is close to zero volts. If the power supply is 4.2V, and the LED voltage drop is 1.2V, then I should get 109mA through the LED with the Darlington, but 136 mA with the NPN - something like a 2N2222A. with good gain. I have more base current than with the Darlington, but still well within the output of the port pin, and I don't dissipate power in the Darlington. So in cases where the I/O current is sufficient to drive an NPN, wouldn't the NPN be the better choice? If I/O current isn't enough, maybe something like oPossum's Darlington-like configuration of two NPNs would still be better than a Darlington. I don't know. Maybe if you reduce the value of the resistor to get the same current, you've just moved some resistance from the resistor into the Darlingon's effective resistance, and you're no worse off other than some of the heat is generated in the Darlington instead of in the resistor. oPossum's circuit: https://www.eevblog.com/forum/projects/questions-about-infrared-leds-for-remotes/?action=dlattach;attach=836991;image |
| Benta:
Do it like this. Simple and reliable and with constant current drive for the IR LED. Use two standard small-signal diodes like 1N4148. It even works with different supply voltages or varying battery voltage. |
| Kleinstein:
It is not the very best way to set the current via the base current, but this is still better than a small series resistor at the collector side with a varying supply. For up to 500 mA-1 A one can get away with a normal transistor (e.g. BC635, BD139 or similar) and set the base current accordingly. So the base resistor may have to go down to some 330 Ohms or the like. The right values depends on the transistor gain and can vary a little. The way with the resistor at the emitter is a valid option, but it looses extra voltage and the supply might be tight: some 1.5 V for the LED (at 1 A), 1-1.5 V from at a Darlington and only some 3.2 V for an rather empty cell. With weak batteries or thin cables the buffer cap should be reasonably low ESR. IT may take way more then 220 µF. For good efficiency the PWM ratio in the 38 kHz modulation should be less than 50%, more like 20-35%. The last bit around 50% has near zero efficiency. If nothing else help, one can consider 2 IR LEDs. Some remotes have quite small opening angles and one may have to really point well to the sensor. With large TVs this means knowing where the sensor actually is. |
| Benta:
Kleinstein, it's unclear to me what you are replying to. If it's this: "Do it like this. Simple and reliable and with constant current drive for the IR LED. Use two standard small-signal diodes like 1N4148. It even works with different supply voltages or varying battery voltage." Then I have the following comments: 1: the BC327-40 is NOT a darlington and has a minimum current gain of 250. 2: it can handle up to 800 mA. 3: the VCE is negligible, being less than the variation in IRLED forward voltage. 4: the emitter resistor drops 0.65...0.7 V, leaving around 2.5 V for the IRLED: plenty. 5: PWM is completely irrelevant, the 38 kHz modulation pattern is the defining factor. 6: 1 A is completely over the top for IRLED current, which is why I suggest around 100 mA. It can be altered easily by changing the emitter resistor (and, if necessary, the base resistor as well). Less current might work just as well, a practical experiment is necessary. |
| oPossum:
--- Quote from: Peabody on September 22, 2019, 07:22:25 pm ---The 22Ω resistor may still be way too high, but I have a question about the use of a Darlington here. It seems to me that while current needed from the I/O port to drive the LED is smaller with a Darlington than with a regular NPN, the total current actually flowing through the LED is less with the Darlingto than with the NPN, provided the NPN's gain is enough to fully saturate it from base current provided by the port. The reason is that the Darlington has a saturation voltage of one junction drop - call it 0.6V, and the NPN is close to zero volts. --- End quote --- You can go lower on the 22 ohm, much lower. You are correct about the vce losses with a darlington. That is why I connect the collector of the first transistor to the supply rather than the collector of the second transistor. |
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