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  1. Problems
  2. /A 9.1 V Zener diode is used in a regulator…
EasyPhysicsHigh School (11–12)Electronics › Semiconductors and DiodesNumerical

A 9.1 V Zener diode is used in a regulator…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A 9.19.19.1 V Zener diode is used in a regulator. The input is 161616 V, the series resistor is 180 Ω180\ \Omega180 Ω and the load is 1.0 kΩ1.0\ \text{k}\Omega1.0 kΩ. Find the current in the series resistor, the current in the load, the current in the Zener diode and the power that the Zener diode uses.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

mA
mA
mA
mW
Need a reminder? The key ideaShowHide

In a Zener regulator, Vout=VzV_{out} = V_zVout​=Vz​, the series current is Is=Vin−VzRsI_s = \dfrac{V_{in} - V_z}{R_s}Is​=Rs​Vin​−Vz​​, the load takes IL=VzRLI_L = \dfrac{V_z}{R_L}IL​=RL​Vz​​, and the Zener diode takes Iz=Is−ILI_z = I_s - I_LIz​=Is​−IL​. The Zener current must stay between Iz,minI_{z,min}Iz,min​ and Iz,max=P/VzI_{z,max} = P/V_zIz,max​=P/Vz​.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The series resistor. Is=16 V−9.1 V180 Ω=38.3I_s = \dfrac{16\ \text{V} - 9.1\ \text{V}}{180\ \Omega} = 38.3Is​=180 Ω16 V−9.1 V​=38.3 mA.

Step 2. The load. IL=9.1 V1000 Ω=9.10I_L = \dfrac{9.1\ \text{V}}{1000\ \Omega} = 9.10IL​=1000 Ω9.1 V​=9.10 mA.

Step 3. The Zener diode. Iz=Is−IL=38.3 mA−9.10 mA=29.2I_z = I_s - I_L = 38.3\ \text{mA} - 9.10\ \text{mA} = 29.2Iz​=Is​−IL​=38.3 mA−9.10 mA=29.2 mA.

Step 4. The power. P=VzIz=9.1 V×29.23 mA=266P = V_z I_z = 9.1\ \text{V} \times 29.23\ \text{mA} = 266P=Vz​Iz​=9.1 V×29.23 mA=266 mW.

Answer: 38.338.338.3 mA, 9.109.109.10 mA, 29.229.229.2 mA and 266266266 mW.

Check. The power is less than the usual rating of 500500500 mW, so the diode is safe. If the load were removed, the Zener diode would take all 38.338.338.3 mA, which is 349349349 mW.

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