Graph 1

 

Half bridge works basically principle

 

Above all block of oversight small size breaks capacitance C_B, criterion N_PThe lower end can regard the join that receives C1 and C2 repeatedly approximately to nod. If the size of C1, C2 is main equal, criterion the in part that the voltage of joint exports voltage for rectification approximately, it is 168V about. Normally the practice is receive equivalence discharge resistor to come each in the two end of C1, C2 balanced and both voltage. Graph 1 medium switch Q1 and Q2 guide by turns understand half cycle. Q1 guides when connecting Q2 to close, the homonymic end of Np (carry a bit) voltage is ＋ 168V, q2 bears voltage is 336V; Manage together, q2 guides when connecting Q1 to close, q1 bears voltage is 336V, np is right now homonymic upright voltage is – 168V.

 

C_Baction

 

Tell C from the principle_BNeed not want, but principle eventually is a principle it is pure Utopian thing, but we are engineers we should design a product to must be contacted actual. The premise that the principle establishs is the voltage on C1 and C2 it is completely equal, but in real half bridge power source scarcely is met equal, why? For instance C1 is mixed C2 although two capacitance are chosen is same type same size but error of total meeting existence, make the voltage of two end of C1, C2 unequal thereby. The voltage 170V of two end of our hypothesis C1, the voltage of two end of C2 is 166V, half bridge develop attacks to be in charge of those who leave a valve to guide telling time is identical, according to volt second balance a cycle comes down to be met 4V×D is much come out answer not broke a balance (this 4V that comes out more is understandable also into dc heft) , core of the magnetism after passing many cycle is saturated.

 

We are analysed again, still be C1 is 170V, c2 is 166V, if have C_Bthe canal on existence guides when connecting, to add in winding the voltage of two end is met by C_BCent is dropped a few (left negative right) won’t achieve 170V, next canals guide when connecting, as a result of C_BExist on left negative right a few voltage, C_Bmeeting overlay is in this voltage over the 166V of C2, add in winding together two end (direction catchs up with of the face contrary) the voltage on winding can exceed 166V, adjusted magnetic flux thereby lopsided, make a system convergent.

 

Notice, there is the element that other very much inaccuracy decides in circuit system, C_BCapacitance is not Zun Zheng certainly right negative, possible also right left negative, no matter be to charge or charge reversely C_BCapacitance is in ceaseless doing the work that maintains a balance.

 

The analysis above has only probably train of thought, have a lot of places not rigorous still forgive please, the friend that have fun at is OK oneself further meticulous analysis.

 

Lie between straight capacitance C_BBe how calculative?

 

Graph 2

 

Set permission fall the quantity is DV, generation should press electric current of falling equivalent and flat-roofed pulse to be I_Pft, and current the time of this electric current is 0.8T/2 (suppose) , needs block breaks capacitance value to be able to use next type to get

 

 

I_Pftcalculation is very simple

 

Set efficiency to be 80 ％ , criterion

 

When power source inputs voltage lowest, power input is equal to primary voltage the least value and the corresponding product with primary average electric current. Namely 1.25P_O=(V_Dc/ 2)(I_Pft) (0.8T/T)

 

 

Among them V_Dc(lower part adds one horizontal stroke) input volts d.c. for lowest.

 

Computation of concrete give an example lies between straight capacitance C_Bcapacity

 

A power is the power source of half bridge switch of 150W, rated dc inputs voltage to be 320V, frequency is 100kHz, the network that sets 15 ％ presses wave motion, voltage of the smallest input is 272V, primary voltage should be ±272/2 ＝ ±136V.

 

The permission of voltage of primary and flat-roofed pulse drops the quantity is 10 ％ about, it is 14V about namely, foregone power is 150W, V_Dc=272V, I_Pft=3.13×150/272=1.73A,

 

By formula

 

 

Can get

 

 

Additionally notable is, C_BCapacitance must choose not to have polarity capacitance.