giftiger_wunsch
New Member
Hi,
I am experimenting with using an infra-red LED and a corresponding photodiode to make use of remote control technology. Right now it's just an experiment, so I have set up a test circuit with both the transmitter and receiver, since I need to get a couple more batteries before I can use separate power supplies. The circuit diagram is attached.
I am using a 555 Timer IC to create a digital wave with a frequency of approximately 32kHz, where the mark time (Tm) is approximately 90% of the wave, also indicated in the attached diagram. I'm using this method to ensure that the LED is off for 90% of the time, so that it does not burn out when provided with an unusually high current (hopefully around 800-1000mA) which should help to boost the signal.
Anyway, the IR LED and photodiode are in very close proximity, so it is difficult to determine the range of the signal, but the red LED lights up as expected, and using a camera with IR capabilities I have also confirmed that the IR LED is coming on properly.
The circuit seems to work with the transmitter and receiver in close proximity, and with no damage to the IR LED, but I inserted an ammeter between R3 and L1 to check whether the current reaching the IR LED is as high as desired, to make sure that the 90% mark-time wave was indeed protecting the LED from a large current, but I found that the current passing through the IR LED was around 30mA, and the voltage across it was as low as 370mV.
Does anyone know why these values are so low? Is there a resistive component inside the IC which is limiting the current? And how might I get around this problem?
Just to rule out the battery as the problem, I measured it at 8.87V and it produced about 2.2A when I connected it directly to the ammeter.
Thanks in advance for any help.
I am experimenting with using an infra-red LED and a corresponding photodiode to make use of remote control technology. Right now it's just an experiment, so I have set up a test circuit with both the transmitter and receiver, since I need to get a couple more batteries before I can use separate power supplies. The circuit diagram is attached.
I am using a 555 Timer IC to create a digital wave with a frequency of approximately 32kHz, where the mark time (Tm) is approximately 90% of the wave, also indicated in the attached diagram. I'm using this method to ensure that the LED is off for 90% of the time, so that it does not burn out when provided with an unusually high current (hopefully around 800-1000mA) which should help to boost the signal.
Anyway, the IR LED and photodiode are in very close proximity, so it is difficult to determine the range of the signal, but the red LED lights up as expected, and using a camera with IR capabilities I have also confirmed that the IR LED is coming on properly.
The circuit seems to work with the transmitter and receiver in close proximity, and with no damage to the IR LED, but I inserted an ammeter between R3 and L1 to check whether the current reaching the IR LED is as high as desired, to make sure that the 90% mark-time wave was indeed protecting the LED from a large current, but I found that the current passing through the IR LED was around 30mA, and the voltage across it was as low as 370mV.
Does anyone know why these values are so low? Is there a resistive component inside the IC which is limiting the current? And how might I get around this problem?
Just to rule out the battery as the problem, I measured it at 8.87V and it produced about 2.2A when I connected it directly to the ammeter.
Thanks in advance for any help.
Attachments
Last edited: