## 3 Comparative Analysis  
The use of fewer (four) switching transistors and diodes (six) decreases the volume and cost of the converter. The control signal generated by any digital controller (microcontroller or processor) is not able to directly govern the switching states (on-to-off and off-to-on) of a switching transistor as their voltage level is as low (normally 0 to 5 V) as that of the power circuit. Thus, electrical isolation of the high-voltage side to a low voltage level is mandatory. For this purpose, the turn-on and turn-off control of a switching transistor requires the use of one gate-controlling (driving) circuit and one isolated DC source. The realization of a gate-controlling circuit is achieved by employing a hybrid chip along with many passive components and microelectronic devices. In the same way, the implementation of an isolated DC source needs a low-frequency step-down transformer, a bridge rectifier, and filtering components to smooth the output. Therefore, the size and cost of the gate-controlling circuit and isolated DC source are much larger than those of a switching transistor. The total volume and cost of the converter heavily depend on the number of switching transistors. In addition, the proposed topology requires fewer diodes as well. Thus, the elimination of diodes and switching transistors is the major achievement of the proposed circuit topology. A bipolar voltage gain boost converter, as reported in [20], is realized with the use of six switching transistors (six gate-controlling circuits and six isolated DC sources) and eight diodes. However, the developed circuit can be used to realize all outputs of [20] with the help of four transistors (four controlling circuits and isolated DC sources) and six diodes only. Therefore, the realization of the proposed topology drastically reduces the overall volume and cost. It can be seen for any of the conducting intervals in all operating modes of Figures 8 and 9 that the operation of two switching transistors and three diodes realizes the required result. There is no change in the average conduction losses of the developed circuit and the circuit in [20]. Equation (17) illustrates the average conduction losses caused by one switching transistor and one diode. Here, '\(V_{d}\)', '\(R_{d}\)' and '\(R_{m}\)' represent the forward voltage of the diode, forward resistance of the diode, and on-state resistance of the switching transistor, respectively. \[P_{cond(diade+transistor)}=\frac{V_{d}I_{o(p)}}{\pi(1-k)}+\frac{I_{o(p)}^{2}}{4 (1-k)^{2}}R_{d}+\frac{I_{o(p)}^{2}}{4(1-k)^{2}}R_{m} \tag{17}\]In one cycle operation of the output voltage, four transistors and six diodes are to be conducted to realize the output. The total average conduction losses may be expressed as  
\[P_{cond}\,=\,\frac{6V_{d}I_{o(p)}}{\pi(1-k)}+\frac{6I_{o(p)}^{2}}{4(1-k)^{2}}R_{ d}+\frac{I_{o(p)}^{2}}{(1-k)^{2}}R_{m} \tag{18}\]  
By employing the approach as used in [20], the average switching losses of a transistor and diode are calculated with the value of switching frequency (\(f_{s}\)) and reverse recovery charge (\(Q_{RR}\)). \[P_{sw(transistor+diode)}\,=\,\frac{V_{o(p)}I_{o(p)}f_{s}(t_{r}+t_{f})}{4(1-k)}+ \frac{Q_{RR}V_{o(p)}f_{s}}{\pi} \tag{19}\]  
It can be observed that switching losses are directly related to the switching frequency, so the switching losses of the low-frequency-operated diodes are negligible. In the proposed topology and the circuit in [20], there are only four conduction diodes (two for each half-cycle) at the high switching frequency.