Chapter 14 AC-DC power supply front stage circuit note I

Charge and discharge stage

• Give an input voltage after diode rectification $v(t)=\sqrt{2}{V}_{ac}\ast |cos(2\pi ft)|$

• The function of the large capacitor behind the rectifier bridge is when its input voltage is lower than the electrolytic voltage , The role of supplying energy to the subsequent circuit . Well, of course , For a while $t$ Inside , Maintain input power $P_{in}$ The constant source of energy is capacitance .

• Definition , Capacitance from peak value $\sqrt{2}{V}_{ac}$ discharge , after $t$ Time （ The power grid recharges the capacitor ）, Discharge to $v_{cap}(t)$, According to the capacitance energy storage relationship , Available ：$$\frac{1}{2}C[(\sqrt{2}{V}_{ac}{)}^2-v_{cap}^2(t)]=P_{in}t$$ Simplification , have to ：$$v_{cap}(t)=\sqrt{2}\ast \sqrt{V_{ac}-\frac{P_{in}}{C}t}$$

• Now we have a capacitor voltage versus time t The relation of , When the capacitor voltage is discharged to the lowest value , What kind of relationship is it ？ Analyze the instantaneous steady state relationship at this time
  

• As shown in the figure , Note that the starting discharge time of the capacitor is $t_1=0$, The time to discharge to the minimum voltage is $t_2$, The corresponding instantaneous voltages are respectively $\sqrt{2}{V}_{ac}$、$V_{sag}$, Set the discharge time as $t_{SAG}$, Charging time is $t_{CON}$.

• We already know the equation of these two curves , According to the knowledge of Mathematics , The simultaneous equations can solve the value of the intersection ？ Let's try ：$$\sqrt{2}{V}_{ac}\ast |cos(2\pi ft)|=\sqrt{2}\ast \sqrt{V_{ac}^2-\frac{10^6}{(uF/W)}t}$$

• here , We found a problem , Time t It is unknown. ,$uF/W$ It's also unknown . Two unknowns have only one equation , How ？

• Use the omnipotent assumption , We assume that the capacitor is not discharged , That is, the voltage at both ends of the capacitor remains unchanged （ Reality is out of reach ） At the time of the $uF/W$, This is an ideal value , Then the discharge time of the capacitor is $$t_2\approx t_{CON},t_2-t_1=\frac{1}{2f}$$

• Simplify the above equation , Available ：$$cos(2\pi ft)\approx \sqrt{1-\frac{10^6}{2\ast f\ast (uF/W)\ast V_{ac}^2}}=\pm A$$

• This square root always has two solutions , One is one minus one. . A solution is estimated $t_{SAG}$, Corresponding to the lowest voltage $V_{sag}$; A solution is estimated $t_{CON}$. The sum of the two must be equal to half a period $1/(2f)$. Introduce here A, It is convenient to calculate .

• Then there are ：$$t_{CON}=\frac{arccos(A)}{2\ast \pi \ast f}$$$$t_{SAG}=\frac{arccos(-A)}{2\ast \pi \ast f}=\frac{1}{2f}-t_{CON}$$$$A=\sqrt{1-\frac{10^6}{2\ast f\ast (uF/W)\ast V_{ac}^2}}$$

• here $t_{SAG}$ The corresponding minimum voltage is ：$$V_{sag}=\sqrt{2}\ast \sqrt{V_{ac}^2-\frac{10^6}{(uF/W)}{t}_{SAG}}$$

• utilize mathcad, Assume that the input voltage is 120V,50Hz, Available
  

• so , When x=8uF/W when , The minimum voltage of capacitor discharge is approximately equal to the peak voltage , In the vernacular, it means , Capacitive filtering , The filter is as flat as the peak . But the problem with this is , The capacitance is very large , Suppose you do 120V10W Products , You need to put one behind the bridge 80uF/160V Electrolysis of , Big size .