# Diodes In Parallel

If the load current is greater than the current rating of a single diode, then two or more diodes can be connected in parallel (see Figure 1) to achieve a higher forward current rating. Diodes connection in parallel do not share the current equally due to different forward bias characteristics. The diode with the lowest forward voltage drop will try to carry a larger current and can overheat. Figure 2 shows the V-I characteristics of two diodes. If these two diodes are connected in parallel at a given voltage, a different current flow in each diode. The total current flow is the sum of *I _{D1}* and

*I*. The total current rating of the pair is not the sum of the maximum current rating for each but is a value that can be just larger than the rating of one diode alone.

_{D2}Parallel diodes can be forced to share current by connecting a very small resistor in series with each diode. In Figure 3, the current sharing resistor R establishes values of *I _{D1}* and

*I*that are nearly equal. Although current sharing is very effective, the power loss in the resistor is very high. Furthermore, it causes an increase in voltage across the combination. Unless using a parallel arrangement is absolutely necessary, it is better to use one device with an adequate current rating.

_{D2}The value of the current sharing resistor can be obtained as follows.

*V = V _{D1} + I_{D1} x R = V_{D2} + I_{D2} x R*

Solving for R,

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

The power dissipated in R is

*PR = I ^{2}_{D1} x R + I^{2}_{D2} x R*

The voltage across the diode combination is

*V = V _{D2} + I_{D1} R = V_{D2} + I_{D2} R*

**Example**

Two diodes having the characteristics shown in Figure 3 are connected in parallel. The total current through the diodes is 50A. to enforce current sharing, two resistors are connected in series. Determine:

- The resistance of the current sharing resistor, so that the current through any diode is no more than 55% of I
- The total power loss in the resistors
- The voltage across the diode combination (V)

**Solution:**

**a. With forced current sharing, such that**

*I _{D1} = 55% x 50 = 27.5 A*

*I _{D2} = 50 – 27.5 = 22.5 A*

We obtained from Figure 2

*V _{D1} = 1.3 V*

*V _{D2} = 1.6 V*

*V = V _{D1} + I_{D1} x R = V_{D2} + I_{D2} x R*

*= 1.3 + 27.5 x R = 1.6 + 22.5 R*

Solving for R,

R = 0.06 Ohm

**b. The power dissipated in R is**

*P _{R} = I^{2}_{D1} x R + I^{2}_{D2} x R = 27.5^{2} x 0.06 + 22.5^{2} x 0.06 = 75.8 W*

**c. The voltage across the diode combination is**

*V = V _{D1} + I_{D1} R = V_{D2} + I_{D2} R*

= 1.3 + 27.5 x 0.06 = 1.6 + 22.5 x 0.06

= 2.95 V

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