# Explaining the current flow of the positive shunt clipper (diodes)

I am currently studying clippers or parallel limiters in the Navy and I was wondering if someone could clear up a few things for me.

Here is a picture that fits my description: A clipper has a diode in parallel with the output signal across from it and this output signal contains a load resistor. This is a positive clipper / positive parallel limiter. Assuming that the current flows from the left as an input, the current goes through the resistor and the diode. The current is forward biased so the diode acts as a short. This happens only for the positive part of the signal (I'm still curious why).

What I'm confused about is the reverse biased case when the current doesn't go through the diode and only through the load resistor parallel to the branch containing the diode. The diode is said to be open. When we talk about reverse bias, are we saying that the current is going from the battery Vdc to the resistor? What direction does the current flow for reverse bias?

• If your Vdc is meant to be a battery that is charged by the Vin sine wave, then you have the battery (or diode) backwards. The diode will conduct with a forward (positive) voltage across it. That will be the sum of the sine wave and the Vdc of the battery.) When the voltage across the diode is negative, then no current will flow. With no current there is no current direction :^) (Mind you I'm using a very simple model of a real diode.) – George Herold Nov 6 '14 at 15:37

For simplicity, I'm going to assume a 0.7V diode drop in the ON state. In your diagram, you can call the low node ground (0V), in that case, you have one of two situations:

• The diode is on (closed switch)

In this state, current can flow through the resistor and therefore, through the diode. In this case, 0.7V is dropped across the diode and the remaining voltage is dropped across the single resistor. In this case, Vout will just be 0.7V higher than the "clipped" voltage set by the power supply; i.e: $V_o = 0.7 - V_{dc}$

You remain in this state, as long as the diode is forward biased, i.e: $V_{in} > 0.7 - V_{dc}$, which of course, is set by your input signal. As soon as the voltage from the signal drops below this threshold... :

• The diode is off (open switch)

In this state, no current can flow through the diode, so no voltage can be dropped across the resistor, therefore, the output voltage is just equal to the input voltage.

Hope that helps :)

From the information you have put in your question I think that in the reverse bias case the 'diode is open' means that it behaves like an 'open circuit' or 'open switch' and no current passes through it and thus no current goes through the battery. The load resistor has current passing through it that goes to circuit/components not shown on the right hand side of the diagram. So in the reverse bias case no current passes through the battery and diode and it is as if they were a switch which was 'open'.

I am not really familiar with this circuit, but from the information you have given this looks like the answer to me.

The diode has a central barrier at the PN junction which allows charge to flow across it only when a sufficiently positive voltage is applied to the P part with respect to the N part so that the barrier potential is overcome.In this case the diode is said to be forward biassed and conducts current.At all times when the P part has a voltage less than the voltage at the N part the barrier potential is reinforced,the diode is said to be reversed biassed and charges cannot flow across it. When the diode is forward biassed and conducting it has almost zero resistance and acts as a short circuit and hence all the current prefers to take this path(of least resistance) than other parallel paths of higher resistance. When the diode is reverse biassed and non conducting it acts as an open circuit and hence the current takes the parallel path having lesser resistance. In the circuit the N part of the diode is clamped to -Vdc.The diode will be forward biassed for all voltages which are more positive than -Vdc.This is easy to understand for the positive part of the cycle.Even when the voltage at P starts becoming negative for all negative voltages applied at P having magnitude less than Vdc the diode remains forward biassed and conducting and the output Vo is -Vdc.When the magnitude of the negative voltage applied to P becomes more than magnitude Vdc the diode becomes reverse biassed and stops conducting.The battery Vdc is not measured at the output and Vo becomes same as the Input sine voltage. Hope this helps.