Where does the 0.7 V diode cut-in voltage come from?
What the 'turn-on voltage' really is, from simulated and measured diode data.
Silicon diodes are said to "turn on at 0.7 V". Plot the current against voltage yourself, though, and there is no sudden corner at 0.7 V. Fitting simulated data from a PSpice diode model to an exponential shows why.
In this guide 6
1. Data and an exponential model
| Diode voltage | Current | Model |
|---|---|---|
| 300 mV | 4.91 µA | 4.92 µA |
| 400 mV | 34 µA | 34.4 µA |
| 450 mV | 91 µA | 91 µA |
| 520 mV | 357 µA | 355 µA |
| 580 mV | 1.14 mA | 1.14 mA |
| 620 mV | 2.52 mA | 2.49 mA |
| 650 mV | 4.48 mA | 4.46 mA |
| 680 mV | 8.05 mA | 7.99 mA |
| 700 mV | 11.8 mA | 11.8 mA |
| 720 mV | 17.4 mA | 17.4 mA |
| 740 mV | 25.5 mA | 25.7 mA |
| 760 mV | 37.3 mA | 37.9 mA |
All twelve points fit this single equation to within 1.6%.
A log-linear fit gives nVT = 51.4 mV (ideality factor n ≈ 1.99) and IS = 14.4 nA.
2. Why there is no corner
Because it is exponential, every 118 mV multiplies the current by 10, at 300 mV just as at 700 mV. On a linear plot the curve seems to bend near 0.6–0.7 V only because that is where the current reaches milliamps and becomes visible. The same data drawn on both kinds of axis makes it plain (dots: PSpice and bench, line: exponential model).
3. The cut-in voltage depends on the current
| Current you call 'on' | Voltage at that current |
|---|---|
| 10 µA | 336 mV |
| 100 µA | 455 mV |
| 1 mA | 573 mV |
| 10 mA | 692 mV |
| 100 mA | 810 mV |
Call 1 mA "on" and the diode turns on at about 573 mV; call it 10 mA and it is about 692 mV. The textbook 0.6–0.7 V corresponds to the few to tens of milliamps typical circuits draw. Higher current or a colder diode pushes it a little higher.
Try it: one curve, two axes
The exponential model above with the PSpice and bench points, on a linear or a log axis. Pick the current you call 'on' to read its voltage.
- Voltage at that current (model)
- 573 mV
- Voltage at that current (bench)
- 600 mV
- Voltage per tenfold current
- 118 mV
- Where the linear plot seems to bend
- 609 mV
Things to try: on the linear axis, change the full scale and the 'corner' moves with it. Switch to log and the corner disappears into a straight line. Step the 'on' current by ten and the voltage rises by the same amount each time.
4. Connecting it to the half-wave rectifier
In the same lab's half-wave rectifier simulation, a 5.00 V input peak gave a 4.35 V output peak, so 0.65 V was lost in the diode. With 4.35 mA peak through a 1 kΩ load, the model gives a diode voltage of 649 mV. That is why 0.65 V is a sensible diode drop for rectifier calculations. Adding a smoothing capacitor raises the charging current and the drop a little. Carry on in the rectifier ripple calculator.
5. Measuring a real diode
The same experiment was repeated on the bench, reading the diode voltage, the voltage across the 100 Ω resistor R1 and an ammeter.
| Target | Actual VD | Ammeter | VR1 ÷ 100 Ω | PSpice (at target) |
|---|---|---|---|---|
| Target300 mV | Actual VD300.0 mV | Ammeter720 nA | VR1 ÷ 100 Ω730 nA | PSpice (at target)4.91 µA |
| Target400 mV | Actual VD408.9 mV | Ammeter11.8 µA | VR1 ÷ 100 Ω11.9 µA | PSpice (at target)34 µA |
| Target450 mV | Actual VD450.6 mV | Ammeter34.8 µA | VR1 ÷ 100 Ω35.6 µA | PSpice (at target)91 µA |
| Target520 mV | Actual VD524.8 mV | Ammeter203 µA | VR1 ÷ 100 Ω197 µA | PSpice (at target)357 µA |
| Target580 mV | Actual VD581.9 mV | Ammeter681 µA | VR1 ÷ 100 Ω659 µA | PSpice (at target)1.14 mA |
| Target620 mV | Actual VD625.8 mV | Ammeter1.7 mA | VR1 ÷ 100 Ω1.64 mA | PSpice (at target)2.52 mA |
| Target650 mV | Actual VD647.3 mV | Ammeter2.68 mA | VR1 ÷ 100 Ω2.6 mA | PSpice (at target)4.48 mA |
| Target680 mV | Actual VD681.1 mV | Ammeter5.54 mA | VR1 ÷ 100 Ω5.41 mA | PSpice (at target)8.05 mA |
| Target700 mV | Actual VD700.9 mV | Ammeter8.52 mA | VR1 ÷ 100 Ω8.34 mA | PSpice (at target)11.8 mA |
| Target720 mV | Actual VD720.9 mV | Ammeter13.3 mA | VR1 ÷ 100 Ω13 mA | PSpice (at target)17.4 mA |
| Target740 mV | Actual VD739.7 mV | Ammeter20.3 mA | VR1 ÷ 100 Ω19.9 mA | PSpice (at target)25.5 mA |
| Target760 mV | Actual VD759.6 mV | Ammeter32.5 mA | VR1 ÷ 100 Ω31.7 mA | PSpice (at target)37.3 mA |
- The slope is nearly the same: above 0.2 mA the measured curve rises 98 to 110 mV per tenfold current, close to the model's 118 mV. The exponential law holds for the real part.
- It takes a little more voltage: 600 mV at 1 mA (model 573 mV) and 708 mV at 10 mA (model 692 mV). A different part and temperature easily account for a few tens of millivolts.
- R1 was not quite 100 Ω: in the milliamp range the ammeter read 2 to 4% above VR1 ÷ 100 Ω. VR1 ÷ ammeter gives about 97.6 Ω, inside the ±5% tolerance. The ammeter guide shows how to split up differences like this.
The half-wave rectifier was measured too. A 5.065 V input peak gave a 4.24 V output peak, losing 0.83 V, more than the simulation's 0.65 V. The part difference above (tens of millivolts) does not explain all of that, so the measurement setup is worth suspecting as well: where the input was probed, or the signal generator's output resistance (usually 50 Ω).
The more interesting result came with the input turned down to 488 mV: the output was only 40 mV. The diode kept 448 mV, and the table shows only 34.8 µA flowing at 450 mV. A single diode barely rectifies signals of half a volt or so, which is why small signals are rectified with an op-amp precision rectifier instead.
6. In your lab report
- Quote a cut-in voltage together with its current, for example "0.57 V at I = 1 mA".
- Plot current on a log axis and the curve becomes a straight line whose slope gives the ideality factor n, which makes for a much more concrete discussion.
LEDs work the same way, which is why each color has a different forward voltage. Size an LED resistor with the LED resistor calculator, and see the zener diode calculator for diodes used in reverse breakdown.