Step-up cycloconverter, four times frequency

A step-up cycloconverter gives an output frequency higher than the input ($f_o = n f_i$). Since each thyristor must be turned off in the middle of a supply half cycle, while it is still forward biased (before its natural current zero), forced commutation is needed. Mid-point circuit:

          a o----+---P1->|---+
            |    +---|<-N1---+
  ~ vs    Tr|                |
  (input) O o----[ Load ]----+  (common point K)
            |                |
          b o----+---P2->|---+
                 +---|<-N2---+

• In the positive input half cycle ($a$ positive): P1 gives a positive load voltage $+v_{aO}$; N2 gives a negative load voltage.
• In the negative input half cycle ($b$ positive): P2 gives a positive load voltage; N1 gives a negative load voltage.

Operation for $f_o = 4f_i$ (50 Hz to 200 Hz)

Each input half cycle (10 ms) must contain 4 output half cycles, i.e. two full output cycles. So each input half cycle is divided into 4 equal intervals of 2.5 ms:

At the end of each interval, the conducting SCR is turned off by a forced-commutation circuit and the next one is fired.

 vs   _____
     /     \
 ---/-------\---------/--
             \_______/
 vo  _   _      _   _
    | | | |    | | | |
 ---+-+-+-+----+-+-+-+---
      |_| |_|    |_| |_|
    P1 N2 P1 N2 P2 N1 P2 N1

The output consists of slices of the input sine wave with alternating sign; its fundamental is 200 Hz. The output amplitude follows the input envelope, so the harmonic content is high. Step-up cycloconverters are used less often than step-down types; typical uses are induction heating and high-speed drives.