# Four-Channel Buck-Type LED Driver with Automatic Current Sharing and Soft Switching
## 2 Proposed LED Driver
### Output Voltage Derivation  
Before the output voltage is derived, the voltages on the three energy-transferring capacitors in the circuit should be obtained first, which can be acquired by using the potential mean value of each point in the circuit:  
\[V_{\text{C1}}=\frac{1+D}{2}\times V_{in} \tag{66}\]  
\[V_{\text{C2}}=\frac{V_{in}}{2} \tag{67}\]  
\[V_{\text{C3}}=\frac{D}{2}\times V_{in} \tag{68}\]  
Afterwards, the output voltage of the proposed circuit is derived. First, in each state, the voltage is multiplied by the current, and the input power \(P_{in}\) is the average of the sum of all the corresponding calculated values shown in (79):  
\[\begin{array}{rl}P_{in}&=\frac{1}{T_{s}}\int\limits_{t_{0}}^{t_{0}+T_{s}}V_{ in}\times i_{in}(t)dt\\ &=V_{in}f_{s}\times\left[\begin{array}{c}\int\limits_{t_{0}}^{t_{1}}i_{\text {C1}}(t)dt+\int\limits_{t_{1}}^{t_{2}}i_{\text{C1}}(t)dt+\int\limits_{t_{2}}^{ t_{3}}i_{\text{C1}}(t)dt+\int\limits_{t_{3}}^{t_{4}}i_{\text{S1}}(t)dt\\ +\int\limits_{t_{4}}^{t_{6}}0 dt+\int\limits_{t_{8}}^{t_{9}}i_{\text{S1}}(t)dt+ \int\limits_{t_{9}}^{t_{0}+T_{s}}i_{\text{C1}}(t)dt\end{array}\right]\end{array} \tag{69}\]  
Since the current in the energy-transferring capacitor \(C_{1}\) rises from \(-I_{o}\) in state 10 and to \(I_{o}\) at the end of state 2, the sum of the corresponding average values is zero, as shown in (70):  
\[\int\limits_{t_{0}}^{t_{1}}i_{\text{C1}}(t)dt+\int\limits_{t_{1}}^{t_{2}}i_{ \text{C1}}(t)dt+\int\limits_{t_{9}}^{t_{0}+T_{s}}i_{\text{C1}}(t)dt=0 \tag{70}\]  
Moreover, the currents in the switch \(S_{1}\) in states 4 and 9 are \(0.5I_{o1}\) and \(-0.5I_{o1}\), respectively, and the elapsed times of states 4 and 9 are the same, so the sum of the corresponding average values is zero, as shown in (81):  
\[\int\limits_{t_{3}}^{t_{4}}i_{\text{S1}}(t)dt+\int\limits_{t_{8}}^{t_{9}}i_{ \text{S1}}(t)dt=0 \tag{71}\]  
Therefore, (69) can be rewritten as  
\[P_{in}=V_{in}f_{s}\times\int\limits_{t_{2}}^{t_{3}}i_{\text{C1}}(t)dt \tag{72}\]  
According to (15), (72) can be rewritten as  
\[P_{in}=V_{in}f_{s}\times I_{o}\times(t_{3}-t_{2}) \tag{73}\]  
By assuming that the input power \(P_{in}\) is equal to the output power \(P_{o}\), the output voltage can be obtained as follows:  
\[P_{in}=V_{in}f_{s}\times I_{o}\times(t_{3}-t_{2}) \tag{74}\]  
\[P_{o}=4\times V_{o}\times I_{o} \tag{75}\]  
\[V_{o}=\frac{t_{3}-t_{2}}{4\times T_{s}}\times V_{in} \tag{76}\]The elapsed time experienced by state 3 is  
\[\begin{array}{ll}t_{3}-t_{2}&=DT_{s}-(t_{2}-t_{0})\\ &=DT_{s}-\frac{t_{2}\times L_{1}}{V_{m}-V_{C1}}+\frac{iC_{1}(t_{0})\times L_{1} }{V_{m}-V_{C1}}\\ &=DT_{s}-\frac{t_{2}\times L_{1}}{V_{m}-V_{C1}}-\frac{t_{2}\times L_{1}}{V_{m}- V_{C1}}+\left(T_{d}-\frac{V_{m}\times C_{S2}}{t_{0}}\right)\\ &=DT_{s}-\frac{2\times L_{s}\times L_{1}}{V_{m}-V_{C1}}+\left(T_{d}-\frac{V_{m }\times C_{S2}}{t_{0}}\right)\end{array} \tag{77}\]  
By ignoring the blanking time \(T_{d}\), the output voltage representation or the output current representation can be obtained to be  
\[\begin{array}{ll}V_{0}=\frac{D}{4}\times V_{in}-\frac{f_{\times}\times L_{ 0}\times L_{1}}{1-D}\\ \Rightarrow I_{0}=\frac{1-D}{L_{1}\times f_{s}}\times\left(\frac{D}{4}\times V _{in}-V_{o}\right)\end{array} \tag{78}\]  
According to (78), the output voltage \(V_{o}\) is determined by the duty cycle \(D\), the resonant inductor \(L_{r1}\), the output current \(I_{o}\) and the switching frequency \(f_{s}\). In this paper, the change in \(D\) is used to adjust \(I_{o}\) to achieve LED dimming.