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Thyristor V-I, avalanche and forward breakdown

Short answer

Avalanche breakdown occurs in reverse bias when voltage reaches VBR, causing destructive carrier multiplication at junctions J1 and J3. Forward breakover voltage VBO is the forward voltage at which the middle junction J2 breaks down by avalanche, turning the thyristor on without gate current.

Draw the V-I characteristics curve of thyristor. What is an avalanche breakdown and forward breakdown voltage of thyristor? Mention with its symbolic diagram.

The thyristor (SCR) is a four-layer PNPN device with anode A, cathode K and gate G. Its V-I characteristic shows anode current IAI_A against anode-cathode voltage VAKV_{AK}.

  Symbol          Structure
                     A
     A               |
     |             +---+
    _V_            | P |
     |\            +---+ J1
     | \ G         | N |
     K             +---+ J2
                   | P |---- G
                   +---+ J3
                   | N |
                   +---+
                     |
                     K
              IA
               ^
               |  |<- ON state (VT = 1-2 V)
               |  |
         IL ---|--|
         IH ---|--+.
               |    `.   Ig2 > Ig1 > Ig=0
               |      `-.___      ___
     -VBR      |            `----'   |
  -----+-------+---------------------+--> VAK
       |       |  forward blocking  VBO
       |       |  (leakage current)
       | reverse blocking
       | (leakage current)
       v reverse avalanche
  • Avalanche (reverse) breakdown: in reverse bias, J1J_1 and J3J_3 block. When the reverse voltage reaches the reverse breakdown voltage VBRV_{BR}, carriers gain enough energy to knock out more carriers (avalanche multiplication), and the reverse current rises sharply. This normally destroys the device, so it is operated below VRRMV_{RRM}.
  • Forward breakover voltage VBOV_{BO}: in forward bias with gate open, J2J_2 blocks. VBOV_{BO} is the forward voltage at which J2J_2 breaks down by avalanche and the SCR switches from forward blocking to the ON state without any gate signal. With gate current, the breakover voltage becomes smaller.

✓ Answer checked by Claude Opus 5.5 (high effort)

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