Let's have a look at the horizontal deflection circuit.

When a positive voltage is applied to the base, the Horizontal Output Transistor (HOT) turns on. This means that the full power supply voltage will be across the deflection coil Ld. This will cause the current in Ld to rise according to the law dI=(V/Ld)dt. In our case V = 150V and Ld = 1mH, so the current in Ld will rise by 150000A per second! The electrons will now move from the center to the right. When the electrons arrive at the right end of the screen, the transistor must be turned off again. The on-time of the transistor is equal to half scan time. Or actually, it's somewhat less than half scan time, since there's also some time needed to move the electron beam from the right side to the left for the next line. This time is called the flyback time, and costs about 10us, leaving 54us to write one line. This means that the HOT must be turned off after 54us/2 = 27us. The coil current has inceased to (150V/1mH)27us=4.05A. After turning off the transistor, there will still be a current flow in Ld. This curent cannot flow through the transistor anymore, nor can it flow through the reverse biased diode. The current can only flow into capacitor Cfb. When Ld has transferred all its energy to Cfb, the current will be 0.

Let's calculate the voltage across Cfb. The energy in a coil equals
0.5LI^{2}. And for the energy in a capacitor we
can write: E = 0.5CV^{2}. Since all energy is
tranferred from the coil to the capacitor, we can write:
0.5LdI_{Ld}^{2} =
0.5V_{Cfb}^{2}. The voltage
across the capacitor will therefore be: V_{Cfb,max} =
I_{Ld}√(Ld/Cfb). Of course, we also need to add the
150V supply voltage: V_{Cfb,max}=150 +
I_{Ld}√(Ld/Cfb) = 150 + 4.05√(1m/10n) = 1431V.

The capacitor will now discharge through the coil. The current flow will be in opposite direction and will therefore be negative. When Cfb has transferred all its energy to Ld, the voltage across Cfb will be zero. At this point, the current in Ld will be -4.05A (assumed that all components are ideal). Ld wants to transfer its energy to Cfb again. The voltage across it will therefore be negative. This will cause the diode to become forward biased, so the voltage across it will be (almost) 0V. Again, a fixed voltage exists across the coil, and the current will start to increase (become less negative) according to the law dI=(V/Ld)dt. By the time the current becomes 0, the HOT has been turned on again. So, the voltage across the deflection coil remains 150V and everything starts all over again.

An appropriate value for Cfb can be easily calculated. We already
saw that Cfb and Ld make a tuned circuit. A half period takes 10us. So the
period time is 20us. We know that f = 1/(2π√(Ld∙Cfb)), so T = 2π√(Ld∙Cfb).
Hence, Cfb =
T^{2}/(4π^{2}Ld). In our
case Cfb = 10nF.

If all components were ideal and the picture tube would be a perfect sphere, the deflection circuit would be as simple as described above. Of course, reality is different.