Function of a voltage clamp

[shortstuff410]

Homework Statement

I am given a drawing of a circuit with a v(in), a resistor, and then two diodes pointing in the same upward direction, with a 5v and -5v limit put next to each diode respectively. This then connects to V out. It looks something like what I have posted below, but with the vertical part shifted over to the other side of the resistor...sorry I couldn't get it to format here.

. 5v
. _
. ^
. |
. |
v(in)----^^^-----------v(out)
. |
. |
. _
. ^
. -5v

The question is to explain the operation of the diode clamp. What happens when v(in) is greater 5v, or less than -5v. What happens when it is in the bounds?

Homework Equations

The Attempt at a Solution

It seems to me that this would simply work as an open circuit when it is outside the bounds, and function normally when it is within the bounds, but this seems to easy. Am I on the right track?

Thanks so much!

 

[anathema]

Hi there,

You are correct in your understanding of the diode clamp circuit. When the input voltage (v(in)) is greater than 5v, the diode connected to the 5v limit will become forward biased and allow current to flow through it, effectively clamping the output voltage (v(out)) at 5v. Similarly, when the input voltage is less than -5v, the diode connected to the -5v limit will become forward biased and clamp the output voltage at -5v.

When the input voltage is within the bounds (-5v to 5v), both diodes will be reverse biased and act as open circuits, allowing the circuit to function normally and the output voltage to be determined by the input voltage and the resistor.

This circuit is commonly used to protect electronic components from overvoltage, as the diodes will limit the voltage that can be applied to the circuit. It is important to note that the diodes have a voltage drop of around 0.7v, so the output voltage will never be exactly at the limit voltage.

I hope this helps! Let me know if you have any other questions.