# Resistance of a semiconductor diode

When we calculate the dynamic resistance $r=(\frac{dv}{dI})$, for any n-p junction, how is it different from the normal resistance $R=\frac VI$? Does the equation for the voltage drop (The fermi potential drop, and not the absolute Galvani potential) work if we use the dynamic resistance with the instantaneous current ($V=Ir$)? Does the power dissipation relation, $P=I^2r$ hold in case of dynamic resistances? If it does, is power dissipated as heat even in case of the n-p junction? I think it is unlikely, as the hole-electron recombinations are the dominant phenomenon here, and I am unsure whether those can produce heat.

• The dynamic resistance depends of the general forward characteristics of the diode (or other power semiconductor), which is a complex logarithmic function with at least 3 constants. So to determine the mean power loss we must have an accurate instantaneous forward voltage and the related instantaneous current. An approximate slope of the resistance Rt and threshold voltage we can get from curve simplification (consider two straight lines joined at knee point Vto). With this method, and with average current known plus the conduction angle, we are able to calculate the Average Power Loss – GR Tech Jan 28 '14 at 8:04

For the ideal resistor, the voltage across is proportional to the current through and thus, their ratio is the constant $R$:

$$\frac{v_R}{i_R} = R$$

For the ideal (semiconductor) diode, we have

$$i_D = I_S(e^{\frac{v_D}{nV_T}}-1)$$

Inverting yields

$$v_D = nV_T\ln (1 + \frac{i_D}{I_S})$$

thus, the diode voltage is not proportional to the diode current, i.e., the ratio of the voltage and current is not a constant.

$$\frac{v_D}{i_D} = \frac{nV_T}{i_D}\ln (1 + \frac{i_D}{I_S}) \ne R$$

Now, the small-signal or dynamic resistance is just

$$\frac{dv_D}{di_D} = \frac{nV_T}{I_S + i_D} \approx \frac{nV_T}{i_D}$$

how is it different from the normal resistance

As shown above, the diode static resistance (ratio of the diode voltage and current) differs from and is, in fact, larger than the diode dynamic resistance by the factor of $\ln (1 + \frac{i_D}{I_S})$

$$\frac{v_D}{i_D} = \frac{dv_D}{di_D} \ln (1 + \frac{i_D}{I_S})$$

which is to say that, in typical operating ranges, the diode dynamic resistance is much smaller than then diode static resistance.

Does the power dissipation relation, $P=I^2r$ hold in case of dynamic resistances?

The instantaneous power associated with the diode is

$$p_D = v_D i_D = nV_Ti_D\ln (1 + \frac{i_D}{I_S}) \ne i_D^2\frac{nV_T}{i_D} = nV_Ti_D$$

Since the power associated with a circuit element is always the product of the voltage across and current through, one would not use the dynamic resistance but, rather the static resistance.

• Then, the static resistance at a particular value of voltage, can be used pretty much in the same way we would use for a resistor? ($P=I^2R,V+IR$)? – Satwik Pasani Jan 28 '14 at 14:57
• @SatwikPasani, see the update to my answer. – Alfred Centauri Jan 28 '14 at 15:00

For any two-terminal device, or for any two terminals of any device, we can graph current vs. voltage. For a purely resistive device, this is a straight line passing through the origin, and the slope is the inverse of resistance. For a non-linear device, like a diode, it's not a straight line (that's what not linear means). Example:

At any point, the slope of this line is the dynamic conductance; the inverse of that is the dynamic resistance. For example, in the reverse region, the line has a very low slope, a very low conductance, or a very high resistance. In the forward region, high slope, high conductance, low resistance.

$P=I^2r$ does not hold if $r$ is the dynamic resistance. $P=IE$ does, however.

The reason $P=I^2R$ works is because a resistive device obeys Ohm's law, $E=IR$. From this, we can calculate the power from any one of voltage or current, because although we need both for power, we can calculate one from the other:

\begin{align} P&=IE \\ E&=IR \\ P&=I(IR) \\ P&=I^2 R \end{align}

Because non-linear devices don't obey Ohm's law, $P=I^2 R$ does not apply for them. $P=IE$ does, however, and if you can come up with some other equation which relates current to voltage for that device and substitute it into $P=IE$, you could come up with an equation which calculates power from just voltage, or just current, for that device.

Phil Frost's answer is great. I'd just like to add that as a rough approximation, it's often possible to model a diode (or other junction drop, like a BJT or IGBT) as a voltage source in series with a resistor.

simulate this circuit – Schematic created using CircuitLab

So for large-signal purposes, you can estimate the losses through the diode as: $$P=I_{mean}V_{D} +I_{RMS}^2R_D$$

Whether this estimate gets you close enough or not depends entirely on your domain, but I've had good success with it for narrowing down my component selection in switching power supplies.

how is it different from the normal resistance R=VI?

It is different because it is dependent of the current we are letting through the diode, which gives different apparent values of resistance. The normal resistance R = VI will appear always same when you measure it on a resistor ( apart from minor deviations ).

Does the equation for the voltage drop work if we use the dynamic resistance with the instantaneous current (V=Ir)?

It does, because 'r' is defined such that it shows the apparent resistance of diode, with regard to given current and voltage. However, for the real picture, looking at the V/I characteristic of a diode would be a better choice.

Does the power dissipation relation, P=I^2*r hold in case of dynamic resistances? If it does, is power dissipated as heat even in case of the n-p junction?

It holds, but as I changes, P will change swiftly too, because it is in quadratic proportion with r, which includes I in a derivative equation.

Think about Ohm's law: if you have a circuit of two series resistors (1k and 2k, U=3V), you could find equivalent resistance ($R_e=R_1+R_2=3k$), calculate current ($I=\frac{U}{I}=1mA$) and then work out resistors' voltage drop ($U_d=IR$; $U_1=1*1=1V$; $U_2=1*2=2V$). Nice. The maths do not break too: $R_2$'s current is $I_{R2}=\frac{U-U_1}{R_2}=\frac{3-1}{2k}=1mA$.
Unlike resistors, which have constant resistance, diodes are defined by constant voltage drop! (yep, resistors also have capacitance and inductance, but in most cases these are not important, so let's consider that to be true). So all you need to do here is to change your variables accordingly: for resistors you know resistance, for diodes - voltage drop.
Back to power dissipation: $P=I^2R$ only works because we can do variable substitution, but in the end the power dissipation is given by $P=UI$. The question is why do we use the first form for traditional resistors? Because we know circuit current and resistance are too lazy to work voltage drop out. And for diodes we already know that. So you can't use $P=I^2R$ for diodes because it is a derived form of $P=UI$ when $R(U)=const$. $$P=UI \ \ \ \ \ <= \text{works even on Mars}$$ And if you are really unsure if diodes can produce heat just disassemble any high power LED torch - you will find that powerful LED happily sitting on top of aluminium radiator. It is there for a reason, right?

In addition of my comment above, let me give a short explanation why there is a non-linear behavior of diode. So at the time that the polarity of the applied voltage is inverted (from reverse to forward), the initial resistance in the diode is high due to the lack of carriers in the semiconductor structure. As the forward current increase the number of carriers builds-up and resistance goes down. This high value of resistance brings an overshoot in the forward voltage from which the voltage in the steady-state value, and the characteristic “knee” established. This phenomenon last some time and highly depended by temperature, that’s why the dynamic resistance of the diode is calculated in both environment and high virtual temperatures. This forward voltage value called “Forward Recovery Voltage” and in some diodes can be high (30Vor more). In power engineering, this transient “conductivity modulation” often neglected and the equation is linear (like MOV). But in small signal application the transition capacitance of the junction as well as the stored charge equivalent capacitance should included in model analysis.

Regarding the last question:

[...] is power dissipated as heat even in case of the n-p junction? I think it is unlikely, as the hole-electron recombinations are the dominant phenomenon here, and I am unsure whether those can produce heat.

When an electron and hole recombine, the extra energy is emitted in the form of a photon. When a photon collides with other particles of matter, it causes increased particle movement, which in turn results in an increase in temperature. So there is heat involved isn't it?