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I am tasked to find the small signal resistance of a diode operating in reverse breakdown region (not forward bias).

Method 1: Is using the small signal resistance formula correct even in the reverse breakdown region?

$$r_d = \frac{nV_T}{I_D}$$


Method 2: The I-V curve in reverse breakdown region seems to be steeper than the curve in forward bias region. If I use the derivative of current w.r.t voltage, then the resistance is close to zero due to the steep gradient.

Which is correct, or are they both wrong? If so, how do I go about solving this?

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  • \$\begingroup\$ to find ....reverse breakdown region OK, it is unclear if this is just a theoretical exercise or that you also will measure a diode and determine the value. Or perhaps both? Your first step is to think about what that small signal resistance actually is. Why would Method 1's formula be correct? Does that formula still apply? I think your Method 2, looking at the slope of the I-V curve makes more sense. Indeed the current will increase dramatically when breakdown occurs. It might be easier to not force a voltage but use a current measure the V-I curve. \$\endgroup\$ – Bimpelrekkie Feb 6 at 7:37
  • \$\begingroup\$ Almost a duplicate of electronics.stackexchange.com/questions/420681/… although that question also has no answers. \$\endgroup\$ – Elliot Alderson Feb 6 at 13:26
  • \$\begingroup\$ @Bimpelrekkie Thanks for the thought. I think you are right. I have checked with my professor and that was what he said. \$\endgroup\$ – GordonJun Feb 7 at 13:00

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