Devices and Applications
Ctec 201.
Zener Diodes
Supplement
Prepared by Mike Crompton. (Rev. 9 March 2005)
Zener Diodes
When a normal silicon diode is in the reverse bias condition, so little current flows that it
is usually ignored and it is said that no current flows. However, if the reverse bias voltage
exceeds the Peak Inverse Voltage (PIV) rating of the diode a large amount of current will
flow (Avalanche Current). The heat generated by this Avalanche Current will very
quickly destroy the diode.
By increasing the amount and distribution of the P and N doping in the diode, the point at
which Avalanche Current starts to flow can be controlled and varied significantly. The
increased doping also allows relatively large amounts of current to flow without
generating excessive heat, and maintains a fairly constant reverse bias voltage drop
across the diode over a wide range of current values.
Such a device is known as a Zener
Diode. Those unique characteristics
make it particularly suitable as a
voltage regulator or a ‘clamping’
device. (Clamping a point in a
circuit to a particular voltage level).
I Forward
PIV Level.
Rev Bias
Volts
Fwd. Bias Volts
Avalanche
Current
Zener Diode Symbol.
I Reverse
The point at which avalanche current flows in the ‘reverse direction’ is known as the
‘Zener’ or ‘Breakdown’ voltage. Zener diodes are available with breakdown voltages
from 1.8 volts to several hundred volts. The current carrying capabilities can also vary
widely. The Zener performs in a similar manner to a normal diode if connected in
forward bias configuration, however this would never be done, as it does not take
advantage of the Zener’s characteristics and a normal silicon diode would suffice.
A typical application for a Zener diode would be to maintain a constant voltage across a
load of varying resistance when the power source is unregulated, or the voltage level of
the source is varying. This is accomplished by the use of a series resistor that will always
drop sufficient voltage to maintain the breakdown voltage across the Zener. The Zener in
fact, acts like a variable resistor that adjusts the total current flow to give exactly the
correct voltage drop across the series resistor. The load would normally be connected
across, or in parallel with the Zener. A circuit to demonstrate this is shown over the page.
2
Vsupply +12V
RS
100
ZD1
Vz
RL
1k
Equivalent Circuit
The 1N-4733 is a 5.1V Zener, that is, as long as the diode is working within it’s limits the
voltage across it will remain relatively constant at 5.1V. Note that it is in the reverse bias
configuration. Note also that it is a simple series/parallel circuit. (See equivalent circuit).
If the voltage across the Zener (VZ) and the 1k load (VRL) is 5.1V, then the voltage
across the 100 series resistor (VRS) is:
VRS = VSUPPLY – VZ (or VRL):
VRS = 12V – 5.1V = 6.9V.
Total series current (ITOTAL) is VRS / RS:
ITOTAL = 6.9V/100 = 0.069A or 69mA.
Some of this is flowing through the 1 k load and the remainder through the Zener.
Current through the 1000 load (IL ) is VL/RL:
IL =5.1V/1000 = 0.0051A or 5.1mA
Therefore the current through the Zener (IZ) is ITOTAL - IL:
IZ = 0.069A – 0.0051A = 0.0639A or 63.9mA.
The resistance of the Zener (RZ) will therefore be VZ / IZ:
RZ = 5.1V/0.0639A = 79.8
If the 1k load resistor is changed to
250, the voltage across the Zener (and the
load) remains at 5.1V.
The Voltage drop across the 100 series
resistor remains the same at 6.9V, therefore
the total series current (ITOTAL) will remain
the same at 0.069A. What will have
changed, is the ratio of current flowing
through the load resistor and through the
Zener. i.e. The Zener adjusted it’s
resistance when the load resistor changed.
3
To calculate this change, first calculate the new current through the load (IL) as VL/RL:
IL = 5.1V/250 = 0.0204A or 20.4mA
The new current through the Zener (IZ) is ITOTAL - IL:
IZ = 0.069-0.0204 = 0.0486A or 48.6mA
The resistance of the Zener (RZ) will therefore be VZ / IZ:
RZ = 5.1V/0.0486A = 104.9
Similarly, if VSUPPLY dropped to 9V another change of Zener resistance (R Z) would occur
to maintain the Zener voltage at 5.1V.
First we calculate the new voltage drop across RS as
VSUPPLY – VZ (or VL): VRS = 9V – 5.1V = 3.9V
The new series current (ITOTAL) is VRS / RS:
ITOTAL = 3.9V/100 = 0.039A or 39mA.
The current through the 250 load RL will remain the
same at 0.0204A or 20.4mA and the new Zener current IZ will be ITOTAL - IL :
IZ = 0.039A - 0.0204A = 0.0186A or 18.6mA.
The new resistance of the Zener (RZ) will therefore be VZ / IZ:
RZ = 5.1V/0.0186A = 274.19
The above calculations demonstrate that the Zener diode acts in a similar fashion to a
variable resistor. The variations in resistance created within the diode are accomplished
by the depletion zone simply adjusting it’s size, as occurs in a normal silicon diode. It
should be noted that all the calculations relied on nothing more than Ohm’s Law coupled
with the laws of series/parallel circuits.
To determine the circuit values that will allow the Zener to operate at optimum efficiency
and within it’s normal parameters is also an exercise in Ohm’s Law. For example, what is
the minimum value of load resistor that could be used in the sample circuit? Obviously
the lower the load resistor the more the current will increase and the greater the voltage
drop across the 100 resistor. At some point that voltage drop will increase to the point
where there will no longer be sufficient voltage across the Zener to cause breakdown. In
the case of the 1N-4733, a minimum of 5.1V. Looking at our last circuit where VSUPPLY is
9V, it is obvious that a voltage drop of 3.9V (9V – 5.1V) is the maximum allowable drop
across RS that will still produce the required 5.1V across the Zener // R L parallel circuit.
From the last circuit we determined the ITOTAL that will produce this 3.9V drop:
ITOTAL = 3.9V/100 = 0.039A or 39mA.
Therefore the value of RL with 5.1V across it and with 0.039A flowing is VRL / IRL:
RL = 5.1V/0.039A = 130.77
4
Any value of RL below 130.77, and the Zener will not breakdown resulting in loss of
regulation.
Another parameter of interest is the maximum amount of current that a Zener can
tolerate. In the specification sheet for many Zeners this information is given as a
maximum power figure rather than current. For the 1N-4733 this is listed as 1 Watt, and
since the Zener voltage is 5.1V the maximum current would be 1Watt/5.1V = 0.196A.
Knowing this it can be determined what the maximum VSUPPLY would be for a given
series resistor (RS)/load resistor (RL) combination, or what size RS would be required for
a given VSUPPLY.
In the last example, RL was 250 and RS was 100. What would be the maximum level
of VSUPPLY that would not exceed a Zener current (IZ) of 0.196A?
The current through RL will remain at 5.1V/250 = 0.0204A or 20.4 mA. The maximum
Zener current is 0.196A giving a total current (ITOTAL) of IRL + IZ:
ITOTAL = 0.0204A + 0.196A = 0.2164A or 216.4mA.
If we now calculate the voltage drop across the 100 RS, and simply add it to the 5.1V
that is across the Zener/ RL combination, we will have the maximum value of VSUPPLY.
VRS = 0.2164A x 100 = 21.64V
Max. VSUPPLY = 21.64V + 5.1V = 26.74V.
This last calculation does not take into account the fact that the load resistor may become
open circuit leaving the Zener in series with RS. If this were the case the current in the
circuit would exceed the maximum allowable and burn out the Zener. The actual amount
of current would be: Max. I = (Max.VSUPPLY - VZ)/ RS:
Max. I = (26.74V – 5.1V)/100 = 0.2164A or 216.4mA.
Obviously this exceeds the Zener’s maximum current specification of 0.196A. The
solution is to restrict Max. VSUPPLY by calculating using Max. IZ in place of ITOTAL:
VRS = 0.196A x 100 = 19.6V
Max. VSUPPLY = 19.6V + 5.1V = 24.7V.
Choosing the minimum value of RS that will protect the Zener in the case of an open
circuit load and VSUPPLY being at the maximum, is simply a matter of subtracting VZ from
Max. VSUPPLY and dividing the result by Max. IZ.
Min. RS = (Max. VSUPPLY - VZ)/ Max. IZ.
Using the last values to prove this:
Min. RS = (24.7V – 5.1V)/0.196A = 100.
Had the first value of Max. VSUPPLY (26.74V) been used, the value of Min. RS would have
risen to:
Min. RS = (26.74V – 5.1V)/0.196A = 110.4.
5