2 Zcap Calculation of Hybrid Capacitors

Two different types of capacitors in parallel are common in application designs, so two types of capacitors in parallel are the main focus of discussion here. The simplified equivalent circuit of hybrid output capacitors is shown in Figure 2-1. C1 is the MLCC with small ESR and the C2 is the capacitors with a high capacitance and large ESR, such as electrolytic capacitor or polymer capacitors.

Figure 2-1 The Simplified Equivalent Circuit of Hybrid Output Capacitors

The impedance of a hybrid output capacitor network can be given by Equation 1.

Equation 1.

Z_{c a p} (s) = (r_{1} + \frac{1}{{sC}_{2}}) ∕∕ (r_{1} + \frac{1}{{sC}_{2}}) = \frac{(1 + {sC}_{1} r_{1}) (1 + {sC}_{2} r_{2})}{s (C_{1} + C_{2}) [1 + s (r_{1} + r_{2}) \frac{C_{1} C_{2}}{C_{1} + C_{2}}]}

Where, r1 is the ESR of C1; r2 is the ESR of C2

The results show that the hybrid capacitor network introduces an additional zero ω_{z_C2} and pole ω_{p_C2} compared to just one type of MLCC capacitor network.

Equation 2.

ω_{z_C 2} = \frac{1}{C_{2} r_{2}} ω_{p_C 2} = \frac{1}{(r_{1} + r_{2}) \frac{C_{1} C_{2}}{C_{1} + C_{2}}}

Figure 2-2 D-CAP3™ Functional Block Diagram With Hybrid Capacitor Network

The only difference between the D-CAP2 and D-CAP3 converter is that the D-CAP3 has an error amplifier (EA) to eliminate static errors of output voltage, while the D-CAP2 converter does not have an EA block. Since the EA does not affect the loop analysis with the hybrid capacitor network, the D-CAP3 is used as an example in the next analysis. Figure 2-2 shows a simplified DCAP3 functional block diagram with hybrid capacitor network. The D-CAP2™ Frequency Response Model based on frequency domain analysis of Fixed On-Time with Bottom Detection having Ripple Injection application note, builds on the D-CAP2/3 small-signal model and provides the transfer function from duty to Vout with an MLCC capacitor network. Using a hybrid capacitor network for output in combination with Equation 1 yields the Gdv(s) in Equation 3.

Equation 3.

G d v (s) = \frac{V i n \times (1 + \frac{S}{ω_{z_C 1}}) (1 + \frac{S}{ω_{z_C 2}})}{[1 + 2 σ \frac{S}{ω_{0}} + {(\frac{S}{ω_{0}})}^{2}] (1 + \frac{S}{ω_{p_C 2}})}

where

Equation 4.

σ = \frac{\sqrt{L ∕ C_{0}} + R_{L} (r_{L} + r_{c}) \sqrt{L ∕ C_{0}}}{{2 R}_{L} \sqrt{1 + r_{L} ∕ R_{L}}}

C₀ is total output capacitance

ω₀ is the double pole:

Equation 5.

ω_{0} = \sqrt{\frac{1}{L (C_{1} + C_{2})}}

ω_{z_C1} is the zero generated by C1:

Equation 6.

ω_{z_C 1} = \frac{1}{C_{1} r_{1}}

ω_{z_C2} is the zero generated by C2:

Equation 7.

ω_{z_C 2} = \frac{1}{C_{2} r_{2}}

ω_{p_C2} is the pole generated by the hybrid capacitors network:

Equation 8.

ω_{p_C 2} = \frac{1}{(r_{1} + r_{2}) \frac{C_{1} C_{2}}{C_{1} + C_{2}}}