The Diode and the 0.7 V Rule

Set a supply, a resistor and a temperature. The page finds where the diode actually sits, walks Newton's method there from the usual 0.7 V guess, and holds the answer against the rule of thumb.

Worked example — The Diode and the 0.7 V Rule

Thermal voltage	25.693 mV
Volts for each tenfold rise in current	59.16 mV
Voltage across the diode	0.6883 V
Current round the loop	4.3117 mA
Current if the diode took exactly 0.7 V	4.3000 mA
How far out the 0.7 V answer is	-0.27%
Newton steps from 0.7 V to within 1 µV	4

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Diode and the 0.7 V Rule

A diode is a junction that conducts easily one way and hardly at all the other. Over its working range the current is an exponential of the voltage, the Shockley equation: I = I_S(e^V/(nV_T) − 1). Here V_T = kT/q is the thermal voltage, 25.69 mV at 25 °C. In a circuit the diode's voltage is not given in advance: it is wherever the diode's curve crosses the line of everything the rest of the circuit allows.

What each symbol means

V_T: the thermal voltage kT/q, in volts. 25.69 mV at 25 °C, and proportional to absolute temperature
I_S: the saturation current, in amperes. Tiny, 10⁻¹⁴ A by default here, and the quantity that decides where along the voltage axis the current becomes large. It grows steeply with temperature
n: the ideality factor, between 1 and 2. It multiplies every voltage step: at n = 2 a tenfold rise in current needs twice the millivolts
E_g: the band gap of silicon, 1.11 eV in this model. It sets how fast I_S grows with temperature, and it is why the drift depends on how far V sits below it

Assumes: The ideal Shockley equation and nothing else: no resistance inside the diode, which in a real part bends the curve over at large currents; no reverse breakdown; steady DC. The saturation current and n are model choices, not the data of a named part. The saturation current's growth with temperature follows the SPICE diode model, with a band gap of 1.11 eV and an exponent of 3, so the drift figures on this page are that model's.

There is no voltage at which a diode switches on 🖖

Press Log above the first panel. The curve becomes a straight line from a picoamp to an amp, and the current is ten times larger every 59.16 mV. Nothing happens at 0.7 V. This diode carries 58 nA at 0.4 V, 2.83 µA at 0.5 V and 0.139 mA at 0.6 V. Now press 100 µA, 1 mA, 10 mA and 100 mA in turn and follow the diamond, which marks where the curve has climbed a tenth of the way up. It lands at 0.532, 0.592, 0.651 and 0.710 V: same diode, same equation. Pick a different fraction and the four numbers change, but the gap between them stays 59 mV for each tenfold change in the axis. On this diode, "0.7 V" is the knee of a graph drawn to 68 mA.

Minus 2 mV per kelvin belongs to one current 🖖

Hold the current fixed and warm the diode, and its voltage falls. While the current is far above I_S, differentiating the model gives the rate as (V − E_g − 3V_T)/T, which depends on where the diode is sitting rather than on which diode it is. The last table works it out row by row at 25 °C: −2.39 mV/K at 1 µA, −2.00 at 100 µA, −1.60 at 10 mA, −1.40 at 100 mA. Each tenfold rise in current takes 0.198 mV/K off the drift. The rule of thumb is right for a diode sitting at 0.591 V, which on this page means about a tenth of a milliamp. Press 5 V at 75 °C: the diode in the 5 V circuit falls 84 mV over 50 kelvin, 1.68 mV/K.

When 0.7 V is close enough 🖖

In series with a resistor, assuming the diode takes 0.7 V gets the current wrong by exactly (V − 0.7)/(V_S − V). The top of that fraction is how far the diode really sits from 0.7 V. The bottom is the voltage across the resistor, and that is the part you control. With 1 kΩ the rule is never more than 3.11% out for any supply from 2 V to 30 V, and 1.18% at 3 V. At 1 V it is 19.89% low, and at 0.7 V it predicts nothing while 0.1067 mA flows. The resistor moves the top: 0.7 V is exactly right only where this diode carries 6.80 mA. Through 100 kΩ the diode sits near 0.54 V and the error at 2 V is 10.83%; through 100 Ω at 5 V it sits at 0.747 V and the rule overestimates by 1.11%.

Problem solved in full

Reading a temperature off two currents 6 steps
A diode is run at 10 µA, then at 100 µA. A meter reads 0.477278 V and then 0.541398 V. Nothing is known about the diode except that n = 1. How warm is it?

Open this problem in the tool
1. Either voltage on its own is no use. It depends on the saturation current, which you do not know, and which in this model is 36 times larger at 50 °C than at 25 °C.
  
  $\frac{I _{S} ( 323 . 15 K )}{I _{S} ( 298 . 15 K )} = (\frac{323 . 15}{298 . 15})^{3} e^{\frac{E _{g}}{k} (\frac{1}{298 . 15} - \frac{1}{323 . 15})} = 36$
2. Subtract the two readings instead. Each is nV_T ln(I/I_S), so the difference is nV_T ln(I₂/I₁) and the saturation current cancels, as long as both currents are far above it.
  
  $V = n V_{T} ln \frac{I}{I _{S}} \Rightarrow V_{2} - V_{1} = n V_{T} ln \frac{I _{2}}{I _{1}}$
3. The difference is 64.120 mV and the current ratio is 10, so V_T = 64.120 mV ÷ ln 10 = 27.847 mV.
  
  $0.541398 - 0.477278 = 0.064120 V, V_{T} = \frac{0 . 064120}{ln 10} = 0.027847 V$
4. Since V_T = kT/q, the temperature is T = qV_T/k = 323.15 K, which is 50.0 °C.
  
  $T = \frac{q V _{T}}{k} = \frac{1 . 602176634 \times 1 0 ^{- 19} \cdot 0 . 027847}{1 . 380649 \times 1 0 ^{- 23}} = 323.15 K$
5. Check it on the page: set the temperature to 50 °C. The row for volts per tenfold rise reads 64.12 mV, and the 10 µA and 100 µA rows of the last table read 0.4773 and 0.5414.
  
  $0.5414 - 0.4773 = 0.0641 V$
6. The difference grows by only 0.198 mV per kelvin, about a tenth of what either voltage moves on its own. A difference good to a tenth of a millivolt pins the temperature to half a kelvin, which is why the readings here are given to the microvolt.
  
  $\frac{d ( V _{2} - V _{1} )}{d T} = \frac{k ln 10}{q} = 0.198 mV/K$
Answer

T = 323.15 K, or 50.0 °C, from two readings and n alone. The difference between two junction voltages is proportional to absolute temperature whatever the saturation current is, provided both currents are far above it, so a silicon chip can measure its own temperature with nothing calibrated on it but the ratio of two currents.

$T = 323.15 K = 50.0^{\circ} C$

Learning path

What is not a resistor

By the 42Math team · Reviewed against published sources. Corrections welcome.