# Are the two resistors R1 and R2 in parallel?

D1 and D2 are ON (short)
D3 is OFF (open)

My goal is to find a relation between Vin and Vout without using currents in the equation. I thought of applying KVL on a loop that includes the upper wire of the small loop (on which R1 resides), and ended up with:

-Vin +Vout+Vout=0

While a KVL around a loop containing the lower wire of the small loop gives:

-Vin +VB +Vout +Vout=0

Apparently, the two relations are different which is inconsistent. Then I thought about uniting the small loop to be just on one wire then finding a relation between Vout and Vin using KVL. Here I was wondering whether or not R1 and R2 are in parallel. They have one common node on the right of the loop, yet I'm not sure about the left node. The final answer I thought might be right is to do source transformation of (VB and R2) and make them as a (current source in // with R2). This way R1 and R2 will share two nodes and become parallel. Then this small loop will have Req= R1 // R2 and a current source parallel to Req. After that, I can do another source transformation and make my latter small loop as a wire of a voltage source that is in series with Req.

Is my thinking right?

Thank you

• I wish I could know why someone would give me -1 for my question while my reputation is 1. Like she/he won't gain anything but just do evil. – Joe Jan 24 '18 at 5:07

$R_1$ and $R_2$ are not in parallel. They are connected to one common node, but $V_B$ is 'in the way' of the other node. If $V_B$ were replaced with a short circuit then $R_1$ and $R_2$ would be in parallel.

Your loop equations do not make sense. (How did you get through Vout twice in one loop?)

I changed some of the notation to make it easier to follow. I follow the convention of subtracting a voltage when going from + to -. Following these loops counterclockwise:

Loop 1: $-v_{in}+v_{out}+(I_1*R_1)=0$

Loop 2: $-v_{in}+v_{out}-(I_2*R_2)+V_B=0$

You will have to expand the expression for $v_{out}$ in terms of of $R_3$ and the current that flows through it to solve these equations. It is unavoidable.

• Thank you. Can I find the Norton equivalent of VB, R2 which yields Req= R1 // R2 and IB = VB/R2 , where IB source is // to Req. Then find the Thevenin of IB, Req and make it as a wire then do KVL. ? – Joe Jan 23 '18 at 5:09
• @Joe - I don't think so. To find R_Norton, you short v_in and find the resistance seen by the terminals that connect to V_B and R2, which is R1//R3//[R4+a diode...?]. These conversions only work with linear circuits. You could use them in a small signal model of this circuit that converts the diode to a differential resistance, but not when solving for the DC operating point. – vofa Jan 23 '18 at 5:44

Are they in parallel, not really, but if you convert the loop R1 R2 VB into the Norton equivalent, the resistance term has the value of them in parallel. (the voltage term will be less than Vb)

You can then add Vin and solve the right hand part of the circuit.