## How does zener diode work in the circuit Moderators: adafruit_support_bill, adafruit

How does zener diode work in the circuit

As we all know,The zener diode breakdown voltage (stable voltage) test needs a small current voltage source which is higher than the breakdown voltage. The tester uses a 555 chip to form an oscillator, and then the output pulse is boosted to more than 100 volts for testing.The circuit is added at the attactachment.
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Using a single 555 timer IC and a small transformer to generate high voltage, the circuit will test the zener diode with a voltage rating up to 50VDC. The 555 timer is used in the unsteady mode, and the output of pin 3 drives a small audio transformer, such as LT700. This has the primary impedance of 1K and the secondary impedance of 8 ohms. It is used to reverse the AC voltage to about 120 volts. It is rectified by a 1N4004 diode and filtered by a 2.2U capacitor to get about 150 volts DC. Connect the measured zener diode with the DC voltage meter of the multimeter, as shown in the figure. Load current switches enable zener diodes to switch at 1 or 2 Ma DC.
But now I am confusing about the following:
I am researching the zener diode--BZX84C7V5,( BZX84C7V5 PDF),and I would like to know that what they're used to work at constant voltage, but I don't know how they work. Actually, I kind of get that too, I know that voltage is constant until breakdown occurs (avalanche breakdown), but I don't get why a minimum current is needed to maintain a constant voltage, which is what the textbooks say. Why is there a minimum? I thought any current below the breakdown voltage was constant. I don't see why the voltage would vary through the diodes if it's below a set amount of current, yet remain constant between voltages x-y?

Any help will be appreciated!

Amandali

Posts: 2
Joined: Fri Jun 24, 2016 4:59 am

Re: How does zener diode work in the circuit

Devices called 'Zener diodes' can have two different operating mechanisms: avalanche breakdown, and the actual Zener effect.

As a quick recap of diodes for the sake of defining vocabulary, silicon has four free electrons and four empty orbitals in its valence shell. In a silicon crystal, each atom shares an electron with four neighbors, and thinks it has eight electrons and no empty orbitals. The resistance of raw silicon crystal is about 100 ohms per centimeter of thickness, and the crystal is otherwise electrically boring. We change the conductivity by implanting atoms called 'dopants' with five valence electrons or five empty orbitals, forcing the crystal lattice to be unbalanced. Electrons will hop from atom to atom because there's nothing to make them stay with any particular silicon atom. Crystal with extra electrons is called n-type, crystal with extra empty orbitals (called 'holes') is called p-type.

If you put a piece of n-type silicon in contact with a piece of p-type silicon, the extra electrons from the n-layer get pulled into the extra holes in the p-layer. Left to itself, entropy likes to see things more or less equally distributed. Electrically, we call that effect 'diffusion current' because it forces electrons to move in a specific direction.

As diffusion current moves electrons out of the n-layer, it leaves dopant atoms that still want to have five electrons. Those atoms end up with a positive charge. The diffusion current also pushes electrons into orbitals around atoms that want to have five empty orbitals, so those pick up a negative charge. That separation of charge creates what's called the 'built-in electric field' between the n-layer and p-layer.

The built-in field pulls electrons back toward the n-layer by the conventional means called 'drift current'.

The diffusion current gets stronger when most of the electrons are still in the n-layer (a sharp difference in electron/hole concentrations) and gets weaker as more electrons move into the p-layer (a more even mix). The built-in electric field does the opposite: it gets weaker when most electrons are in the n-layer (close to their original positions) and gets stronger as more electrons move into the p-layer (far from their original positions).

The built-in field expands until drift current pulling electrons back to the n-layer is just as strong as the diffusion current pulling electrons toward the p-layer. When both kinds of current are equal, the net current through the built-in field is zero, so we refer to that as a 'depletion region' where there are no free carriers to make current go either way.

If you apply an external electric field that cancels the built-in field (positive voltage on the p-layer, negative voltage on the n-layer) it pushes electrons toward the depletion region in the n-layer and pulls them away from the depletion region in the p-layer. The overall effect is to make the depletion region narrower, which makes the drift current weaker and the diffusion current stronger. The diffusion current pulls more electrons through the depletion region, and the external power source pulls them out of the p-layer while pumping new ones into the n-layer. We call that a 'forward biased diode', and the amount of current that flows through depends on how much the external field changed the size of the depletion region. The voltage-to-current ratio is exponential: increasing the forward bias voltage by about 60mV increases the current by a factor of 10.

If you apply an external electric field that makes the built-in field stronger (positive voltage on the n-layer, negative voltage on the p-layer), the depletion region gets wider. The diffusion current drops to almost nothing, and the drift current can't pull electrons through the depletion region for low external voltages. That's called a 'reverse biased diode'.

If you increase the reverse bias voltage far enough, one of two things will happen:

1) The external voltage pushes an electron into the depletion region so hard that it hits a silicon atom and knocks a second electron free (mechanical ionization). Both of those electrons hit new atoms and knock two more electrons free, etc. That's avalanche breakdown, and the amount of current that gets knocked free depends on the strength of the external electric field. That voltage-to-current ratio is also exponential, but doesn't change as dramatically as the forward-biased current.

2) The energy levels of valence electrons in the p-layer (ones shared between silicon atoms) become higher than the energy level of free electrons in the n-layer. If both layers are heavily doped, the depletion region will be narrow enough that some of the p-layer's valence electrons can hop to the n-layer through a mechanism called 'quantum tunneling'.. ending up on the other side of a barrier while provably never passing through the barrier (quantum mechanics is weird). That's the Zener effect, and its voltage-to-current ratio is also exponential at a rate roughly the same as avalanche breakdown.

Either way, breakdown diodes behave like nonlinear resistors: putting more voltage across them sends more current through them. The ratio isn't as simple as V=IR, but if you know the voltage-to-current function of a breakdown diode you can calculate the current from voltage, or the voltage from current.

We normally use breakdown diodes to create reference voltages, so it makes sense to calculate the voltage in terms of the current.

We know how to build circuits that produce a specific amount of current across a wide range of voltages, so if we know a diode's reverse voltage to current curve is within 1% of 25V when the current flowing through it is 10mA, we can build a circuit to deliver that 10mA. Running the current through the diode gives us a 25V +/-1% voltage reference that's reasonably stable and predictable.

That isn't the only point on the diode's voltage-to-current curve though. At 5mA, the reverse-bias voltage might be 20V. At 15mA, the voltage might be 35V.

We generally prefer to operate avalanche-breakdown and Zener diodes at the point where their voltage-to-current curve has a slope close to 1. That way small errors in the current create small errors in the voltage, but we can also be confident that 10,000 diodes operating at the same current will conduct well enough to put them all near the same voltage. Historically, we've designed diodes so that occurs when the current is near 5mA or 10mA. That way a voltage buffer with 10k to 100k of input impedance won't pull enough current away from the diode to change the reference voltage much.