Bulk capacitor hold up time calculation

[Akanimo]

CORRECTION:

The expression that I marked in red assumes that R is constant just like we have a constant R in series R-C circuit. Here, we have a different case actually. Our R keeps changing (decreasing actually) as we keep pulling energy out of the bulk capacitor.

Normally, the efficiency of the converter varies as Vin changes. But we will assume that the efficiency of our converter remains the same throughout the range of Vin that we consider in this analysis.

Realize that in our case, after we lose the supply to our converter, as every switching cycle passes our converter draws energy from the bulk capacitor, and as our converter draws energy from the bulk capacitor, the voltage across our bulk capacitor drops. Because this voltage drops, the converter has to draw more current from the bulk capacitor to be able to provide for the required power at the output of our converter. If we check R = V_bulk/I_in we see that as V_bulk goes down and I_in goes up our R goes down. This decreasing R will not allow our decaying capacitor voltage to take that famous exponential capacitor discharge curve.

So that expression is ruled out. The correct expression is given in green font.

When you apply AC voltage through a bridge rectifier to the bulk capacitor, you get a waveform across the capacitor. In normal operation, the waveform rises to the peak of the applied sinusoidal voltage as the cap charges and then sags to a level and gets charged again. This happens at a frequency that is twice the line frequency. The idea is that you have to determine that voltage level that it sags to and use that as your initial voltage. You will need another voltage (the level that this initial voltage will not have to decay to within the hold up time), and that will be your final voltage. You will need resistance value for your calculation. This resistance is Vinitial*Vinitial/Pin or Vinitial/Iin_ave. Of course you already have the hold up time.

Your expression will now be Vfinal = Vinitial*e^(-t/tau).

Now,
Time constant, Tau = R*C and t = holdup time = 1 sec.

You now have Vfinal/Vinitial = e^(-t/(R*C)) and you solve for C.

Now, the question is how to find Vinitial. This is why I referred you to the textbook in Post #11.

This voltage is determined at worst case. For flyback, this is at low line and maximum load. Note that that valley voltage, Vinitial, is not the peak voltage mains voltage at low line (i.e. Vinitial is not equal to Vac*SQRT(2)).

Determining that valley voltage is not trivial because the mains voltage is sinusoidal.

For reliability:
Note that C is the capacitance of the bulk capacitor at the End-of-Life (EOL) of your converter.

At the EOL of your your capacitance will have dropped from its Beginning-of-Life (BOL) value by 20%. So to select the bulk capacitor, we will have to compensate for this. For instance, an EOL value for a capacitor with (C = 1uF) means it has degraded by 20% in capacitance already, such that this capacitance C is 80% or 0.8 (i.e. 100% - 20% or 1-0.2) of the BOL value. So to obtain the BOL value, we have [1uF*1/(1-0.2) = 1.25*1uF]. So generally, this means 1.25*C.

We now consider capacitor tolerances. Capacitors have tolerances: 20%, 10%, 5%...and so on. At worst case, a 1uF cap with 20% tolerance will have capacitance of (100%-20%)*1uF = (1-0.2)*1uF = 0.8*1uF. So to be sure we get a 1uF capacitor we have to select a 1uF*1/(1-0.2) = 1.25*1uF. So, generally, we compensate for this with the formula C*[1/(1-tolerance)].

We then combine the two results for capacitor lifespan degradation and tolerance to select the capacitance of the bulk capacitor at BOL. Thus, C_bulk = C*1.25*[1/(1-tolerance)].

When you apply AC voltage through a bridge rectifier to the bulk capacitor, you get a waveform across the capacitor. In normal operation, the waveform rises to the peak of the applied sinusoidal voltage as the cap charges and then sags to a level and gets charged again. This happens at a frequency that is twice the line frequency. The idea is that you have to determine that voltage level that it sags to and use that as your initial voltage. You will need another voltage (the level that this initial voltage will not have to decay to within the hold up time), and that will be your final voltage. You will need resistance value for your calculation. P_in is P_out/efficiency. Of course you already have the hold up time.

Your expression will now be 2*Pin*t_holdup/(V_initial² - V_final²).

Now, let me explain how we come by the formula.
Energy stored in bulk capacitor at V_initial - Energy left in bulk capacitor at V_final = Energy drawn out of the bulk capacitor during hold-up time.

0.5*C_bulk*V_initial² - 0.5*C_bulk*V_final² = Pin*T_s*m

where: T_s is the switching period (i.e. 1/f_sw)
m is the number of switching cycles that there is in the desired hold-up time such that t_holdup = T_s*m

then you solve for C_bulk.

Now, the question is how to find Vinitial. This is why I referred you to the textbook in Post #11.

This voltage is determined at worst case. For flyback, this is at low line and maximum load. Note that that valley voltage, Vinitial, is not the peak voltage mains voltage at low line (i.e. Vinitial is not equal to Vac*SQRT(2)).

Determining that valley voltage is not trivial because the mains voltage is sinusoidal.

For reliability:
Note that C is the capacitance of the bulk capacitor at the End-of-Life (EOL) of your converter.

At the EOL of your your capacitance will have dropped from its Beginning-of-Life (BOL) value by 20%. So to select the bulk capacitor, we will have to compensate for this. For instance, an EOL value for a capacitor with (C = 1uF) means it has degraded by 20% in capacitance already, such that this capacitance C is 80% or 0.8 (i.e. 100% - 20% or 1-0.2) of the BOL value. So to obtain the BOL value, we have [1uF*1/(1-0.2) = 1.25*1uF]. So generally, this means 1.25*C.

We now consider capacitor tolerances. Capacitors have tolerances: 20%, 10%, 5%...and so on. At worst case, a 1uF cap with 20% tolerance will have capacitance of (100%-20%)*1uF = (1-0.2)*1uF = 0.8*1uF. So to be sure we get a 1uF capacitor we have to select a 1uF*1/(1-0.2) = 1.25*1uF. So, generally, we compensate for this with the formula C*[1/(1-tolerance)].

We then combine the two results for capacitor lifespan degradation and tolerance to select the capacitance of the bulk capacitor at BOL. Thus, C_bulk = C*1.25*[1/(1-tolerance)].