Optimization of Charge Pump Based on Piecewise Modeling of Output-Voltage Ripple

Abstract

This work proposes a piecewise modeling of output-voltage ripple for linear charge pumps. The proposed modeling can obtain a more accurate design expression of power-conversion efficiency. The relationship between flying and output capacitance, as well as switching frequency and optimize power-conversion efficiency can be calculated. The proposed modeling is applied to three charge-pump circuits: 1-stage linear charge pump, dual-branch 1-stage linear charge pump and 4× cross-coupled charge pump. Circuit-level simulation is used to verify the accuracy of proposed modeling.

1. Introduction

Charge pump is an important circuit in many applications such as energy harvesting, micro-sensor, flash memory and the step-up part after rectifier of wireless power transfer. There exist many variants of charge pump, for example, linear, dual-branch linear, cross-coupled, exponential, Fibonacci and Cockcroft–Walton charge pumps [1,2,3,4,5,6]. The main purposes to develop these structures are to improve the gain and power-conversion efficiency (PCE), as well as to reduce the number of switches and flying capacitors.

Studies of PCE of charge pump is an important topic and have been carried out for many decades [7,8,9,10,11]. Flying and output capacitances are the considerations of optimization of PCE because smaller chip size can be achieved upon minimizing on-chip flying capacitances and production cost can be reduced when smaller and fewer off-chip flying capacitors are used. Switching frequency impacts PCE and output-voltage ripple. There are many factors towards PCE optimization. Modeling of output-voltage ripple for PCE optimization is one of the approaches [7]. However, the modeling is not accurate enough and has room to further enhance. In this paper, a piecewise modeling of output-voltage ripple of linear charge pump is proposed. Based on the proposed piecewise modeling, the expressions of PCE of 1-stage linear charge pump, dual-branch 1-stage linear charge pump and 4× cross-coupled charge pump are found. The PCE expressions can be used to select appropriate flying and output capacitances, as well as switching frequency.

2. Methods

2.1. Proposed Piecewise Modeling of Output-Voltage Ripple for 1-Stage Linear Charge Pump

The two-phase switching operation of a 1-stage linear charge pump is shown in Figure 1. Two complementary clock signals Φ1 and Φ2, which have approximately half of a switching period (T), are used to control the ON and OFF of switches. Deadtime is inserted in transitions between Φ1 and Φ2 to avoid short-circuit loss of the switches. V_i, V_o and I_o represent input voltage, output voltage and output current. C_f and C_o are flying and output capacitors. Assuming the lump resistances of switches are the same, the lump sum of the resistances from switches, routing and bond-wire could be noted as R_l, which is around two times of a single switch. Figure 1a shows the case for Φ1 = 1 and Φ2 = 0. V_i and C_f are connected in series to generate approximately two times of V_i for V_o. The corresponding modeling is shown on the right-hand side of Figure 1a. Similarly, Figure 1b shows the case for Φ1 = 0 and Φ2 = 1. C_f is charged by V_i, while C_o maintains V_o and supply current to load. The right-hand-side figure of Figure 1b shows the modeling. When T/2 is larger than about 6R_lC_f, the voltage across C_f is close to V_i. However, usually C_f could not be fully charged since a large T leads to large output voltage drop due to the long discharging time of C_o.

Figure 2 shows the output-voltage ripples of a 1-stage linear charge pump. The actual output-voltage ripple is represented by the solid black line. The green line shows the modeling proposed in [7]. There is no charging time of C_f and no charge re-distribution between C_f and C_o in this modeling so that V_o is sharply increased at nT in the n-th switching cycle. Since the average V_o (i.e., $\overline{V_o}$) is used to evaluate PCE in [7], inaccurate modeling of the output-voltage ripples results in inaccurate $\overline{V_o}$ and PCE.

The proposed modeling of the output-voltage ripples is shown by the red line in Figure 2. There are three segments within a switching period, and each segment is represented by a straight line. A short period of T_x is used to model the situation when V_o raises at nT. Comparing the proposed modeling with the actual output-voltage ripple shows that the proposed piecewise method can better represent the ripple voltage at output. Thus, $\overline{V_o}$ and PCE can be predicted more accurately.

Figure 3 shows more details of the output waveform and the corresponding RC circuits (from Figure 1) of a 1-stage linear charge pump. V_o starts at V_o3 at nT and reaches V_o1 within T_x. Then, it decreases from V_o1 to V_o2 within (T/2 − T_x). Finally, it further decreases from V_o2 to V_o3 to complete a cycle. More details will be provided below to investigate the up and down of V_o within a switching cycle to generate the voltage ripple.

In the previous cycle, C_f is previously charged to V_Cf, which is close to V_i, while C_o is lightly discharged by the load and the voltage across C_o is less than 2V_i. At nT, the series-connected combination of the voltage source V_i and C_f has a sum of voltage of V_i + V_Cf. Thus, C_f is discharged itself to provide charges to C_o and the load I_o. As the voltage across C_o is increasing, V_o is increased from V_o3 to V_o1, and the required time to complete this operation is T_x.

From the RC circuits in Figure 3, at nT, the voltage of C_f is V_Cf, and thus the charge of C_f is C_fV_Cf. Moreover, the charge of C_o is C_oV_o3. Then, at (nT + T_x), the voltage across C_f is dropped to (V_o1 − V_i − I_lR_l), so that the charges of C_f is C_f (V_o1 − V_i − I_lR_l). I_R1 has the same value as I_o, since the current going into C_o should be zero at (nT + T_x). Meanwhile, the charge of C_o is C_oV_o1. The charge supplied to the load is I_oT_x. By principle of conservation of charges [7], the following relationship is achieved.

$$C_f{V}_{Cf}+C_o{V}_{o3}=C_f\left(V_{o1}-V_i+I_o{R}_1\right)+C_o{V}_{o1}+I_o{T}_x$$

(1)

V_Cf is the voltage obtained by capacitor C_f when charging with an ideal voltage supply V_i within a time period T/2, and the initial voltage is V_o2. Therefore, V_Cf is given by

$$V_{Cf}=\left(1-2e^{-\frac{T}{2R_1{C}_f}}\right)V_i+\frac{I_o{C}_f{R}_1}{C_f+C_o}{e}^{-\frac{T}{2R_1{C}_f}}+V_{o2}{e}^{-\frac{T}{2R_1{C}_f}}$$

(2)

At (nT + T/2), the voltage of C_o is V_o2. Since the charge redistribution of C_f and C_o is complete, the current passing through C_f and C_o is in constant ratio. Therefore, the current of C_f should be C_fI_o/(C_{f +} C_o). As a result, the following relationship is obtained.

$$C_f\left(V_{o1}+I_o{R}_1\right)+C_o{V}_{o1}=C_f\left(V_{o2}+\frac{I_o{C}_f{R}_1}{C_f+C_o}\right)+C_o{V}_{o2}+I_o\left(T/2-T_x\right)$$

(3)

Between (nT + T/2) and (n + 1) T, the voltage across C_o drops from V_o2 to V_o3. The change of charges of C_o is C_o (V_o2 − V_o3). These charges supply current to the load to give

$$C_o\left(V_{o2}-V_{o3}\right)=\frac{I_{o}T}{2}$$

(4)

Assume that $T=m_1{R}_1{C}_f$, $C_o=m_2{C}_f$, and $T_x=m_3{R}_1{C}_f$. It should be noted that $T/2>6R_1\left(C_f//C_o\right)$, i.e., $m_1>\frac{12m_2}{m_2+1}$, so that the charge redistribution between C_f and C_o is complete. By solving Equations (1)–(4), V_o1, V_o2 and V_o3 can be found, respectively.

$$V_{o1}=2V_i-\frac{m_1{I}_o{R}_1}{1-e^{-m_1/2}}-\frac{I_o{R}_1}{m_2+1}-\frac{m_2{I}_o{R}_1}{{\left(m_2+1\right)}^2}+\frac{\left(m_1-2m_3\right)I_o{R}_1}{2\left(m_2+1\right)}$$

(5)

$$V_{o2}=2V_i-\frac{m_1{I}_o{R}_1}{1-e^{-m_1/2}}-\frac{I_o{R}_1}{m_2+1}$$

(6)

$$V_{o3}=2V_i-\frac{m_1{I}_o{R}_1}{1-e^{-m_1/2}}-\frac{I_o{R}_1}{m_2+1}-\frac{m_1{I}_o{R}_1}{2m_2}$$

(7)

From Equations (5)–(7), as well as the durations of each segment within a switching cycle, the average value of V_o is found and given by:

$$\overline{V_o}=2V_i-\frac{m_1{I}_o{R}_1}{1-e^{-m_1/2}}-\frac{I_o{R}_1}{m_2+1}-\left(\frac{1}{4}+\frac{m_3}{2m_1}\right)\frac{m_1{I}_o{R}_1}{2m_2}-\frac{m_2{I}_o{R}_1}{4{\left(m_2+1\right)}^2}+\frac{\left(m_1-2m_3\right)I_o{R}_1}{8\left(m_2+1\right)}$$

(8)

For m₃ in Equation (8), it can be evaluated by the following. Refer to Figure 4 for the currents and voltages of C_f and C_o during the period from nT to (nT + T_x), it can be found that

$$V_i+v_{Cf}\left(t\right)=i_R\left(t\right)R_l+v_{co}\left(t\right)$$

(9)

By considering the charges in capacitors and differentiating in Equation (9) on both sides with respect to time, it gives

$$\frac{d{V}_i}{dt}+\frac{1}{C_f}\frac{d{Q}_{cf}\left(t\right)}{dt}=R_l\frac{d{i}_R\left(t\right)}{dt}+\frac{1}{C_o}\frac{d{Q}_{co}\left(t\right)}{dt}$$

(10)

where Q_cf(t) and Q_co(t) are the charges stored in C_f and C_o at t. From Figure 4, it can be found that $\frac{d{Q}_{cf}\left(t\right)}{dt}=-i_R\left(t\right)$ and $\frac{d{Q}_{co}\left(t\right)}{dt}=i_C\left(t\right)=i_R\left(t\right)-I_o$. Based on these relationships and substituting into Equation (9), the following expression is obtained. It is noted that $\frac{d{V}_i}{dt}=0$ as V_i is a dc voltage.

$$\frac{d{i}_R\left(t\right)}{dt}=-\frac{i_R\left(t\right)}{R_l}\left(\frac{1}{C_f//C_o}\right)+\frac{I_o}{R_l{C}_o}$$

(11)

Solving the above differential equation, and determining the constant of integration by the initial condition of the circuit, the expression of i_R(t) is given by

$$i_R\left(t\right)=\frac{I_o{C}_f}{C_f+C_o}+\left(I_o-\frac{I_o{C}_f}{C_f+C_o}\right)e^{\frac{-t}{R_l\left(C_f//C_o\right)}}$$

(12)

where $C_f//C_o=\frac{C_f{C}_o}{C_f+C_o}$. The capacitor current of C_o is given by

$$i_C\left(t\right)=i_R\left(t\right)-I_o=\frac{-I_o{C}_f}{C_f+C_o}+\left(I_{Co0}-\frac{I_o{C}_f}{C_f+C_o}\right)e^{\frac{-t}{R_l\left(C_f//C_o\right)}}$$

(13)

where $I_{Co0}=\frac{V_i+V_{Cf}-V_{o3}}{R_l}$ is the initial current of C_o at nT.

Refering to Figure 2, the peak voltage of V_o (i.e., the peak voltage across C_o) occurs at about (nT + T_x). Therefore, T_x can be found by differentiating Equation (11) with respect to time to find the maximum point. As a result, T_x is given by

$$T_x=R_l\left(C_o//C_f\right)\ln \left(\frac{I_{co0}\left(C_o+C_f\right)-I_o{C}_f}{I_o{C}_o}\right)$$

(14)

By solving Equation (14), we have

$$m_3=\frac{m_2}{m_2+1}\left(ln\left(m_1\right)+ln\left(1+\frac{1}{m_2}\right)+ln\left(1+\frac{1}{2m_2}\right)\right)$$

(15)

Figure 5 shows the charge transfer in both phases. In Phase 2, the total charges from C_o to load is Q_a, and so $Q_a=I_o\left(T/2\right)$. Since the output-voltage waveform is periodic, the net charges leaving C_o in Phase 2 equals to the net charges inputted into C_o in Phase 1. Thus, the injected charges to C_o in Phase 1 is also Q_a. Assuming the net charges to load in Phase 1 is Q_b where $Q_b=I_o\left(T/2\right)$, the charges from C_f in Phase 1 becomes (Q_a + Q_b). For the series connections of V_i and C_f in Phase 1, the charges from V_i is also (Q_a + Q_b) in Phase 1. Since, again, the output-voltage waveform is periodic, the net charges leaving C_f in Phase 1 equals to the net charges inputted into C_f in Phase 2. As such, the charges from V_i to C_f in Phase 2 is (Q_a + Q_b). The total charges from V_i is equal to (Q_a + Q_b) in Phase 1 plus (Q_a + Q_b) in Phase 2, which is 2(Q_a + Q_b) = 2I_oT. Therefore, the input current (I_i) from V_i is given by two times of I_o (i.e., I_i = 2I_o), which is two times the load current.

The PCE of a 1-stage linear charge pump is the ratio of output power (P_o) to input power (P_i) and is given by

$$PCE=\frac{P_o}{P_i}=\frac{\overline{V_o}{I}_o}{V_i{I}_i}=\frac{\overline{V_o}}{2V_i}$$

(16)

where $\overline{V_o}$ is the expression shown in Equation (8).

2.2. Proposed Piecewise Modeling of Output-Voltage Ripple for Dual-Branch 1-Stage Linear Charge Pump/Cross-Coupled Voltage Doubler

In this section, the proposed piecewise modeling of output-voltage ripple is applied to dual-branch 1-stage linear charge pump. It is applicable to cross-coupled voltage doubler, since the ON and OFF arrangements of switches of both dual-branch 1-stage linear charge pump and cross-coupled voltage doubler are the same. Figure 6 shows the switching of a dual-branch 1-stage linear charge pump or cross-coupled voltage doubler, where C_fA and C_fB are flying capacitors. Similarly, R_l is used to denote the lump sum of the resistances from switches, routing, and bond-wire. The parallel structure enables the load supplied by the flying and output capacitors simultaneously when Φ1 = 1; Φ2 = 0 and Φ1 = 0; Φ2 = 1, except that only C_o provides charges to the load at deadtime (i.e., Φ1 = 0; Φ2 = 0). In fact, T_d is much shorter than T.

Figure 6a shows the case for Φ1 = 1 and Φ2 = 0. V_i and C_fA are connected in series to generate approximately two times of V_i for V_o, and C_fB is charged by V_i. The corresponding modeling is shown on the right-hand side of Figure 6a. R_l, same as before, is the lump sum of the resistances from switches, routing, and bond-wire. Similarly, Figure 6b shows the case for Φ1 = 0 and Φ2 = 1. V_i and C_fB are connected in series to provide about two times of V_i for V_o, and C_fA is charged by V_i. The right-hand-side figure of Figure 6b shows the modeling. Finally, Figure 6c shows the moment of deadtime (i.e., the case for Φ1 = 0 and Φ2 = 0), where all switches are turned off. Only C_o maintains about 2V_i and supplies charges to the load.

Figure 7 shows more details of the output waveform of one switching cycle and the corresponding RC circuits (from Figure 6) of a dual-branch 1-stage linear charge pump and cross-coupled voltage doubler. In the previous cycle, C_fA is previously charged to V_Cf, which is close to V_i, while C_o is lightly discharged by the load and the voltage across C_o is less than 2V_i. At nT, the series-connected combination of the voltage source V_i and C_fA has a sum of voltage of V_i + V_Cf. Thus, C_fA is discharged itself to provide charges to C_o and the load. C_fB is connected with V_i for re-charging. As the voltage across C_o is increasing, V_o is increased from V_o3 to V_o1, and the required time to complete this operation is T_x. After T_x, where the output voltage achieves the highest value, both C_fA and C_o discharge themselves to provide charges to the load. The duration is (T/2 − T_x − T_d), and V_o drops to V_o2 finally. At (nT + T/2 − T_d), all switches are turned off in the deadtime period. Only C_o supplies charges to the load. Thus, the drop of V_o is more rapid than before. At (nT + T/2), V_o reaches V_o3 to complete half of a cycle. Between (nT + T/2) and (n + 1)T, the operation of the first half switching cycle repeats. The only difference is that another half of the circuit enables C_fB to supply charges to the load. In the above analysis, it is assumed that T/2 is longer than 6R_l(C_f//C_o) (where C_f = C_fA = C_fB) to ensure that the redistribution of C_f (C_fA and C_fB) and C_o is complete when the capacitors are connected.

As shown in Figure 7, the output-voltage waveforms in the first half and second half of a switching cycle are the same. The analysis below takes a period of T/2 into account. C_fA and C_fB are considered to have the same value, such that C_fA = C_fB = C_f. From the RC circuits in Figure 8, at nT, the voltage of C_fA is V_Cf, and thus the charges of C_fA is C_fAV_Cf. Moreover, the charge of C_o is C_oV_o3. Then, at (nT + T_x), the voltage across C_fA is dropped to (V_o1 − V_i − I_oR₁), so that the charge of C_fA is C_fA(V_o1 − V_i − I_oR₁). Meanwhile, the charge of C_o is C_oV_o1. The charge supplied to the load is I_oT_x. By the principle of conservation of charges [7], the following relationships are achieved.

$$C_f{V}_{Cf}+C_o{V}_{o3}=C_f\left(V_{o1}-V_i-I_o{R}_1\right)+C_o{V}_{o1}+I_o{T}_x$$

(17)

V_Cf is the voltage obtained by capacitor C_fA when charging with an ideal voltage supply V_i within a time period T/2, and the initial voltage is V_o2. Therefore, V_Cf is given by

$$V_{Cf}=\left(1-2e^{-\frac{T-2T_d}{2R_1{C}_f}}\right)V_i+\frac{I_o{C}_f{R}_1}{C_f+C_o}{e}^{-\frac{T-2T_d}{2R_1{C}_f}}+V_{o2}{e}^{-\frac{T-2T_d}{2R_1{C}_f}}$$

(18)

At (nT + T/2 + T_x − T_d), the voltage of C_o is V_o2. Since the charge redistribution of C_fA and C_o is complete, the current passing through C_fA and C_o is in constant ratio. Therefore, the current of C_fA should be C_fI_o/(C_f + C_o). As a result, the following relationship is obtained.

$$C_f\left(V_{o1}+I_o{R}_1\right)+C_o{V}_{o1}=C_f\left(V_{o2}+\frac{I_o{C}_f{R}_1}{C_f+C_o}\right)+C_o{V}_{o2}+I_o\left(T/2-T_x-T_d\right)$$

(19)

Between (nT + T/2 + T_x − T_d) and (nT + T/2), the voltage across C_o drops from V_o2 to V_o3. The change of charges of C_o is C_o(V_o2 − V_o3). These charges supply current to the load to give

$$C_o\left(V_{o2}-V_{o3}\right)=I_o{T}_d$$

(20)

Assume that $T=m_1{R}_1{C}_f$, $C_o=m_2{C}_f$, $T_x=m_3{R}_1{C}_f$, and $T_d=m_4{R}_1{C}_f$,. It should be noted that $T/2>6R_1\left(C_f//C_o\right)$, i.e., $m_1>\frac{12m_2}{m_2+1}$, so that the charge redistribution between C_f and C_o is complete. By solving Equations (17)–(20), V_o1, V_o2 and V_o3 can be found, respectively.

$$V_{o1}=2V_i-\frac{m_1{I}_o{R}_1}{2\left(1-e^{-\left(m_1/2\right)+m_4}\right)}-\frac{I_o{R}_1}{m_2+1}-\frac{m_2{I}_o{R}_1}{{\left(m_2+1\right)}^2}+\frac{\left(m_1-2m_3-2m_4\right)I_o{R}_1}{2\left(m_2+1\right)}$$

(21)

$$V_{o2}=2V_i-\frac{m_1{I}_o{R}_1}{2\left(1-e^{-\left(m_1/2\right)+m_4}\right)}-\frac{I_o{R}_1}{m_2+1}$$

(22)

$$V_{o3}=2V_i-\frac{m_1{I}_o{R}_1}{2\left(1-e^{-\left(m_1/2\right)+m_4}\right)}-\frac{I_o{R}_1}{m_2+1}-\frac{m_4{I}_o{R}_1}{m_2}.$$

(23)

From Equations (21)–(23), as well as the durations of each segment within half of a switching cycle, the average value of V_o is found and given by

$$\begin{gathered} \overline{V_o}=2V_i-\frac{m_1{I}_o{R}_1}{2\left(1-e^{-\left(m_1/2\right)+m_4}\right)}-\frac{I_o{R}_1}{m_2+1}-\frac{m_4\left(m_3+m_4\right)I_o{R}_1}{m_1{m}_2}-\frac{\left(m_1-2m_4\right)m_2{I}_o{R}_1}{2m_1{\left(m_2+1\right)}^2} \\ +\frac{\left(m_1-2m_4\right)\left(m_1-2m_3-2m_4\right)I_o{R}_1}{4m_1\left(m_2+1\right)} \end{gathered}$$

(24)

The conditions to evaluate T_x is same as before, and T_x has the same expression as stated in Equation (13). Thus, we have:

$$m_3=\frac{m_2}{m_2+1}\left(ln\left(m_1\right)+ln\left(1+\frac{1}{m_2}\right)+ln\left(\frac{1}{2}+\frac{m_4}{m_1{m}_2}\right)\right)$$

(25)

Similar to the analysis of Equation (16), the input energy is given by a simple expression, as below.

$$E_i=2V_i{I}_o$$

(26)

Finally, the PCE can be easily derived by the ratio of E_o to E_i.

$$PCE=\frac{E_o}{E_i}=\frac{\overline{V_o}{I}_o}{2V_i{I}_o}=\frac{\overline{V_o}}{2V_i}$$

(27)

where ${\overline{V}}_o$ is found as shown in Equation (24).

2.3. Proposed Piecewise Modeling of Output-Voltage Ripple for 2-Stage Cross-Coupled Voltage Doubler

To verify the application of proposed piecewise modeling, the analysis is extended to 2-stage cross-coupled voltage doubler. Figure 8 shows the switching behaviors of a 2-stage cross-coupled voltage doubler. Figure 8a shows the case for Φ1 = 1 and Φ2 = 0, Figure 8b illustrates the condition for Φ1 = 0 and Φ2 = 1 and Figure 8c reveals the situation of deadtime when Φ1 = 0 and Φ2 = 0. Since the number of switches in each branch is different, the lump sums of the resistances from switches, routing and bond-wire are noted as R_l1, R_l2, and R_l3. It is noted again that T_d is much shorter than T. C_fA1 and C_fB1 are the flying capacitors in the first stage, and C_fA2 and C_fB2 are the flying capacitors in the second stage.

Basically, the operations for {Φ1 = 1 and Φ2 = 0} and {Φ1 = 0 and Φ2 = 1} are the same due to the parallel structure. Thus, the corresponding modeling of both cases are the same, except different flying capacitors are used to complete the operation of the circuit. Ideally, from the switching operations, C_fA1 and C_fB1 are charged to V_i, while C_fA2 and C_fB2 are charged to 2V_i. Therefore, V_o is 4V_i theoretically, which is the sum of the source voltage and the voltages across C_fA1 (or C_fB1) and C_fA2 (or C_fB2). During the deadtime, all switches are turned off to disconnect the output from the flying capacitors and V_i. Thus, the load is supplied.

Figure 9 shows the details of the output waveform of one switching cycle and the corresponding RC circuits (from Figure 8) of a 2-stage cross-coupled voltage doubler. The operations at nT and (nT + T/2) are the same, except another half circuit operates alternatively. Thus, the analysis can be conducted for half of a switching cycle. At nT, C_fB1 and C_fB2 are previously charged to V_Cf and about 2V_i, respectively, at ((n − 1)T + T/2) (i.e., the same operation at (nT + T/2). C_fB1 provides charges to C_fA2 to re-charge it to about 2V_i. Similarly, they supply charges to C_o such that the output increases from V_o3 to V_o1. The required time to complete this operation is T_x. Then, at (nT + T_x), the source V_i, C_fA2, C_fB1, C_fB2 and C_o supply charges to the load. The discharges of flying capacitors and output capacitor decrease the output from V_o1 to V_o2. At (nT + T/2 − T_d), all switches are turned off. Only C_o supplies charges to the load, and thus the output drops more rapidly than before from V_o2 to V_o3. In the above analysis, it is assumed that T/2 is longer than 6R_lC_f2 (where C_f1 = C_fA1 = C_fB1 and C_f2 = C_fA2 = C_fB2) to the charge redistribution during T to T_x, T/2 to T/2 + T_x, is complete, while C_fA2 and C_fB2 are charged to about 2V_i at the end of half of a switching cycle.

Based on the RC circuits in Figure 9, at (nT − T_d), the charge redistribution between C_fB1, C_fA2, and C_fB2 is completed. Considering the voltage at the output of the first stage of charge pump is a constant value, the current passing through C_fB2 and C_o is in constant ratio. Hence, the current passing through C_fA2 and C_o can be approximated as C_fA2I_o/(C_fA2 + C_o) and C_oI_o/(C_fA2 + C_o). Similarly, the current passing through C_fB2 and C_fA1 is C_fA2C_fB2I_o/(C_fA2 + C_o)/(C_fB2 + C_fA1) and C_fA2C_fA1I_o/(C_fA2 + C_o)/(C_fB2 + C_fA1). At nT, the charges stored in C_fA2 and C_fB1 are C_fA2(V_o2 + C_fA2I_oR_l2/(C_fA2 + C_o)) and C_fB1V_Cf1, respectively. It is noted that the negative charges at the bottom plate of C_fB2 should also be considered. Assuming that C_f1/C_f2 = C_f2/C_o = m₂, the highest output voltage of the first and second stage of the charge pump achieves at T_x1 and T_x2. According to similar analysis of Equations (14) and (15), it could be seen that T_x1 and T_x2 have close values due to the log relationship of T_x and m₁. Hence, it is reasonable to assume that V_b and V_o1 both achieve at (nT + T_x). Then, at (nT + T_x), the current passing through C_fA2 and C_o is 0. Therefore, the charges remaining in C_fA2, C_fB1 and C_fB2 are C_fA2V_b, C_fB1(V_b − V_i + I_oR_l1) and C_fB2(V_o1 − V_b + I_oR_l2), respectively. By the principle of conservation of charges [7],

$$\begin{gathered} C_{f2}(V_{o2}+I_o{R}_{l2}\frac{C_{f2}}{C_{f2}+C_o}-V_c)+C_{f1}{V}_{Cf1}-C_{f2}\left(V_c+I_o{R}_{l3}\frac{C_{f2}}{C_{f2}+C_o}\frac{C_{f1}}{C_{f1}+C_{f2}}\right) \\ =C_{f2}{V}_b+C_{f1}\left(V_b-V_i+I_o{R}_{l1}\right)-C_{f2}\left(V_{o1}-V_b+I_o{R}_{l2}\right) \end{gathered}$$

(28)

T_x could be approximated by T_x2, which satisfies the following equation.

$$T_x=R_{l2}\left(C_o//C_{f2}\right)\ln \left(\frac{I_{co0}\left(C_o+C_{f2}\right)-I_o{C}_o}{I_o{C}_o}\right),$$

(29)

with $I_{Co0}=\frac{V_{i2}\times 2-V_{o3}}{R_{l2}}$. For simplification of the calculation, the average output voltage value for the first stage V_i2 is approximated as V_c, which could be calculated without T_x, and the related calculation of V_c is given in the following part.

V_Cf1 is the voltage obtained by capacitor C_fA1 when charging with an ideal voltage supply V_i within a time period T/2. Therefore, V_Cf1 is given by

$$V_{Cf}=\left(1-2e^{-\frac{T-2T_d}{2R_{l1}{C}_{f1}}}\right)V_i+V_c{e}^{-\frac{T-2T_d}{2R_{l1}{C}_{f1}}}+\frac{I_o{C}_{f1}{C}_{f2}{R}_{l1}}{\left(C_{f1}+C_{f2}\right)\left(C_{f2}+C_o\right)}{e}^{-\frac{T-2T_d}{2R_{l1}{C}_{f1}}}$$

(30)

At nT, the charges in C_fB2 obtained before the deadtime in the previous half of a cycle is C_fB2[V_c + I_oR_l3C_fA2C_fA1I_o/(C_fA2 + C_o)/(C_fB2 + C_fA1)], and the charges in C_o is C_oV_o3. At (nT + T_x), the charges in C_fB2 and C_o are C_fB2(V_o1 − V_b + I_oR_l2) and C_oV_o1, respectively. The charges to the load are I_oT_x. Thus, the following relationship is obtained.

$$C_{f2}\left(V_c+I_o{R}_3\frac{C_{f2}}{C_{f2}+C_o}\frac{C_{f1}}{C_{f1}+C_{f2}}\right)+C_o{V}_{o3}=C_{f2}\left(V_{o1}-V_b+I_o{R}_{l2}\right)+C_o{V}_{o1}+I_o{T}_x$$

(31)

Between (nT + T_x) and (nT + T/2 − T_d), some charges in C_fA2 and C_fB1 go to C_fB2. Since the overall charges in the connection of three capacitors are constant, the following equation is obtained.

$$\begin{array}{ll}{C}_{f2}{V}_b+C_{f1}& \left(V_b-V_i+I_o{R}_1\right)-C_{f2}\left(V_{o1}-V_b+I_o{R}_2\right)\\ & =C_{f2}\left(V_c+I_o{R}_3\frac{C_{f2}}{C_{f2}+C_o}\frac{C_{f2}}{C_{f1}+C_{f2}}\right)+C_{f1}\left(V_c-V_i+I_o{R}_1\frac{C_{f2}}{C_{f2}+C_o}\frac{C_{f1}}{C_{f1}+C_{f2}}\right)-C_{f2}(V_{o2}\\ & +I_o{R}_2\frac{C_{f2}}{C_{f2}+C_o}-V_c)\end{array}$$

(32)

Moreover, the net charges of C_fA2 and C_o will supply the load so that the following expression is obtained.

$$C_{f2}\left(V_{o1}-V_b+I_o{R}_2\right)+C_o{V}_{o1}=C_{f2}\left(V_{o2}-V_c+I_o{R}_2\frac{C_{f2}}{C_{f2}+C_o}\right)+C_o{V}_{o2}+I_o\left(T/2-T_d-T_x\right)$$

(33)

Finally, during the deadtime, the charges from C_o supplies to the load gives the following relationship.

$$\left(V_{o2}-V_{o3}\right)C_o=I_o{T}_d$$

(34)

Solving Equations (28)–(34), V_o1, V_o2 and V_o3 could be calculated, and the average value of V_o is given by

$$\overline{V_o}=\frac{T_x}{T}\left(V_{o1}+V_{o3}\right)+\frac{T_d}{T}\left(V_{o2}+V_{o3}\right)+\frac{T-2T_x-2T_d}{2T}\left(V_{o1}+V_{o2}\right)$$

(35)

Similar to the analysis of Equation (16), PCE can be easily derived by the ratio of E_o to E_i.

$$PCE=\frac{E_o}{E_i}=\frac{\overline{V_o}{I}_o}{4V_i{I}_o}=\frac{\overline{V_o}}{4V_i}$$

(36)

where ${\overline{V}}_o$ is found as shown in Equation (35).

3. Results

Simple simulations were conducted to prove the analysis. Ideal switch models with non-zero on-resistance are used [12]. The supply voltage V_i is 1.6 V, and the load current I_o is 50 mA.

The values of switching frequency, flying capacitors and output capacitor are chosen according to the load, which are highly related to the ripple voltage at the output. With the condition fulfilled and the assumptions presented, the proposed analysis is appliable to different values of switching frequency, flying capacitors and output capacitor. In the simulations, for 1-stage linear charge pump and dual-branch charge pump, the switch resistance is 1 Ω and the clock frequency is 36 kHz. The capacitance of the flying capacitor C_f and the load capacitor C_o is 4.7 µF and 1.5 µF, respectively. The calculation results according to [7] and this work are shown in

Table 1. The simulation result for 1-stage linear charge pump.

Parameter	[7]		This Work		Simulation
	Result (V)	Error	Result (V)	Error	Result (V)
Vo1	3.02	8.72%	2.77	−0.10%	2.77
Vo2	2.90	5.90%	2.74	−0.05%	2.74
Vo3	2.44	7.11%	2.28	−0.05%	2.28
ΔVo	0.58	16.14%	0.49	−0.34%	0.50
$\overline{V_o}$	2.82	7.98%	2.57	−1.56%	2.61

and

Table 2. The simulation result for dual-branch 1-stage linear charge pump.

Parameter	[7]		This Work		Simulation
	Result (V)	Error	Result (V)	Error	Result (V)
Vo1	3.16	5.78%	2.99	0.00%	2.99
Vo2	3.05	4.08%	2.93	−0.02%	2.93
Vo3	3.05	4.09%	2.93	−0.01%	2.93
ΔVo	0.12	85.07%	0.06	0.48%	0.06
$\overline{V_o}$	3.11	4.66%	2.96	−0.29%	2.97

. The simulation is conducted by Hspice, and the result is also given in Table 1 and Table 2. The error value, which is the percentage of the difference between calculation and simulation over the simulation result, is also given for better insight.

For 4× cross-coupled charge pump, the on-resistance of each switch is 0.5 Ω and the clock frequency is 20 kHz. The capacitance of the flying capacitor C_f1, C_f2, and the load capacitor C_o is 8 µF, 4 µF and 2 µF, respectively. The simulation and calculation results are shown in

Table 3. The simulation result for 4× cross-coupled charge pump.

Parameter	[7]		This Work		Simulation
	Result (V)	Error	Result (V)	Error	Result (V)
Vo1	5.67	1.58%	5.59	0.04%	5.58
Vo2	5.46	1.28%	5.39	−0.08%	5.39
Vo3	4.84	−10.26%	5.39	−0.08%	5.39
ΔVo	0.83	333.78%	0.20	3.44%	0.19
$\overline{V_o}$	5.36	−2.64%	5.49	−0.31%	5.50

.

According to the tables, the prediction error is reduced to below 0.1% for V_o1, V_o2 and V_o3 in all three cases, which shows a more reliable method when choosing optimal capacitor values for charge pumps.

4. Conclusions

By properly modeling the charge pump circuits, this paper demonstrates a better way for analyzing charge pumps. Hspice simulation is conducted to prove the reliability of the analysis method. According to the equation given, the PCE of charge pumps is closely related to the value of clock period, lumping resistance, flying capacitor and load capacitor. Hence, optimization of charge pump is possible based on careful choosing of the component values. Additionally, the analysis could be extended to other charge-pump circuits such as exponential, Fibonacci and Cockcroft–Walton charge pumps.