# Series Operation of Diodes

The maximum power that can be controlled by a single diode is determined by its rated reverse voltage and by its rated forward current. In high power applications, a single diode may have insufficient power handling capability. To increase power capability diodes are connected in series.

## Series Connection of Diodes

In very high voltage applications, the reverse voltage rating of a single diode may not be sufficient. A series connection of two or more diodes (See Figure 1 below), is then used to increase the voltage rating. However, the reverse voltage may not be equally divided between the two diodes, the diode with the lower leakage current can have excessive reverse voltage across it. Even if we use same number type diodes, their V-I characteristics may not be identical, al illustrated in Figure 2. The current rating of the diodes in series is the same as the current rating of one of the diodes. In the reverse direction both series diodes have the same reverse leakage current but as shown, the have different values for reverse voltage. In such a case diode *D _{1}* may exceed its reverse voltage rating.

**Figure 1: Series Connection of Diodes**

**Figure 2: V-I Characteristics**

Forced voltage sharing can be obtained by connecting voltage sharing resistors of appropriate value across each series diode. Figure 3 shows the effect of placing resistors across the diode. To be effective the resistors must conduct a current much greater than the leakage current of the diodes. These sharing resistors will consume power during revere bias operation, so it is important to use as large a resistance as possible.

In addition, there can be excessive reverse voltage across a diode due to different reverse recovery times. A capacitor connected in parallel with each diode (see Figure 4 below) will protect the diode from voltage transients.

The value of the voltage sharing resistor can be obtained as follows:

The source current is

*I _{S} = (V_{D1} / R) + I_{D1} = (V_{D2} / R) + I_{D2}*

Solving for *R*,

*R = (V _{D1} – V_{D2}) / (I_{D2} – I_{D1}) ------------------------------------------------- Equation 1*

The power dissipated in *R* is

*P _{R} = I^{2}_{R1} x R + I^{2}_{R2} x R*

**Figure 3:**

**Figure 4:**

**Example 1**

Two diodes with voltage ratings of 800 *V* and reverse leakage currents of 1 mA are connected in series across an AC source whose peak value is *V _{s(max) }*= 980 V. the reverse characteristics are as shown in Figure 2 Determine

- The reverse voltage across each diode
- The value of the voltage sharing resistor, so that the voltage across any diode is no more than 55% of
*V*_{s(max) } - The total source current and power loss in the resistors

**Solution:**

**A.** With no force sharing, the current through the diodes is the leakage current. Therefore, at 1 mA, from Figure 2

*V _{D1} = 700 V*

*V _{D2} = 280 V*

**B.** With forced voltage sharing, such that

*V _{D1} = 55% x 980 = 539 V*

*V _{D2} = 900 – 495 = 441 V*

We obtain from the graph

*I _{D1} = 0.7 mA*

*I _{D2} = 1.4 mA*

Using Equation 1

*R = (V _{D1} – V_{D2}) / (I_{D2} – I_{D1})*

*R = (539 V – 280 V) / (1.4 mA – 0.7 mA)*

*R = 140 K*

**C.** The current through R is

*I _{R1} = 539 / 140K = 3.85 mA*

*I _{R2} = 441 / 140K = 3.15 mA*

Source current = 0.00385 + 0.0007 = 4.55 mA

Or

Source current = 0.00315 + 0.0014 = 4.55 mA

The power dissipated in R is

*P _{R} = I^{2}_{R1} x R + I^{2}_{R2} x R = 2.1 + 0.44 = 2.54 W*