Lesson 4 - A Simple Voltage Regulator

In this lesson, we are going to design a voltage regulator, the heart of any power supply.

The need for Voltage Regulators

The purpose of Voltage regulators is to provide a steady supply voltage to the circuits you are designing.

They are the most common circuits (every electronic system no matter what it's function, has at least one), and yet they are often neglected because of their utilitarian nature.

We need voltage regulators because the sources of primary power (like regular batteries, or the AC voltage we get from the wall plug) are usually not very stable, or not stable enough to ensure that our circuits are working within their specifications.

For instance, the voltage we get from a car battery can vary between a high of 14.4 V when the engine is running and the alternator is charging the battery, and as low as 8 or 9 V when you crank the engine on a cold morning. Because there may be positive or negative spikes superimposed to the battery voltage due to other equipment, most automotive equipment is designed to operate with voltages as high as 16V. Internally, some circuits need stable voltage for proper operation, such as the microprocessor used to control the radio. Most microprocessors run from a 3 V or 5 V supply, which should be regulated to within a fraction of a Volt. For instance, many chips designed to operate from a 5 V nominal require the voltage to remain between 4.5 and 5.5 Volts.

Voltage Reference

Voltage regulators need a reference to operate. A Voltage Reference is a part or a circuit that provides a stable voltage when outside parameters, such as supply voltage or temperature vary.

The most common voltage reference is the Zener diode ([1]). A Zener diode is a diode where the avalanche reverse breakdown behavior has been optimized and quantified such that the diode can be operated safely in that region.

We can use SwitcherCAD to illustrate the behavior of a Zener diode.

<Zener-1.png>

Create the schematic now, you do not need to enter any value in Source V1 for now. Do not worry about the .DC statement at the bottom of the schematic, it is just a line of text I placed there for reference. When you are done creating the schematic, click on Simulate->Edit Simulation Cmd then select "DC sweep".

Enter the following values:

• Name of 1st Source to Sweep: V1
• Type of Sweep: Linear
• Start Value: -4
• Stop Value: 16
• Increment: 0.1

You should obtain a plot like this:

<Zener-2.png>

At source voltages above about 6 V, the Zener starts conducting current and the voltage across it levels around 6.2 V, which is the rated Zener voltage for that part.

The negative voltage region is interesting in that it shows that a Zener diode is like a real diode when biased in the forward direction. However, we do not intend to use the Zener diode in this region

The most interesting part is the reverse bias region (when voltages from V1 are positive). The Zener effect provides a voltage around 6.2 V that is quite stable, compared to the source voltage.

To find out how stable, let's re-run the simulation but by sweeping the source between 8 and 18V.

<Zener-3.png>

Line Regulation = Delta(V_out) / Delta(V_in)

In this case, the change in output voltage when the input voltage changes from 14 to 16 V (a 2 V change) is 20 mV, so the Line Regulation between 14 and 16 V is 1%.

If we were to replace the source V1 with a car battery, we would expect the regulated Zener voltage to vary between 6.24 and 6.38 V while the battery voltage changes from 8 to 16 V, a considerable improvement.

Let's see the effect of temperature by adding a .STEP statement to the simulation.

Click on the Text icon and enter the following in the text box: ".STEP TEMP LIST 0 25 50" then click "Directive" and "OK" and run the simulation again.

<Zener-4.png>

Shunt Regulators

This type of circuit is called Shunt Regulator because the regulating element is in parallel (as opposed to being in series) with the load. While our schematic does not show a load (for now), the load is any circuit powered from the regulated voltage, that would therefore be placed in parallel with the Zener diode.

A feature of the shunt regulator, which can be an advantage or an inconvenient depending on where and how the circuit is used is that a shunt regulator draws a constant current from the source. The current drawn from the source is the current that flows through the series resistor. Since the current that flows through the series resistor is only a function of the source voltage, the Zener voltage and the value of the resistor, it is constant as long as the source voltage is constant, and it is not a function of the load current.

The advantage is that the source current is independant of the load current.

The disadvantage is that the efficiency of the circuit is very poor at light load currents, so the circuit is not optimized for battery operation.

It is hard to imagine a simpler circuit, it has only two basic components.

On the other hand, the available current is limited. Let's see how much current we can get from this circuit.

Calculating the maximum Load Current

In this modified circuit, I have added resistor R2 to represent a circuit that will use the reference voltage. The resistor has no value yet, it is there to illustrate the point. This resistor constitutes the load and it will draw a certain amount of current. We need to make sure the regulator can deliver the current that is needed by the circuit represented by resistor R2.

<Zener-5.png>>

I_R1 = I_D1 + I_R2

As the value of R2 is reduced, the currenty through it will increase and the current through D1 will decrease by the same amount.

If the load current (current through R2) gets anywhere near 5.66 mA, the Zener diode will be starved (current through it will be very low or null) and it will not do its job of regulating the voltage. Let's find out how much current we can feed through D1 by looking at the specification.

For the complete document, click on the picture.

<BZX84C2V4LT1-D.pdf>

I_max = P_max / V_Zener

If you read the notes on the data sheet, you can see that 225 mW is the absolute maximum rating at 25 degree C ambient. The data sheet also gives you the thermal resistance and the de rating for temperatures above 25 degrees.

Without going into the details of these calculations right now, a good design practice is to limit the maximum current in our circuit to not more than 50% of the absolute maximum rating. That's 18 mA.

If our circuit is such that the load current may vary from zero to some value, we must make sure there is no more than 18 mA flowing through R1.

With the value of R1 we have chosen (somewhat arbitrarily), we will reach 18mA when the voltage from V1 is 6.2 + (1000 * 0.018) = 24.2 V where 6.2 is the nominal Zener voltage, and (1000 * 0.018) is the voltage we need to apply across R1 to cause 18 mA of current to flow through it. So it looks like we have quite a bit of design margin with regard to maximum power dissipation in the Zener.

Now we need to consider what happens when the supply voltage is at the minimum. Taking the example of the car radio, the minimum voltage from the battery can be as low as 8 V. With a supply voltage of 8 V, the current through R1 will be only:

I_R1 = (V_source - V_Zener) / R1

So, if this circuit was used in a car radio to provide a regulated 6.2 V to some sensitive circuits, we could draw up to 1.8 mA without loosing regulation, and without risking to blow the Zener when the battery voltage is maximum.

In practice, just like we de rated the maximum current, we would not want to completely starve the Zener and to make sure the voltage stays in regulation, we should keep a minimum amount of current in the Zener. The data sheet lists the Zener voltage for 3 current values of 1, 5 and 20 mA, so while it is legitimate to interpolate between the values given, it is less recommended to use the part outside the range of values given, so we should keep a minimum of 1 mA though the Zener for it to work well.

That means we have up to 0.8 mA of available current for the load.

Getting more current using a Series Pass Regulator

What should we do if 0.8 mA is not enough?

Well, we could either:

1. Reduce the value of R1. We have seen that with the current value of 1 k-ohm, we would not reach the safe maximum power dissipation until the supply voltage is 24.2 V. We could reduce the value of R1 so that the maximum safe power dissipation is reached at 18 V, which is the maximum supply voltage we need to design for.
2. Redesign the circuit with a higher power rated Zener (and decrease the value of resistor R1 further to cause more current to flow through it), or
3. Add a current amplifier, using one or more transistor(s).

Solution 1 is easy to implement and cost little, but it does not provide much of an improvement. In this case, the maximum Zener current being 18 mA, that's also the maximum possible load current.

In general, solution 2 does not make too much sense, because higher power Zener are harder to get and the circuit quickly would waste a lot of energy. With the trend for battery operated equipment, it is important to be familiar with solutions that do not waste power, or waste the minimum needed to perform the function.

Solution 3 is a little more complex, but offers more flexibility and is more efficient.

So we will try solution 3.

There is a well known circuit that performs the function we need, so without further ado, there it is:

<Regulator-1.png>

Similarly, the current source will generate any voltage needed to draw the amount of current we requested.

You can select the current source from the Component menu, simply locate and click on "current".

Current sources are not as intuitive as voltage sources, so don't be too concerned if the concept seems strange. Just follow what we will do with it and over time it will become familiar to you.

The other thing you may have noticed if you are really observant, is that we have a Zener with a part number of BZX84C5V6L, which was not in the library.

I cheated. I wanted to demonstrate a well known circuit, which is a 5 V regulator. The previous circuit was a 6.2 V regulator which, while sufficient for the purpose of the exercise, is seldom used. 5 V is a much more common voltage, and the 5.6 V Zener is often used in a circuit just like the one I just described. But the SwitcherCAD library did not include a 5.6 V Zener.

If you refer to the Motorola specification (the full pdf document, not the excerpt above), you will see that some part numbers are in bold. The note indicates that these part numbers are preferred, meaning they are much more likely to be in stock. The 5.6V part is in bold, so it is reasonable to assume it should have been in the library. Considering how much we paid for SwitcherCAD, we will forgive Linear Technology for not having included all the possible part numbers.

So how did I get a 5.6 V Zener in SwitcherCAD?

I opened the diode library file, C:\Program Files\LTC\SwCADIII\lib\cmp\standard.dio in a text editor and added the BZX84C5V6L as follows:

.model BZX84C5V6L D(Is=1.66n Rs=.5 Cjo=205p nbv=3 bv=5.6 Ibv=1m Vpk=5.6 mfg=Mot type=zener)

I had to close and reopen SwitcherCAD because the program apparently reads the libraries when the program starts and after I changed the file, it did not automatically reload it.

OK, enough with SwitcherCAD library, the transistor we added to the shunt regulator is in a configuration known as Emitter-Follower. That means the voltage on the emitter follows the voltage on the base (with a small offset typically of 0.6 to 0.7 Volt). The voltage gain of such a circuit is slightly less than 1.

So, if the base voltage is maintained at 5.6 V, the voltage on the emitter will be about 4.9 to 5.0 Volts.

Before going any further, make sure you have programmed V1 to be a 12 V voltage source.

To make the simulation more interesting, we will do a DC sweep on the current.

Click on Simulate->Edit Simulation Cmd and select DC sweep. Enter the values as follows:

• Name of 1st Source to sweep: I1
• Type of Sweep: Linear
• Start Value: 0
• Stop Value0.1
• Increment: 0.001

<Regulator-2.png>>

Load Regulation is expressed in percent of the output voltage, or in absolute value.

If we express it as a voltage change over the current change that caused it, it will be called Output Resistance since the value of a resistance is eual to the ratio of voltage across it versus current through it.

Load Regulation = Delta(V_out) / Mean V_out

Output Resistance = Delta(V_out) / Delta(I_out)

The Load Regulation is 0.04 / 4.92 = 0.81 %

Notice how the voltage creeps up rapidly at low currents (below a few mA). This is due to the fact that at very light load current, the base current, which is = Load Current / Hfe, is so small that the base voltage necessary to generate it becomes very small, much lower than the typical 0.6 to 0.7 V.

I added resistor R2 (100 k ohm) in order to provide a minimum load current and without that resistor, the voltage would creep up even more at light current values of I1. You can try to change R2 to 1000k for instance (1 meg-ohm).

In practice, if the circuit actually had to operate down to such low currents, it would be a good idea to decrease the value of R2 a little bit in order to reduce the voltage rise at light loads.

On the other hand, notice that this circuit now delivers 100 mA nicely while maintaining regulation between 4.85 and 5.05 V for currents between approximately 5 mA and 100 mA.

This would be perfect to drive most 5 V powered microprocessors.

Ripple Rejection

The Ripple Rejection is another measure of the regulator's capability to reject Line Voltage variations. However, Line Regulation, defined above, is measured with static (slow changing) input voltage changes, where Ripple Rejection is measured with rapidly changing input voltage, usually at the line frequency (60 Hz) or it's second harmonic (120 Hz).

If we were to use real instruments, we would measure the Ripple Rejection by superposing a small AC voltage on top of the DC Input Voltage, then measuring the amplitude of the same signal at the output of the regulator and computing the ratio. For instance, we could apply a 1 V AC peak voltage (2 V p-p) because it is well within the regulation range of the regulator and it makes calculations easier.

We can use the same technique with Spice, even though Spice offers another method that we will study in the next lesson. For convenience, we will measure the ripple rejection at 1 kHz.

Set the current source I1 to a fixed value of 50 mA, set the voltage source V1 to be a SINE source with 12 V DC offset, 1 V amplitude and 1 kHz frequency, then edit the simulation command as follows:

• Transient Analysis
• Stop Time: 5 mS
• Time to Start Saving Data: 0

Here is the plot of the output ripple (please note the voltage scale):

<Regulator-3.png

<Regulator-4.png

Exercises

1. How much current can we draw from the regulator before regulation becomes really bad? (you can use SwitcherCAD to experiment).
   What are the limiting factors to getting more current?
2. Plot the voltage on the base of the transistor on the same graph as the output voltage to see the difference. Explain the difference.
3. Compute the Ripple Rejection Ratio in dB. Because ripple is measured in Volts and not in Watts, the equation is 20 * log(V2/V1).
4. Plot the temperature variation of the output voltage at 25, 50 and 75 degree C.

Conclusions of this lesson

• We established that voltage regulators are a necessary part of most modern electronic circuits.
• Voltage regulators need a voltage reference, usually a Zener diode.
• Voltage regulators are characterised by their line and load regulation, ripple rejection and temperature stability characteristics.
• We learned how to use SPICE to obtain these values.

In the next lessons, we will improve the voltage regulator with a gain stage separate from the power stage.

Links

1. Zener Diode
2. .