The mains voltage in many Indian homes wanders through the day. It sags in the evening, when every house lights its lamps and runs its fans, and rises late at night. A phone charger plugged into that socket still gives a steady \(5\,\text{V}\) to the phone. Something inside the charger holds the output fixed while the input moves.
The circuit
A voltage regulator takes an input that changes and gives an output that does not. The simplest one needs only two parts. A resistor \(R_S\), the series resistance, carries the current from the input. A Zener diode, connected in reverse bias, sits across the output. The load, the device being supplied, is joined in parallel with the Zener, so both always have the same voltage across them.
Two rules of circuits are enough to find every current. First, Ohm's law: the voltage across a resistor is its resistance times the current, \(V = IR\). Second, the current law at a junction: the current flowing into a junction equals the current flowing out, since charge does not collect there. Both are developed in the unit on electricity.
How it holds the voltage
While the Zener is in breakdown, the voltage across it is \(V_Z\), and so the output is \(V_Z\). The series resistor has the rest of the input across it, \(V_{in} - V_Z\). That fixes the series current. At the junction above the Zener, this current splits: the load takes what it needs, and the Zener takes the remainder. When the input rises, only the series current and the Zener current grow; the output does not move.
While the Zener diode stays in breakdown, the output voltage equals \(V_Z\). Changes in the input voltage change only the series current and the Zener current. Changes in the load change only how the series current is shared between the load and the Zener.
Try \(10\), \(12\) and \(14\,\text{V}\) in the sim. The output stays at \(5.00\,\text{V}\) and the load current stays at \(20\,\text{mA}\). The series current is \(50\), \(70\) and \(90\,\text{mA}\), and the Zener current \(30\), \(50\) and \(70\,\text{mA}\). Every extra milliampere drawn from the input goes into the Zener.
Now go below \(7\,\text{V}\). At \(7\,\text{V}\) the series current is only \(20\,\text{mA}\), all of which the load needs. Below that the Zener has nothing left, so it leaves breakdown and stops conducting. The two resistors then share the input in proportion to their resistances, like any two resistors in series, and \(V_{out} = V_{in} \times 250/350\). At \(6\,\text{V}\) in, the output is only \(4.29\,\text{V}\).
A regulator has an input of \(12\,\text{V}\), a series resistance of \(100\,\Omega\), a \(5\,\text{V}\) Zener diode and a load of \(250\,\Omega\). Find the series current, the load current and the Zener current, and the power turned into heat in the Zener.
Assume the Zener is in breakdown, so the output is \(5\,\text{V}\). Check at the end that \(I_Z\) is positive.
\(I_S = (V_{in} - V_Z)/R_S\), \(I_L = V_Z/R_L\), \(I_Z = I_S - I_L\), and \(P_Z = V_Z I_Z\).
\(I_S = 7/100 = 70\,\text{mA}\). \(I_L = 5/250 = 20\,\text{mA}\). \(I_Z = 70 - 20 = 50\,\text{mA}\). \(P_Z = 5 \times 0.050 = 0.25\,\text{W}\).
The load gets \(20\,\text{mA}\) at a steady \(5\,\text{V}\); the Zener carries \(50\,\text{mA}\) and must be rated above \(0.25\,\text{W}\). Values used: the regulator of this lesson, \(V_{in} = 12\,\text{V}\), \(V_Z = 5\,\text{V}\), \(R_S = 100\,\Omega\), \(R_L = 250\,\Omega\).
A regulator has \(V_{in} = 9\,\text{V}\), \(R_S = 150\,\Omega\), a \(6\,\text{V}\) Zener and a load of \(1\,\text{k}\Omega\). What is the Zener current, in mA?
When the load changes
Keep the input at \(12\,\text{V}\) and change the load instead. The series current stays at \(70\,\text{mA}\), because the voltage across \(R_S\) is still \(7\,\text{V}\). What changes is how that current is shared. A smaller load resistance draws more current, and the Zener gets less.
The last chip of Fig 2.1 shows the limit. A \(50\,\Omega\) load at \(5\,\text{V}\) would need \(100\,\text{mA}\), but only \(70\,\text{mA}\) can arrive. The formula would give \(I_Z = -30\,\text{mA}\), which a reverse-biased Zener cannot carry. So the Zener stops conducting, and the two resistors share the input: \(V_{out} = 12 \times 50/150 = 4\,\text{V}\). The regulator works only while \(I_L\) stays below \(I_S\).
A regulator works correctly with a \(250\,\Omega\) load. The load is changed to \(125\,\Omega\), and the Zener stays in breakdown. What happens to the series current and the Zener current?
Take the regulator of the example: \(12\,\text{V}\) in, \(100\,\Omega\) and a \(5\,\text{V}\) Zener. What is the smallest load resistance, in ohms, that still leaves the output at \(5\,\text{V}\)?
Two limits on the series resistance
A real Zener diode needs a small least current \(I_{Z,min}\) to stay firmly in breakdown, often a few milliamperes. It also has a rated power \(P_{max}\), which fixes its largest current, \(I_{Z,max} = P_{max}/V_Z\). The series resistance must respect both limits, at the worst moments.
The Zener current is smallest when the input is lowest and the load takes the most. So \(R_S\) must be small enough to pass \(I_{Z,min} + I_L\) at the lowest input. The Zener current is largest when the input is highest and the load takes the least, perhaps nothing when the device is unplugged. So \(R_S\) must be large enough to keep \(I_Z\) below \(I_{Z,max}\) then.
\[ I_S = \dfrac{V_{in} - V_Z}{R_S} = I_Z + I_L \]
Why. the series current, from Ohm's law, splits at the junction into the Zener and the load currents.
\[ \dfrac{V_{min} - V_Z}{R_S} \ge I_{Z,min} + I_{L,max} \]
Why. at the lowest input and the largest load current the Zener must still carry at least \(I_{Z,min}\).
\[ \dfrac{V_{max} - V_Z}{R_S} \le I_{Z,max} + I_{L,min} \]
Why. at the highest input and the smallest load current the Zener must carry at most \(I_{Z,max} = P_{max}/V_Z\).
\[ \dfrac{V_{max} - V_Z}{I_{Z,max} + I_{L,min}} \le R_S \le \dfrac{V_{min} - V_Z}{I_{Z,min} + I_{L,max}} \]
Read it. turn each inequality round to give a bound on \(R_S\).
The input of a regulator varies from \(10\,\text{V}\) to \(14\,\text{V}\). A \(5\,\text{V}\) Zener rated \(0.5\,\text{W}\) needs at least \(5\,\text{mA}\). The load takes \(20\,\text{mA}\), but it may be unplugged. Find the range of \(R_S\), and check the choice \(R_S = 100\,\Omega\).
Largest Zener current: \(I_{Z,max} = 0.5/5 = 100\,\text{mA}\).
Upper limit, at \(10\,\text{V}\) with the load on: \(R_S \le (10 - 5)/(0.005 + 0.020) = 200\,\Omega\).
Lower limit, at \(14\,\text{V}\) with the load unplugged: \(R_S \ge (14 - 5)/0.100 = 90\,\Omega\).
So \(90\,\Omega \le R_S \le 200\,\Omega\). With \(100\,\Omega\): at \(10\,\text{V}\), \(I_Z = 50 - 20 = 30\,\text{mA}\), well above \(5\,\text{mA}\). At \(14\,\text{V}\) with no load, \(I_Z = 90\,\text{mA}\) and \(P_Z = 0.45\,\text{W}\), just inside the rating.
The resistor itself then turns \(9 \times 0.090 = 0.81\,\text{W}\) into heat, so it must be a \(1\,\text{W}\) resistor.
A \(5\,\text{V}\) Zener regulator is fed from a supply that never falls below \(12\,\text{V}\). The load draws \(25\,\text{mA}\), and the Zener needs at least \(5\,\text{mA}\). What is the largest series resistance, in ohms, that keeps the output at \(5\,\text{V}\)?
Put the steps of designing a Zener regulator in order.
The Zener must be connected in reverse bias, with its cathode (the bar) towards the positive side. Turned the other way, it conducts like an ordinary diode at about \(0.7\,\text{V}\), and the output is \(0.7\,\text{V}\) instead of \(V_Z\).
The input of a Zener regulator rises by \(2\,\text{V}\). Explain why the output voltage does not rise with it.
While the Zener is in breakdown its voltage stays at \(V_Z\), and the output is taken across it. The extra \(2\,\text{V}\) appears across the series resistor instead, which then carries more current. The load current is still \(V_Z/R_L\), so all the extra current flows through the Zener.
A modern phone charger does not waste power in a big series resistor. It switches the mains current thousands of times a second. A control circuit adjusts the switching to hold the output at \(5\,\text{V}\). To know what \(5\,\text{V}\) is, the control circuit compares the output with a fixed reference voltage. That reference is a small breakdown diode or a precise relative of it. The idea of this lesson, a voltage held fixed by breakdown, sits at the heart of the charger.
Return to the charger at the start of the lesson. Its input swings with the mains, yet its output stays at \(5\,\text{V}\). In the simplest regulator, a Zener diode holds the output at \(V_Z\), and the series resistor and the Zener current absorb every swing. A charger does the same job more efficiently, but it still measures its output against a breakdown voltage.
- Using the load resistance to find the series current.\(I_S = (V_{in} - V_Z)/R_S\). The load only decides how \(I_S\) is shared.
- Keeping the output at \(V_Z\) when the formula gives a negative Zener current.A negative \(I_Z\) means the Zener is off. Then the resistors share the input, and \(V_{out} < V_Z\).
- Checking the Zener's power at the lowest input.The Zener is hottest at the highest input with the smallest load, or no load at all.