How do you calculate power loss for a diode?

For example, the DSB20I15PA Schottky diode datasheet mentions a threshold voltage of 0.23V and a slope resistance of 7.2mΩ.

I need to calculate the power loss at different voltages, as the battery drains. e.g. 14V down to 10V.

Is it as simple as subtracting 0.23V from your source voltage, or is it more complicated than that? Do you need to factor in the slope resistance too? Does current play a factor? How about diode temperature?

I'm using a 12V battery and a multimeter to measure the drop (by measuring pin to pin on the diode). In my test circuit, the drop appears to be 0.325V when there is a load, but only 0.168V when there is no load. The load is a DC motor.

• The slope resistance shows how much the forward voltage will rise vs. an increase in forward current. Jul 23 at 15:33
• You didn’t mention your current or if the measurements were peak, avg or rms nor the diode temp. Rise. Jul 23 at 16:23
• I was looking for an equation, rather than an exact answer to my specific problem. However, I do need to measure the current and temp as you're alluding to. As an easy solution, I'm thinking about putting the voltage sensor at the battery, before my idiot diode. Jul 23 at 17:38

The two parameters given are a linear approximation of a portion of the V-I curve of the Schottky diode at a certain junction temperature. Vf0 is a theoretical intercept, not something you can measure directly, whereas rF is the average slope of the curve (in the example curve, between the points of about 65A and 210A, so you need to be able to produce those currents to measure it).

Vf = Vf0 + If * rF

The power loss is the product of the instantaneous forward current and voltage.

Here is information on a different device, illustrating the approximation:

The maximum is going to be higher than the typical, as shown in the curve.

The maximum power loss from forward current will be higher at lower junction temperatures since the forward voltage will be higher, but usually that's not a concern.

Electrical power is, in every case, $$\V·I\$$. The power dissipated in the diode is the product of the voltage drop in the diode and the current through the diode; it's linear with current to a first approximation (linearithmic to a second approximation). In the same way, the power dissipated in a resistor is the product of the voltage drop in the resistor and the current through the resistor, and the power consumed by your motor is the product of the voltage across the motor and the current through it (not all of it is dissipated in this case because a significant portion of that power--hopefully!--gets converted to mechanical power; that's the point of a motor).

• I think this answer would be improved by discussing the specific parameters (slope resistance, threshold voltage) that the question is asking about. As it stands, I read it more like a general discussion of power, with some side comments about diode approximation. Jul 24 at 21:38

Well, let's make a mathematical closed solution. I know that this is maybe above the OP's knowledge, but I think it is important to show it in combination with the other answers given.

We know that in a DC-circuit the power is defined by:

$$\text{P}=\text{V}\cdot\text{I}\tag1$$

Where $$\\text{V}\$$ is the voltage across the component and $$\\text{I}\$$ is the current through the component.

The Shockley diode equation, gives the relation between the voltage across and the current trough a diode:

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\tag2$$

Where $$\\text{I}_\text{D}\$$ is the diode current, $$\\text{I}_\text{S}\$$ is the reverse bias saturation current, $$\\text{V}_\text{D}\$$ is the voltage across the diode, $$\\text{q}\$$ is the electron charge, $$\\text{k}\$$ is the Boltzmann constant, $$\\text{T}\$$ is the temperature and $$\\eta\$$ is the ideality factor.

So, the power in a diode is given by:

$$\text{P}_\text{D}=\text{V}_\text{D}\cdot\text{I}_\text{D}=\text{V}_\text{D}\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)=\frac{\text{I}_\text{D}\eta\text{k}\text{T}}{\text{q}}\cdot\ln\left(1+\frac{\text{I}_\text{D}}{\text{I}_\text{S}}\right)\tag3$$

Let's do an example. We are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_1=\text{I}_2+\text{I}_3\tag4$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\ \\ \text{I}_3=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_1}{\eta\text{k}\text{T}}\right)-1\right) \end{cases}\tag5$$

Substitute $$\(5)\$$ into $$\(4)\$$, in order to get:

$$\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}=\frac{\text{V}_1}{\text{R}_2}+\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_1}{\eta\text{k}\text{T}}\right)-1\right)\tag6$$

For the LED, let's use parameters taken from a Luminus PT-121-B LED: $$\\eta=8.37\$$, and $$\\text{I}_\text{S}=435.2\:\text{nA}\$$. (Assume $$\\text{V}_\text{T}:=\frac{\text{kT}}{\text{q}}=\frac{8094745087}{320435326800}\approx0.0252617\:\text{V}\$$, of course.)

Using the known values, we find:

$$\text{V}_1\approx2.27078\space\text{V}\tag7$$

So, for the power we get:

$$\text{P}_\text{D}\approx0.045602\space\text{W}\tag8$$

I solved for all the knowns using Mathematica. The code is given below.

In[1]:=Clear["Global*"];
q = ((1602176634/(10^9)))*10^(-19);
k = ((1380649/(10^6)))*10^(-23);
T = 20 + ((5463)/20);
Is = (4352/10)*10^(-9);
R1 = 320;
R2 = 220;
Vi = 12;
\[Eta] = 837/100;
FullSimplify[
Solve[{I1 == I2 + I3, I1 == (Vi - V1)/R1, I2 == V2/R2,
I3 == (Is*(Exp[(q*V1)/(\[Eta]*k*T)] - 1)),
I1 > 0 && I2 > 0 && I3 > 0 && V1 > 0}, {I1, I2, I3, V1}]]

Out[1]={{I1 -> 23437313/1054687500 + (
752811293091 ProductLog[(
1549482824704 E^(133516268982824704/5774404804959375))/
5774404804959375])/1139325606400000,
I2 -> 5859443/263671875 - (
68437390281 ProductLog[(
1549482824704 E^(133516268982824704/5774404804959375))/
5774404804959375])/71207850400000,
I3 -> -(17/39062500) + (
1847809537587 ProductLog[(
1549482824704 E^(133516268982824704/5774404804959375))/
5774404804959375])/1139325606400000,
V1 -> 257815492/52734375 - (
752811293091 ProductLog[(
1549482824704 E^(133516268982824704/5774404804959375))/
5774404804959375])/3560392520000}}

In[2]:=N[%1]

Out[2]={{I1 -> 0.0304038, I2 -> 0.0103217, I3 -> 0.0200821, V1 -> 2.27078}}

In[3]:=2.270777378007583*(Is*(Exp[(q*2.270777378007583)/(\[Eta]*k*T)] - 1))

Out[3]=0.045602
`
• Fantastic answer. Jul 23 at 17:36
• @NickBolton thank you very much.
– Jan
Jul 23 at 17:38

For the “purposes of computing power dissipation”

$$\P_D= V_{FO}I_F + r_FI_F^2\$$ using the parameters in the datasheet at the max specified temperature Tj=150’C where the Vf vs If curves converge with temperature.

Diodes have inverse exponential resistance with rising current until it reaches about twice their “threshold current”.

Bulk resistance dominates power dissipation more linear with current and temperature effects reduce greatly but now is dissipating near maximum rating. This effect is the balance if the Vf dropping with rising temp (NTC effects) with resistance rising giving a uniform effective resistance.

Compare your Pd {Vf, If, Tj} using Rjc, Tc with the above formula to estimate junction temp and plot your results to determine the accuracy with a spreadsheet. PWM effects will alter the results.

PN junctions operate faster and more efficiently at max. junction temperatures but at the tradeoff of reliability.

In a switching diodes resistance is inversely proportional to conduction so you're probably dissipating an amounts of reverse current as well, as the motor emits back emf. the equation would probably look like the sum of both forward and reverse resistances.

• This is a Schottky diode; it doesn't have any reverse recovery time. Leakage is likely too small to be a significant amount of the loss, even for a Schottky. Jul 24 at 0:55
• Good to know. Incidentally, as well as what Hearth said, the motor is behind a buck module, so I'm pretty sure that catches any reverse current/voltage from getting to my voltage sensor. Jul 24 at 10:43