I am confused with this and the image shown below given about half H-bridge circuits!

Half h bridge

Why are there are two capacitors connected if DC gets blocked?(roughly, I am assuming that, to prevent short from Vcc to ground.) Please Help me to understand this!

EDIT: Consider a dc motor instead of L1

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    \$\begingroup\$ To block the DC? \$\endgroup\$ – Trevor_G Feb 9 '18 at 20:47
  • \$\begingroup\$ Due to an 'AC' signal at L1 a DC motor would not work, but an AC motor will. L1 is typically the primary winding of a SMPS. This design is very common and has lower cost than a full H-bridge. \$\endgroup\$ – VTNCaGNtdDVNalUy Feb 9 '18 at 21:57
  • \$\begingroup\$ @Sparky256 - So with full h bridge configuration, I will be able to drive a DC motor, but not with the half h bridge, right! \$\endgroup\$ – Rajan Feb 9 '18 at 23:24
  • \$\begingroup\$ Only in the sense that the MOSFET's can be used to change motor direction without relays. Those capacitors allow AC operation only, and are typically 470 uF 400 VDC, with a low ESR rating. \$\endgroup\$ – VTNCaGNtdDVNalUy Feb 10 '18 at 0:35
  • \$\begingroup\$ Show your source. Can you find ANY example of a split capacitor half bridge driving a motor? This COULD be done in special cases but strict criteria need to be met. This arrangement takes DC in and applies AC to L1. This is usually used to produce a n AC voltage to be rectified to DC. If you had a motor that was appropriate to be driven by AC in place of L1 then it could be done, but, that is probably not what you are looking at. Show your source - with link and or whole circuit. \$\endgroup\$ – Russell McMahon Feb 10 '18 at 7:22

Vcc is DC.
You should provide a link and/or reference to the source document to help people help you.
You should have a good idea that Vcc is DC from the context in which you found the diagram.

The two capacitors act similarly to two switches.
Current flows alternately in either direction through them, so that their net DC current is zero - so they pass alternating current.

It is best but not necessary to assume C1 fully discharge and C2 fully charged, as will be the case at the start of the cycle below. If desired you can start with C1-C2 centre point at V22/2 which is the starting condition.

Q1 on.
Current flows Vcc-Q1-L1-C1-ground.
C1 charges and C2 discharges (as its voltage decreases) so C1-C2 centre point voltage rises towards Vcc.

When V C1-C2 approaches Vcc the 1/2 cycle is complete.

Q1 off. Q2 on Current flows Vcc-C2-L1-Q2-ground. Thge C1-C2 centre point falls towards ground.
Current passes through L1 in the opposite direction to the previous cycle.


Note that as Andy says in his answer, C1 & C2 may be sized such that the center point voltage vary little over a half cycle. You still get bidirectional AC current in L1 - which is the object.

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  • \$\begingroup\$ I still don't get it. When Q1 conducts it will open a free path for Vcc to flow through L1 and then through the capacitor but if Vcc is Vdc then capacitor should block and no current flows further to the ground and thus no motor (L1) is on, isn't it right? \$\endgroup\$ – Rajan Feb 9 '18 at 21:09
  • \$\begingroup\$ @Rajan You need to read answers step by step and be sure that you understand what they say. If you glance at them as a whole you will often miss the relevant detail. I said : " ... C1 charges and C2 discharges (as its voltage decreases) so C1-C2 centre point voltage rises towards Vcc." <- Vrising- same as part of an AC cycle. Then I said " ...When V C1-C2 approaches Vcc the 1/2 cycle is complete." <- you seem to be assuming that because the supply is DC that it is always applied everywhere. The job of the switches is to change its direction as time varies . ... \$\endgroup\$ – Russell McMahon Feb 9 '18 at 21:26
  • \$\begingroup\$ @Rajan - ... I suggest that you read again step by step. Draw a diagram showing current flows at eachj step. Then report back. \$\endgroup\$ – Russell McMahon Feb 9 '18 at 21:27
  • \$\begingroup\$ @ Russell McMahon - If Q1 charges C1 a little that means the amount of time PWM signal is applied to Q1 (For example pulse active time is 50ms in applied PWM) will only conduct about 50ms and Vcc will flow through the motor (L1) through C1 and now C1 has one end connected to ground and other end will be positive thus it will charge for 50ms and meantime C2 will discharge a bit in that in that 50ms. Thus when Q2 will on and the motor will rotate reverse and the junction point between C1-C2 will have a negative voltage.If this how it works than would motor spin at full speed in either direction? \$\endgroup\$ – Rajan Feb 9 '18 at 23:18

The circuit is likely part of a schematic of a medium power switching and isolating power supply where only the primary is shown. It would normally be fed from a 50% duty cycle square wave from the half bridge mosfets. The centre voltage of the two capacitors will show signs of ripple voltage but this will be small due to the filtering effect of the primary inductance and those two capacitors.

The important point is that the voltage at the common node of the two capacitors is half the DC power rail and it is kept this way providing the Half bridge duty cycle remains at 50%.

There are similar designs where two mosfets are used to drive a centre tapped primary. Both achieve the same thing namely providing a secondary voltage that is isolated and at a lower or higher output level (due to the turns ratio).

Q1 charges C1 up a little each cycle and this also discharges C2 a little. Q2 does the opposite and, roughly speaking, the junction remains at half the power rail plus a bit of ripple.

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  • \$\begingroup\$ I would post an answer but the OP seems lost. Existing answers are correct, but the OP does not 'see' that. \$\endgroup\$ – VTNCaGNtdDVNalUy Feb 9 '18 at 21:31

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