Input Voltage vs Output Voltage for a Shunt Positive Clipper with Bias Voltage Assume voltage drop across diode is 0.7V in this case.

From what I understand

When V(input) is positive, but less than 0.7+V OR When V(input) is negative AKA When the diode is acting as open circuit:

V(output) will equal some fraction of V(input), depending on the value of R and the value of the Load Resistance. For example if R=RL, V(output) will equal 1/2 of V(input). And if R is very small compared to RL, V(output) will be approximately equal to V(input). This makes sense to me intuitively.

My confusion arises from the information I've read from varying sources, which all seem to indicate that V(output) will always be equal to V(input), when the diode is acting as an open circuit.

When V(input) is positive, and greater than 0.7+V AKA When the diode acts like a jumper wire

Vo is equal to 0.7+V. This makes sense because Vo is the voltage parallel to 0.7+V.

I tried simulating a circuit like this in falstad, and noticed that Vo will get clipped when Vo>(0.7+V), instead of when Vi>(0.7+V). I also noticed that Vo does not equal Vi when the diode is open circuit, Vo = 1/2Vi (I made the load resistance and the other resistor equal) But every source I've looked to says that Vo will always be equal to Vi AND that Vo will be clipped when Vi tries to go above 0.7+V.

My question is, what am I misunderstanding? Is Vo clipped when Vi is greater than 0.7+V or is Vo clipped when Vo is greater than 0.7+V. Is Vi always equal to Vo? The only situation in which I can think of Vi being equal to Vo for a circuit like this, is when R is small compared to RL, as mentioned earlier.