The forward voltage for an IR LED is much lower than for a visible light LED, typically around 1.3 V, but rising if you push real high currents through them, like > 100 mA. There seems to be no reason why you couldn't place two of them in series, especially if your Vcc would be 5 V. If your Vcc comes from a pair of AA batteries though, the voltage drop of two LEDs + the transistor's saturation voltage may come close to Vcc and that could limit the output current.
The two outputs to drive the four LEDs are to avoid overloading the microcontroller's output. Or better, should avoid overloading. A 120 Ω resistor means 35 mA base current per transistor, and that's too much already for the AVR, let alone the 70 mA which it will draw now.
The 2N3904 is not a good transistor for this either: it's only rated at 100 mA and the low hFE necessitates the high base current. A BC337-40 has an hFE of minimum 250 at 100 mA collector current, then 5 mA base current should be enough to drive it. A base resistor of 820 Ω will allow you to drive all four resistors from 1 pin. The BC817 is also rated at 500 mA.
Alternatively you could use a FET to drive the LEDs. A PMV20XN can handle several amperes and has an on-resistance of only 25 mΩ so it will dissipate hardly any power. 1.5 V gate voltage is sufficient for 2.5 A.
edit
A note about current limiting. Usually we'll have a resistor in series with the LED for that, but if you look at the schematic of a commercial remote control that resistor is often missing, because they count on the batteries' internal resistance for that, and then they save another 0.001 dollar per remote controller.
This is not a good idea if you power from a mains powered voltage regulator. That will limit the current, but at a too high level, and if it doesn't destroy the LED immediately it will severely limit its lifetime. So a small series resistor is recommended. At 5 V supply and 2 LEDs in series you'll have a voltage drop around 2.9 - 3.0 V, so for 100 mA you need a 30 Ω resistor. Peak power will be 300 mW, but at a 50 % duty cycle average power is only 150 mW, then a 1/4 W resistor will do.
Here you will find a good explanation using a PNP transistor instead of a NPN, but the way it works is the same. In what follows I will describe the general idea.
To understand how the feedback occurs look at the picture below. The red U is the fork (for illustrative purpose). Each leg of the fork has a permanent magnet that interacts with the drive coils and the feedback coils. When it is oscillating around its equilibrium position, the maximum speed will be achieved when the legs are passing through that equilibrium position, like a mass on a spring. It is during maximum speed that the induction will be maximum.
Now suppose that the capacitor is initially discharged, then the voltage from Vdc will bring the base voltage above the emitter voltage and the transistor will turn on. The current that will flow through the drive coils D1 and D2 will be enough to start the oscillations. Meanwhile, the base current will start to charge the capacitor. In steady state the voltage induced in the feedback coil F1 will be turning the transistor on and off, with the capacitor barely discharging during the off cycles (C*R >> 1/(frequency of oscillations ).
Suppose the the fork is going through the point of equilibrium, then it will induce a voltage in F1, with the sign depending on the direction of movement. Only one of those peaks of voltage will turn the transistor on, the one which brings the base voltage above the emitter's. When that happens the current through D1 and D2 will sustain the fork oscillation by creating a magnetic field that 'pushes' the permanent magnets in the fork.
Next, suppose that the oscillations grow larger in amplitude and let's see how the circuit corrects it. First, a higher voltage will be induced in F1, but that shouldn't matter when the transistor is fully on (saturated). On the other hand, a higher voltage will also be induced in D1 and D2 and it will be such that it will tend to oppose the current drawn by the transistor through the collector. The result is that the driving current (collector current) decreases, and that means a smaller magnetic field and consequently a smaller "push" to the fork. Therefore the oscillation amplitude will decrease.
Another way by which the circuit corrects itself is through the capacitor C. If the oscillations get larger then a higher voltage will be induced in F1, but if the transistor is not fully saturated then a higher current will flow through the base in response to the higher voltage in the base with respect to the emitter. That current will charge C during the ON cycle, which will begin to discharge during the OFF cycle through R. So a higher base current means a higher voltage in the capacitor at the end of the ON cycle, which means that it will be harder for the next peak of voltage to overcome that new capacitor voltage and turn on the transistor in the next cycle. Remember the RC constant is relatively big, so the charge time of C is larger than the discharge time.
I think that's all. If you are wondering how a vibrating fork could replace the escapement of a mechanical watch watch this video.
![enter image description here](https://i.stack.imgur.com/vXmB8.gif)
Best Answer
That is called an "H-Bridge."
It is often used to drive motors forwards as well as backwards.
In your case, it allows you to generate a magnetic field whose polarity and intensity you can vary using "control signal 1" and "control signal 2."
When both are high (or both are low,) no current flows through the coil.
If one is high and the other is low, then current will flow in a particular direction.
If you swap the high and lows, it will flow in the opposite direction.
Now, if you hold one steady and pulse the other you will get a pulsed current through the coil. It will be smoothed (somewhat) by the coil to a steady magnetic field whose strength is propotional to the duty cycle of the pulses.
Switching the polarity of the current also changes the polarity of the magnetic field.
That is very much a simplified description, but I think it contains enough key words that you should be able to locate more details on your own.
It is a common circuit with many uses - and plenty of tricks and traps that go into making, using, and controlling it.
A bit more on how it operates:
The key to the whole thing is how pnp and npn transistors function.
When the voltage on the base of an npn transistor is more than 0.7 volts above the voltage on the emitter, then current will flow through the collector to the emitter.
When the voltage on the base of a pnp transistor is more than 0.7 volts below the voltage on the collector, then current will flow through the collector to the emitter.
So, looking at the H-bridge, putting a high signal on one of the control signals will turn off the pnp and turn on the npn - that side of the bridge is connected to the positive supply voltage.
Now, if you put a low signal on the other control line the npn transistor will turn off and the pnp will turn on. That side of the bridge is connected to the ground.
Current can now flow from V+ on one side of the bridge, through the coils, to ground on the other side of the bridge.
So, which control signal is high and which is low dictates the direction of the current flow through the load in the middle of the bridge.
You also asked it is possible for both transistors on one side to turn on and cause a short circuit.
It can happen, and is called shoot through. Part of the design and operation of an H-bridge goes into making sure that it doesn't happen.
In the design you've posted, I don't think it can happen.
It looks to me like the transistors on each side can never be on at the same time. But, I'm not an engineer and may well have overseen something (though Tony is an engineer and doesn't think it can happen with this circuit.)