difference between reverse saturation and leakage current

Question

I searched a lot but I didn't find any answer

For a PN junction diode is the leakage current when the diode is reverse biased

Equal to the reverse saturation current??

If they are not equal so where does the excess current come from?

Answer 1

In an ideal diode, they are the same. The diode equation is

$$I = I_s \left[\exp\left( \frac{qV}{n k_B T}\right) - 1\right]$$

so if you apply a strong reverse bias (\$v \ll 0\$), then the reverse current will be very close to \$I_s\$, the saturation current.

In a real diode, there may be other leakage paths aside from through the PN junction itself that allow current to pass, so the leakage current may be greater than the saturation current. For a diode mounted on a PCB, there could be additional leakage due to surface contamination.

Answer 2

Real diodes have various deviations from the ideal diode equation. Let me explain the forward imperfections first, then the reverse ones.

At low forward biases (say < 300 mV for Si at room temperature), additional recombination effects (Shockley-Reed-Hall recombination) dominate, and the exponential term is more like exp(q.V/(2.k.T)).

At very high forward bias, charge neutrality (basically that the carriers injected across the junction don't affect the background (majority) carrier concentration) also means the exponential term has a /2 factor.

Series resistance also adds to the effective forward voltage.

Devices with 'short' (short compared to diffusion length) P or N regions will have a lower VF than otherwise. In addition, the depletion region width changes with forward bias, thus further changing the effective device 'size'.

In reverse, at moderate bias (say to a few V reverse), the exponential equation is valid (and since the exponential term swamps the '-1' term at more than about -100 mV, it can't be noticed.

Any light shining on the junction (light can sometimes get through the plastic moulding in a diode package) will generate carriers near the junction and significantly increase the current.

As the reverse V increases, two different effects can dominate -- Zener breakdown, or avalanche breakdown.

Zener breakdown occurs when the E field in the depletion region is so high that carriers can directly tunnel from the N to the P (or P to N for holes) region. Zener breakdown typical occurs for very highly doped diodes with a BV less than about 5V; some diodes have 'zener' voltages of only 3 V.

Avalanche occurs in lighter doped devices where the high E field in the depletion region directly causes electrons to be stripped from their lattice Si atom and travel across the junction. Each (energetic) electron has a probability of hitting another atom and causing it to free an electron, thus leading to a very sharp increase in total current as reverse voltage increases. Because the avalanche 'multiplication' depends on doping (and depletion region width), and on the actual field-vs-position, a simple equation is not possible, but generally some type of exponential-type behavior is observed (until the device is destroyed by the high power dissipation).

All the above assume that the device is at a constant temperature; however power dissipation in the diode will modify the temperature and thus the 'apparent' equations.

Imperfections in diodes, particularly surface states can cause leakages that are more like a large resistance in parallel with the junction.

Some diodes can be doped so heavily that they are already in breakdown with no applied voltage -- these can conduct better in reverse than in the forward direction, at least until a few 100 mV. These are called backwards diodes, or tunnel diodes.

So, the classical exponential equation is convenient, but (as usual) there can be many caveats and it really only applies well in a small region. in a modern integrated circuit, the (forward-bias) equation may actually apply very well over 5-6 decades of current, but that is only ~360 mV -- so perhaps from 200 mV to 560 mV forward across the junction (any therefore typically only for currents from <<< nA to uA levels.

Answer 3

You asked about leakage current and saturation current. If the voltage over the diode is negative, "reverse biased", the current is not exactly zero, but a small current is flowing, called the leakage current.

The leakage current depends on the value of the negative voltage in a somewhat proportional way, the larger the voltage, the larger the current. But the leakage current goes asymptotically to a maximum value, this is called the saturation current. But as the I-V curve for negative voltages greater than a few tenths of volt is pretty much flat, you don't really have to consider infinite, or large, negative voltages for getting the saturation current.

As the saturation current is a particular kind of leakage current, there is no such concept as "excess current".

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(this answer is edited to more closely correspond to the actual question)

Answer 4

A quick answer to your question is no they are not the same. On a IV Characteristic curve of a PN junction diode, you will generally have a very low reverse bias current (Leakage current). This current is said to be 1uA in the most extreme conditions for a silicon small signal diodes.

The 'Reverse Saturation Current' is also called the 'Zener Breakdown' or the 'Avalanche Region'. This current will drastically increase as the Reverse Breakdown Voltage is achieved and most likely destroyed your diode unless you have a high resistance in series to limit the current flowing through the diode in breakdown conditions.

If you are looking at a specifications for a PN junction diode, I would assume the reverse saturation current is specified as a maximum. This will give you guidance on what resistor you should put in series with the diode to protect your circuit.

Answer 5

On a standard diode. When its reverse bias the depletion region expands. Tis effect causes the diode to become a capacitor (there are special diodes that are enhanced to work like this and used as a variable capacitor). The reverse bias leakage current is the insulation breakdown of this diode in its capacitance state. The depletion breakdown voltage is the voltage potential at which the diode will conduct in reverse bias. Since the standard pn junction diode is not constructed like a zener diode (zener diodes actually have pn-pn arrangements on its die internally), it will short out in its typical circuit design due to the lack of current limiting. Zener diodes have to be current limited (usually by a resistor in series) so it can maintain this reverse bias voltage potential without the current running away causing the diode to short.