3-unit string with guard ring: redistribution of voltage

A guard ring adds capacitance between each link pin and the line conductor. This line capacitance partly cancels the charging current lost to the tower, so the voltage spreads more evenly across the discs.

Circuit and junction equations

Number the discs 1 (tower end) to 3 (line end), with voltages $V_1, V_2, V_3$. Let $V$ be the voltage of the line to earth. Each link pin has $0.2C$ to earth and $g\,C$ to the line.

 Tower ─[C]─ A ─[C]─ B ─[C]─ Line (V)
             |       |
   earth ─0.2C     0.2C
   line  ─gA·C     gB·C

Kirchhoff's current law at a junction (current entering from the line side = current leaving towards the tower + current to earth):

$$V_{k+1} = V_k + m\,(V_1+\dots+V_k) - g\,[\,V-(V_1+\dots+V_k)\,]$$

Case 1: line capacitance 0.1C at both pins

Junction A ($m=0.2,\ g=0.1$):

$$V_2 = V_1 + 0.2V_1 - 0.1(V - V_1) = 1.3V_1 - 0.1V$$

Junction B:

$$\begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.1(V - V_1 - V_2)\\ &= 1.99V_1 - 0.23V \end{aligned}$$

Since $V_1+V_2+V_3 = V$: $4.29V_1 - 0.33V = V$, so $V_1 = 0.3100V$.

$$\eta = \frac{V}{3V_3} = \frac{1}{3\times 0.38695} = 86.14\ \%$$

Case 2: guard ring raises the lower pin's line capacitance to 0.3C

Junction A is unchanged: $V_2 = 1.3V_1 - 0.1V$.

Junction B ($g = 0.3$):

$$\begin{aligned} V_3 &= V_2 + 0.2(V_1+V_2) - 0.3(V - V_1 - V_2)\\ &= 2.45V_1 - 0.45V \end{aligned}$$

Sum: $4.75V_1 - 0.55V = V$, so $V_1 = 0.3263V$.

$$\eta_2 = \frac{1}{3\times 0.34947} = 95.38\ \%$$

Answer: Without the larger guard ring capacitance the discs carry 31.00 %, 30.30 % and 38.69 % of $V$, and the string efficiency is 86.14 %. With 0.3C at the lower pin the voltages become 32.63 %, 32.42 % and 34.95 % of $V$, and the efficiency rises to 95.38 %. The guard ring takes stress off the line-end disc.

(For comparison, with no line capacitance at all the same string would carry 26.0 %, 31.3 % and 42.7 %, giving an efficiency of 78.0 %.)