IR Remote Control Test Circuit

Circuit Description:

The IR detector circuit is an example of a narrow band tuned radio frequency receiver (TRF) and can be used to test the keys of a common IR remote control. In operation, the suspect remote control is pointed toward the IR LED which receives the IR signal and then lights a white LED. It doesn't decode the IR message and only indicates that an IR signal at 36KHz is present. The circuit has a limited range of about 2 feet.

The circuit consists of broad band first stage, followed by a tuned narrow band second stage, followed by a diode rectifier and final output stage that lights the white LED. The overall gain of the circuit is about 60dB so that just a couple millivolts from the IR LED is needed to dimly light the white LED. The IR LED will produce about 60 millivolts when the remote control is very close within a couple inches.

There is another IR remote control test circuit in the LED section which is much simplier, but it employs a IR receiver module which does most of the work and has a long range. The circuit here was designed to use a common IR LED and then amplify the tiny signal to drive an indicator. The range could be extended with subsequent tuned stages but the construction is tricky since high gain circuits tend to oscillate if things aren't laid out just right. I found I could build this circuit on a plug-in board without regard to the placement of the parts. Be even so, it will oscillate with a long wire connected to the input, or if you touch the base of the first transistor.

The first stage was chosen to operate at about 1 milliamp with a collector voltage about half of the supply voltage. So, the collector resistor will be 4.5/.001 = 4500 ohms. I rounded that off to 3900 since I had one laying around. The gain of the transistor was assumed to be around 100 so the base current will be about 10 microamps. The voltage from collector to base is about 4 volts so the base resistor is about 4/.00001 = 400K. I rounded that off to 330K since I had one of those laying around. The actual collector voltage measured 4.6.

The second stage is a narrow band tuned stage using a parallel LC circuit tuned to 36KHz and operates at about 3mA. The resonant frequency can be found with formula Fo = 2*pi*sqrt (LC), The actual resonant frequency of the LC parts (6.8mH and 2700pF) works out to 37144. The capacitor could be a little higher at 2874pF to move the frequency to 36K but I had a 2700pF laying around and used that. I did add another 200pF in parallel but it didn't have much effect. Other values of inductance and capacitane could be used but the rule of thumb is for the reactance to be about 1K to 2K. Inductive reactance is 2*pi*F*L or in this case 6.28 * 36000 * .0068 = 1537 ohms.

Most IR remotes operate at 36KHz where each bit is transmitted with maybe 30 cycles of the 36KHz carrier followed by some off time to separate the bits. Using a tuned LC circuit allowes the receiver to reject most of the noise and only respond to the desired data within a narrow band. The bandwidth is the range of frequencies that the LC circuit will respond while maintaining at least 70% of the max amplitude. So, if the tuned center frequency is 36KHz and the amplitude falls off to 70% at 34KHz and 38KHz, the bandwidth will be 38-34= 4KHz. The quality factor or Q is the center frequency divided by the bandwidth or Q = 36 / 4 = 9.

There is a compromise between the Q of the circuit and the response time. The LC circuit needs about the same number of cycles as the Q to charge to full amplitude, so if the Q were 100 and the transmitter only sent 30 cycles, the circuit would not work since it needs more than 100 cycles to charge up to full amplitude. Since we only have about 30 cycles to work with, the Q should be quite a bit less than 30.

The unloaded Q of the LC circuit itself, when not connected to anything can be found from the ratio of inductive reactance to the resistance of the inductor wire, or Q = XL /R = 1538/60 = 25.6. The actual Q will be less when the circuit is loaded and is fairly complicated to figure out. It depends on the transistor parameters, bias current and other things. I found a Q calculator that works out the Q based on LC values, inductor resistance and generator resistance. The actual Q worked out to about 9 with the assembled circuit. Using the calculator, the Q will be 9 when the generator resistance is about 22K. One methode of working out the transistor output impedance is to measure a change in collector current for a change in CE voltage and then divide the voltage change by the current change to get the output impedance. For example, if the collector current changes by 42uA when the CE voltage changes by 1 volt, Zo will be 1 / .000042 = 23.8K.

The plot at the top of the page illustrates the bandwidth between the two 1/2 power points. The left side shows the attenuation in dB where -3dB is the 70% amplitude point, or 1/2 power point. The 2 verticle white lines mark the lower and higher frequencies. The center frequency is shown at 37Khz since I used a 2700pF cap which was slightly low. Adding another 200pF in parallel will bring the center frequency closer to 36Khz.

Q Calculator for Parallel LC Circuits

The third stage is just a switch to drive the white LED when there is a charge on the 2.2uF cap. The negative half cycles at the collector of the tuned stage are rectified by the diode which charges the 2.2uF cap to the negative peak of the sine wave. The 100K resistor sets the base current at about 50 microamps when the cap is charged to 6 volts. This results in about 5 milliamps of LED current assuming the transistor gain is 100. The RC time of the cap and resistor is .000022 X 100K = 220 milliseconds which maintains the light for a couple RC times or maybe half a second after the signal ends. I didn't see any need for a series resistor with the indicator LED since the hFE gain of the PNP transistor would have to be about 250 to get 20 milliamps into the LED. But you can add a resistor if you like.

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