diodes - Solving Clipper Circuit containing battery and resistance in series

I have learned to solve clipper circuit with and with out battery in parallel to the output node. But have no idea what to do if a resistance also connected with it in series. There is nothing said about it in my book. How can I solve this circuit?

^{simulate this circuit – Schematic created using CircuitLab}

A general procedure for solving this type of problem will be helpful.

Answer

Steps

1. Consider the circuit in three cases as shown below.



2. In each case, replace diodes with their equivalent model: reverse biased diode with open circuit and forward biased diode with a voltage source (drop) of 0.7V (+ve terminal to anode side).


3. Use node or mesh analysis to find the output.

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case1: only D1 conducts. (\$V_i > 5.7 V\$)

$$V_{R_1} = (V_i - 5.7)\frac{R1}{R1+R}$$ $$V_o = V_{R_1} + 5.7$$

case2: only D2 conducts.(\$V_i < -5.7 V\$)

$$V_{R_2} = (V_i + 5.7)\frac{R2}{R2+R}$$ $$V_o = V_{R_2} - 5.7$$

case3: Neither D1 nor D2 conduct. (\$-5.7 < V_i < 5.7 \$)

$$V_o = V_i$$

Note: The equivalent model of diode considered is piece-wise linear, an ideal diode in series with a voltage source (drop). You can add a resistance in series also to include the forward resistance of diode. I assumed silicon diode. If using any other diode, replace 0.7V with the cut-in voltage of that diode.