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When I put my IR remote near any AM radio and push any button on the remote, I can hear a sound from the radio speaker (like beeping). This phenomenon is very weird for me because the radio has no IR receiver inside.

On the other hand, the frequency of the AM radio is more than 530 kHz but the frequency of the IR remote is usually only 30 to 38 kHz.

Furthermore, the human ear can't sense frequencies higher than 20 kHz but the frequency of the IR remote is more than 30 kHz.

So, I'm wondering why do AM radios react to IR remotes?

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    \$\begingroup\$ Man you should try putting your radio next to a calculator or computer! I used to do that all the time as a kid. \$\endgroup\$ – MooseBoys Aug 29 at 18:46
  • \$\begingroup\$ You can transmit music from your computer to a nearly AM radio by running a carefully programed sequence of code designed to generate modulated electromagnetic interference. retrocomputing.stackexchange.com/questions/9634/… \$\endgroup\$ – 比尔盖子 Sep 4 at 9:52
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This IR signal is indeed ignored by the AM radio. However, an AM radio is very sensitive the radio waves (yeah, DUH! ;-) )

When the IR remote operates (you push a button) the chip in the remote will switch on a clock resonator circuit which it needs to generate the IR signals. I have seen most IR remotes using a 455 kHz resonator. This is simply used because it is cheap.

The IR remote chip has a circuit to divide down this frequency to get the 38 kHz it needs. A division by a factor 12 would do as 455 kHz / 12 = 37.9 kHz. Yes that is "close enough" as the IR receivers aren't that accurate, they cannot distinguish between 38 kHz and 37.9 kHz. Also, that isn't needed, the 38 kHz is just a "carrier" it does not contain information.

So we now have 38 kHz which is a signal that has a square wave shape when it comes out of the IR remote chip. This is because this is simple (logic circuitry works with square wave signals) and the IR LED needs to be on or off. So there is no need for "in between" levels.

Now a property of a square wave signal is that it does not only contain a single frequency (like 38 kHz), it also contains many multiples (mostly uneven harmonics) of that frequency as well so: 2 x 38 kHz = 76 kHz, 3 x 38 kHz = 114 kHz, ... 14 x 38 kHz = 532 kHz. There you go, the 14th harmonic is already on a frequency the AM radio can receive!

Never underestimate the harmonic content of switching and square wave signals. I once worked on a product where the 238th harmonic of a DCDC converter running at 600 kHz was disturbing the receiver which was working at 142.8 MHz!

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    \$\begingroup\$ Additionally, the actual data modulated onto the 38kHz comes framed at a quite low, audible frequency, which would contribute to the beepy sound of it: Example frame timing. \$\endgroup\$ – Richard the Spacecat Aug 28 at 14:35
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    \$\begingroup\$ And IR signals have a tolerance of 10%. I've worked with universal remotes and it is possible to tweak the carrier frequency to fall within the tolerance of different brands and have the same remote control both devices. \$\endgroup\$ – Nelson Aug 29 at 0:47
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    \$\begingroup\$ It could also simply be a 455kHz oscillator in the IR transmitter getting picked up in the (probably 455kHz) IF stage of the AM rx. \$\endgroup\$ – peeebeee Aug 29 at 13:55
  • \$\begingroup\$ A square wave contains no 14th harmonic. \$\endgroup\$ – richard1941 Aug 30 at 19:07
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    \$\begingroup\$ @richard1941 A pure 50% duty cycle square wave indeed does not contain the 14th harmonic, but what about a 49.99% duty cycle square wave? I challenge you to find a device/circuit which produces such a pure square wave that the 14th harmonic does not exist. In my opinion, such a device/circuit only exists in theory. In practice there will always be some 14th harmonic. But if you can prove me wrong please do :-) \$\endgroup\$ – Bimpelrekkie Aug 30 at 20:07
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Most likely, your radio is picking up unintended EM radiation from the remote's circuitry. You mention that it operates between 30 and 38KHz, but the IR probably uses square wave modulation, so you'll still pick up the harmonics. Of course, it could be some other signal than the LED drive getting picked up.

Once you have a signal or harmonic near the frequency your radio is tuned to, the radio will heterodyne it down into the audio band. Try it with a calculator, those can be even more amusing if you have a noisy one.

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  • \$\begingroup\$ Back in the early days of personal computers, we put AM radios next to them and generated (crappy) music using a similar process. \$\endgroup\$ – Barmar Aug 29 at 14:28
  • \$\begingroup\$ @Barmar: do you have any links about this? Did it have a name? \$\endgroup\$ – stib Aug 31 at 10:55
  • \$\begingroup\$ @stib The only thing I can find is a random comment on retrocomputing.se: retrocomputing.stackexchange.com/questions/9634/… \$\endgroup\$ – Barmar Sep 1 at 0:48
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you've got 2 nanosecond edges inside the Remote.

2nanosecond edges are so fast, they serve as FINE IMPULSES to most circuits.

Thus the AM radio circuits are getting hit with tiny lightning bolts, and ringing, and you hear that.

"its safe to say they do not contribute with any EMI" though clearly the impulses do contribute, because the activity can be heard. An AM radio with 10KHz bandwidth (double sideband) has noise floor of -174dBm/rootHz + 10dB Noise Figure in frontend transistors + 40dB boost in noise floor from noise power being proportional to bandwidth, = -174 + 50 == 124 dBm. With 0dBm across 50 ohms being 0.632 volts PP, and -120dBm being 1 million times lower in voltage, the floor of detectability is about 0.6microVolts. Or 0.0000006 volts; now you want to bet on 5 volt MCU logic transitions NOT being detected by an AM radio, these receivers being notorious for static susceptibility.

So now we have some science, some actual math and physics, behind why the IR REMOTE can be detected by an AM RADIO. Neat, eh?

Now for some details on coupling between the IR Remote and the AM radio:

The remote will have several centimeters of PCB trace from the MCU to the LED driver transistor, which spits out 0.1amp or 0.2 amp currents for the LED, limited by 5 ohm or 10 ohm resistor. Into the transistor base will be 10mA with 2nanoSecond edges. From the collector will be 100mA (SWAG) with fast fall and slow rise (as the transistor exits saturation slowly). These currents may magnetically couple into ANY circuit loop inside the AM radio.

However, lets just think about capacitive coupling.

The AM radio is of non-zero size and we will assume several centimeters of PCB trace that are capacitively coupled to the IR remote.

So lets model these PCB traces: 2cm long 1mm wide, 2cm apart.

C = Eo * Er * Area/Distance = 9e-12 Farad/meter * 1(air) * (2cm * 1mm)/2cm

C = 9e-12 * 1mm = 9e-15 ~~ 1e-14 farads. [this ignores fringing & alignment]

Now lets compute a displacement current (the current generated by charging and discharging, by changing the electric field flux), between IR remote and the AM radio.

Q = C * V; and we differentiate to get dQ/dT = dC/dT * V + C * dV/dT

now assume constant C (thru the air) and we have dQ/dT = C * dV/dT = Icurrent

Our injected (by changing electric field) current is

I == 1e-14 Farad * 3 volts / 2 nanoseconds

I ~~ 1e-14 * 1/nano == 1e-5 amp = 10 microAmps injected into the AM radio

Assume the impedance of the node is 1,000 ohms. Use Ohms Law, and you get

10uA * 1Kohm = 10 milliVolts.

And either the AM tuned circuits can ring, with this 2 nanosecond impulse, or a higher harmonic (per Bimpelrekkie) may enter thru the antenna.

================== Now for magnetic coupling ===========

2 nanosecond edges are plenty fast for skin-effect in copper planes to cause some magnetic shielding and thus attenuation of the induced voltage.

We'll assume there is NO attenuation by planes, and just compute the worst-case induced voltage in the AM radio circuits.

As with the Efield coupling, assume 2 centimeter spacing between aggressor and victim. And assume the victim (the AM radio) has 2cm by 2mm loop. And assume worst-case alignment.

The relevant equation (ignoring some natural-log terms for easy math) is

Vinduce = [ MUo * MUr * Area/(2*pi*Distance) ] * dI/dT

where we'll assume dI/dT = 10 milliAmps / 2 nanoSeconds

Using MUo = 4*pi*1e-7 Henry/meter and MUr = 1(air, copper,FR-4, etc) we have

Vinduce = 2e-7 * Area/Distance * dI/dT

Vinduce = 2e-7 * (2cm * 2mm)/2cm * 0.01amp/2nanoSecond

Vinduce = 2e-7 * 0.002 * 0.01/2nano

Vinduce = 2e-7 * 2e-3 * 1e-2 * 0.5 * 1e+9

Vinduce (I don't have a clue how big/small this will be, until math is done)

= 4 * 0.5 * 1e(-7-3-2+9) = 2e(-12+9) = 2e-3 = 2 milliVolts magnetic coupling

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    \$\begingroup\$ I didn't downvote, but the currents involved in those ns switching times are so low, and the traces so insignificant in size, that it's safe to say they do not contribute with any EMI. \$\endgroup\$ – a concerned citizen Aug 29 at 14:15
  • \$\begingroup\$ microcontrollers will switch 0.1 amps in that 2 nanoseconds. And the detector for the impulse is........................ a radio. Nothing more sensitive than a narrow-band radio that is hit by an impulse. \$\endgroup\$ – analogsystemsrf Aug 29 at 17:50

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