# Clipper circuits, voltage across voltage divider

I understand that using diodes we can control the range of voltages across a voltage divider. However, following this tutorial I noticed he drew a smooth linear transition for voltages 0-6.7 in the bottom right corner.

I would've thought that once the ideal diode reaches its forward bias voltage and connecting the voltage source, the voltage across resistor R_L would immediately be 6.7V because it is in parallel like so:

Can someone please clear this up for me?

• Where did the tutorial come from? Sep 12, 2020 at 23:08

The two resistors form a voltage divider - the voltage at the junction between them will be half the supply voltage, unless the battery and diode have an effect.

Since the supply voltage is a 10 V peak sine wave, the junction of the two 10K resistors cannot rise above 5 volts, so the 6 volt battery and diode will have no effect on the operation.

The voltage across Rl will be a 5 V peak-to-peak sine wave.

The voltage of RL must reach 6.7V before there is 0.7V over the diode, as there is the extra 6V supply that must be exceeded too. Use a circuit simulator (falstad, spice, etc) to verify this.

Lets break the problem down. First we know that input signal here is pure sine wave. So output sine wave across RL will be pure sine wave except when current has low impedance path to flow from alternate branch which is branch with diode. Cathode of diode is at 6V with respective to ground(ground is not marked) Forward voltage of diode is 0.7V Therefore, voltage at Anode of diode must be 6.7V or mode for current to flow from this branch. Until input voltage reached 6.7V, all the current will flow through RL. Hence RL curve will follow input. Once voltage reached 6.7V current will flow through Diode branch which will prevent any voltage increment across RL.

Let $$\V_L\$$ be the load voltage.

Without the diode you have $$\V_L = {1 \over 2} V_{in}\$$.

With the diode, we have $$\V_L = {1 \over 2} V_{in}\$$ as long as $$\V_L \le 6.7\$$ (or $$\V_{in} \le 13.4\$$) and $$\V_L = 6.7\$$ if $$\V_{in} \le 13.4\$$.

In other words, $$\V_{L} = \max(6.7, {1 \over 2} V_{in})\$$.

If $$\V_{in}(t) = 10 \sin \omega t\$$ then $$\V_{L}(t) = 5 \sin \omega t\$$ as Peter pointed out.