Remote control 4: the emitter
== 4 == BUILD THE EMITTER
a – Pick the headphone cable
b – connect one led between left and right channel (don’t care GROUND)
c – connect the other led between left and right channel, but the opposite direction
Your emitter should look like this:
Please note the opposite orientation of the two leds!
You should connect them to your headphone cable this way:
Ok, your emitter is ready.
Unfortunately, as it is it will be very weak, and it will have just a few centimeters range! To get a suitable range you need a suitable power; and to get more power than the phone can provide, you need an amplifier.
An amplifier is an electronic circuit which uses an external power source to add some power to a weak electric signal; you can build a simple one using just a few components: 5 resistors and 1 transistor:
The resistors are the component which have a “xxx kOhms” label; the transistor (here, actually, a “double transistor”, properly a “darlington”) is in the circle.
The external power here comes from a 9V battery (bottom left in the picture) .
Let’s divide the circuit in parts to better understand it.
- We have a voltage divider: this set of components divides the voltage in parts, depending on how it is built, according to formula:
Vout = Vin * R2/(R1+R2)
If R1 = R2, Vin is exactly splint in 2 equal parts: Vout = Vin / 2
This means that in our circuit the voltage in Vout will be 4.5V if we use a 9V battery.
Why do we need this voltage divider? Because the input signal coming from phone to our amplifier will have positive and negative values (its carrier is a sinusoid), so if the base of the transistor was biased at 9V, transistor wouldn’t be able to amplify positive parts of the wave, being it at higher voltage than the battery; if it was biased at 0V, negative parts of the input wave would be lost; we need “the right in the middle value”: 4.5V
- Now we can add the transistor, which will get Vout as input to its base:
- In case the current/signal coming from the phone was too high, we need a limiting resistor on the input; I don’t know how actually high can the phone output voltage be, anyway the current that will flow trhough transistor will be Iin = Vphone/R3. This current will just activate and drive the transistor, it does not flow through the speaker:
- The current which flows in the speaker is determined by the battery voltage, not by the phone voltage; the power which the transistor must handle due to surrounding components is given by P = V^2 / Rs, where Rs is the internal resistance of the speaker, which we do not know:
but we know from datasheet that the MPSW45A cannot tolerate more than 1W (Pd value); to get Pd<=1 we must have V^2/Rtot <=1 ,i.e 81<= Rtot , where Rtot is the total resistance given by sum of speaker resistor and additional R4: we get that it must be Rtot = Rs + R4 >=81 ohm , and Iout would be Iout = V/Rtot <= 9/81 = <=111 mA:
for this current flowing through the transistor we get around the maximum amplification, as we can see from datasheet:
In “phone world”, 5V is a most common voltage (USB charger voltage); what does it happen by applying 5V to this circuit rather than 9V?
The original site the schematic comes from does not explore this case… but I assume that we would get:
- Rtot >=V^2 –> Rtot >=25
- Vbase = 2.5V
- P = 25/Rtot
- 25/81= 0.3 = 1/3
This means that by sure the transistor won’t get damaged, and it also probably means that we’ll get an output power wich is 1/3 of what we get using 5V, assuming we use same R4. To get same power of the 9V-powered amplifier, I think we’d need a total resistance Rtot >= 25 Ohm; R4 value is given by: R4 = Rtot – Rs = 25-Rs.
>> NEXT- REPLAY THE WAVEFORM: (in this same page)
a – Plug the emitter into HEADPHONE output of your audio card
b – Position the two leds just in front of your device
c – Press PLAY in Audacity: your device should react to the command you previously sampled.