**2. Traditional Unbalanced Electric Bridge DC-IM Method and Analysis of the Influence on GC**

The unbalanced electric bridge DC-IM topology circuit is shown in Figure 1, where *v*dc is the DC voltage; *Rf*<sup>2</sup> and *Rf*<sup>1</sup> are the positive and negative insulation resistances, respectively; *R<sup>a</sup>* and *R<sup>b</sup>* are the bridge resistances, *R<sup>a</sup>* = *R1*, *R<sup>b</sup>* = *R1*||*R2*, so *R<sup>a</sup>* > *R<sup>b</sup>* . The unbalanced electric bridge method works in two phases, namely, *M*<sup>1</sup> and *M*2. In the *M*<sup>1</sup> phase, *Q*<sup>1</sup> turn-on and *Q*<sup>2</sup> turn-off, the positive half-bridge resistance is *Ra*, the negative half-bridge resistance is *R<sup>b</sup>* , the negative and positive half-bridge voltages are *v*<sup>11</sup> and *v*12, respectively, and the ground current is *i*1, as shown in Figure 1a. In the *M*<sup>2</sup> phase, *Q*<sup>1</sup> turn-off and *Q*<sup>2</sup> turn-on, the positive half-bridge resistance is *R<sup>b</sup>* , and the negative half-bridge resistance is *Ra*. The negative and positive half-bridge voltages are *v*<sup>21</sup> and *v*22, respectively, and the ground current is *i*2, as shown in Figure 1b. The conventional DC-IM method can be expressed as Equation (1) by Kirchhoff's law.

**Figure 1.** The circuit of unbalanced bridge. (**a**) The circuit of *M*<sup>1</sup> phase. (**b**) The circuit of *M*<sup>2</sup> phase. (**c**) The equivalent circuit of *M*<sup>1</sup> phase. (**d**) The equivalent circuit of *M*<sup>2</sup> phase.

$$\begin{cases} \begin{array}{l} \mathcal{R}\_{f1} = \frac{\upsilon\_{dc}(\mathcal{R}\_{d} - \mathcal{R}\_{b}) - (i\_{1} - i\_{2})(\mathcal{R}\_{d}\mathcal{R}\_{b})}{i\_{1}\mathcal{R}\_{b} - i\_{2}\mathcal{R}\_{d}}\\ \mathcal{R}\_{f2} = \frac{\upsilon\_{dc}(\mathcal{R}\_{d} - \mathcal{R}\_{b}) - (i\_{1} - i\_{2})(\mathcal{R}\_{d}\mathcal{R}\_{b})}{i\_{1}\mathcal{R}\_{d} - i\_{2}\mathcal{R}\_{b}} \end{array} \tag{1}$$

When the DC system has GC, the capacitance value of the DC negative pole to the earth is *C*1, and the capacitance value of the DC positive pole to the earth is *C*2. Thus, the circuits in Figure 1c,d can be changed to those shown in Figure 2a,b. To facilitate calculation and analysis, the equivalent resistance of the two working modes is assumed to be what is shown in Equation (2). Figure 2a,b can be simplified as Figure 2c,d.

$$\begin{cases} R\_{11} = \frac{R\_b R\_{f1}}{R\_b + R\_{f1}}, R\_{12} = \frac{R\_b R\_{f2}}{R\_a + R\_{f2}}\\ R\_{21} = \frac{R\_a R\_{f1}}{R\_a + R\_{f1}}, R\_{22} = \frac{R\_b R\_{f2}}{R\_b + R\_{f2}} \end{cases} \tag{2}$$

**Figure 2.** Equivalent circuit of unbalanced bridge with GC. (**a**) The circuit of *M*<sup>1</sup> phase. (**b**) The circuit of *M*<sup>2</sup> phase. (**c**) The equivalent circuit of *M*<sup>1</sup> phase. (**d**) The equivalent circuit of *M*<sup>2</sup> phase.

The following parameters are set.

$$\begin{cases} X\_{11} = \frac{R\_{11}}{R\_{11} + R\_{12}}, X\_{12} = \frac{R\_{12}}{R\_{11} + R\_{12}}\\ X\_{21} = \frac{R\_{21}}{R\_{21} + R\_{22}}, X\_{22} = \frac{R\_{22}}{R\_{21} + R\_{22}} \end{cases} \tag{3}$$

According to Figure 2, the time constants of the *M*<sup>1</sup> and *M*<sup>2</sup> phases are defined as

$$\tau\_1 = \frac{(\mathbb{C}\_1 + \mathbb{C}\_2)\mathbb{R}\_{11}\mathbb{R}\_{12}}{(\mathbb{R}\_{11} + \mathbb{R}\_{12})} \text{ and } \tau\_2 = \frac{(\mathbb{C}\_1 + \mathbb{C}\_2)\mathbb{R}\_{21}\mathbb{R}\_{22}}{(\mathbb{R}\_{21} + \mathbb{R}\_{22})} \tag{4}$$

When the two phases switch with each other, the charging process of GC belongs to the first-order circuit full response process, and the curvilinear function Equation (5) can be obtained, where *v*110*, v*120*, v*210, and *v*<sup>220</sup> are the initial voltage of the full response processes of *v*11, *v*12, *v*21, and *v*22, respectively.

$$\begin{cases} v\_{11} = v\_{dc}X\_{11} + (v\_{110} - v\_{dc}X\_{11})e^{-\frac{t}{\tau\_1}} \\ v\_{12} = v\_{dc}X\_{12} + (v\_{120} - v\_{dc}X\_{12})e^{-\frac{t}{\tau\_1}} \\ v\_{21} = v\_{dc}X\_{21} + (v\_{210} - v\_{dc}X\_{21})e^{-\frac{t}{\tau\_2}} \\ v\_{22} = v\_{dc}X\_{22} + (v\_{220} - v\_{dc}X\_{22})e^{-\frac{t}{\tau\_2}} \end{cases} \tag{5}$$
