AC-DC Linear Power Supply & Voltage Regulation

 Background

    One of the most common sources of electrical power is an electrical outlet. This power is in the form of AC. To use it in smaller devices or to charge a phone, it must be converted to DC. Next, it must be regulated to a steady voltage that can supply a specified amount of current. This two-step process is known as rectification and voltage regulation. One of the most common and simplest design for voltage regulation is by using a linear voltage regulator. It has a drawback of always being on and losing the unused power as heat, but the design process is simple which is why it's still used even today.

Specifications

    In this post I will document the process of rectifying an AC voltage, regulating the output and then simulating it and testing it.

The specifications for the regulated power supply are as follows:

• 9V @750mA
• Input voltage = 120Vrms AC 

There are two stages of the design. First the regulated part and then the unregulated part which is the rectification of AC to DC. I will begin with the voltage regulator part first. 

Voltage Regulator Design

Choosing a Linear Voltage Regulator

    There are many readily available ICs that can regulate voltages to any common voltage level in the market. To avoid simply using a specially-made one for 9V , I will be using a 5V regulator IC to demonstrate the design process for a different voltage.

    I will be using the LM7805 voltage regulator which supplies 5V at its output and a maximum of 1.5A which is well-within the required specifications. 

Design calculations

First, the dropout voltage of the regulator must be considered in the design as the minimum input to the regulator is calculated as follows:

Where Vd is the dropout voltage across the LM7805. Here Vd = 2V according to the datasheet, so that means the minimum input voltage

Vin = 9V + 2V = 11V

Now we can begin calculating the resistor values based on the expected voltages and currents and lay down the schematic below.

To simplify calculations, I assume the leakage current of I_ref = 0A. The reason I make this assumptions because I_ref << I_resistors. 

Next, I set the current going through the resistors to be 100mA since this value is less than 20% of the target output current of 750mA. That means I_sistors = 100mA.

Resistor Power Calculation

    When designing power supplies, it is important to ensure that each component can tolerate the power dissipated. Normally resistors of 1/4 Watt tolerance are used for most electronics applications, but for a linear voltage regulator the power not used will be dissipated through the resistors and it must be calculated to know what type of resistors to use. Even though this voltage regulator is designed to output a current less than 1A, it is still useful to know how much power will be dissipated ineach resistor.

The power dissipated in the 50 Ohm and 40 Ohm resistors can be calculated with P = IV. 

• For the 50 Ohm: P = 5V * 100mA = 500mW
• For the 40 Ohm: P = 4V * 100mA = 400mW

This means that regular 1/4 Watt resistors won't be fit to use in this application. I will use resistors rated for 5W for both resistors to be safe. 

Rectifier Design

    Now it's time to design the unregulated part of the circuit which is the rectifier. The power that comes from the mains power grid here in the US is 120Vrms 60Hz AC, so we must step down the voltage to an acceptable input voltage and then rectify it to a DC voltage before we use it on our regulator.

    The minimum voltage for our regulated part must be 11V and the maximum according to the datasheet must be 25V so our ideal voltage would be anything slightly above 11V.

Choosing rectifier and transformer tap

    Linear power supplies are wasteful by nature of their design, so to reduce wastefulness I decided to go with a full-wave bridge rectifier with no center tap to use the complete wave of the AC signal. 

  I will make use of a full-wave bridge rectifier with a filter capacitor such as the one shown below.

    To be able to accurately simulate it in OrCAD PSpice a resistance must be added to the transformer coil to prevent zero resistance loop. I set this value to be in the milliohm range. The primary coil always has a greater resistance than the secondary, so the secondary resistance can be neglected. 

The characteristics table of a full-wave bridge rectifier that aid in picking components are the following:

In the lab, I am limited to using either a 24V or 12V step-down transformer. 

Using the characteristics table above, the minimum step-down voltage Vs can be calculated:

By assuming Vd = 0.7V, we can solve for Vs

So as long as the step-down voltage is greater than 8.77V, the transformer can be used. The 12V step-down transformer satisfies this condition, so I chose to use it. 

Filter capacitor

To determine the minimum capacitance for the design,  the nominal voltage must be calculated first using the characteristics table.

This nominal voltage allows us to determine the maximum allowable ripple that will maintain the input at the regulator to 11V.

The maximum allowable ripple should be no greater than 4.57V. 

Now to calculate the capacitance based on our ripple we use the following formula:

By plugging in our voltage ripple and known frequency of 60Hz we determine the minimum capacitance.

Ratings of capacitors and diodes

    Similar to calculating the power rating for the resistor, the ratings of capacitors and diodes used in the circuit must be calculated to avoid breakdown of components which could result in unpredictable behavior. 

    First we can calculate the peak-reverse voltage(PRV) rating of the diodes that will be used in the rectifier circuit. This is calculated as follows: 

    The average diode current would simply be half the DC load current in this configuration as only 2 diodes are on at a time. The total current going in to the circuit is the sum of all the currents required in our design of the regulated part. This means the total DC load current is the sum of the current going through the resistors and through the load.

•  DC load current =  100mA + 750mA = 850mA.

Now we can just calculate the average diode current:

And also calculate the peak diode current for the full-wave rectifier

The capacitor voltage rating is straightforward as it should be greater than the nominal voltage. 

I chose to use a rectifier IC package with the following ratings:

This rectifier safely met all ratings so it was appropriate to use in this application. 

Further Design Choices

In the lab I was limited to capacitor ranges that met the minimum requirement. The only capacitor value that met the minimum ratings was a 2.2mF one. We just used two in parallel to increase capacitance.

The final design implemented is shown in the PSPICE schematic below:

Unregulated Part Testing

I recorded the DC output voltage and pk-pk ripples values for different loads and put them in the table below:

The oscilloscope shots for each No load, 33% load and 100% load are shown below in their respective order:

No Load

33% Load

100% Load

Next, I used a VARIAC between the AC line and the input to the power supply to simulate a change of AC line voltage in the input voltage that varied from 105V to 155V to 125V for a load of 33% and recorded the DC output voltage as follows:

Since the DC voltage output even at the lowest input voltage of 105.6Vrms resulted in a value that was greater than 11V, the power supply meets the criteria needed for the input stage. 

Final Design 

After verifying functionality of both the unregulated part and regulated part independent of each other, it was now time to put them together. The schematic for the whole power supply is shown below:

Power supply testing

Finally, after assembling it on the breadboard I started testing different loads by using power resistors as a load. I measured the output voltages with the different loads in the following table:

And once again, here are the oscilloscope shots for No load, 33% load and 100% load respectively:

No Load

33% Load

100% Load

Simulation of different loads

First, the max load was simulated with the following schematic. To get max load, the load resistor value had to be 12.03 Ohms. 

Simulation setup for max load

Running a transiet simulation gave a stable voltage of 9.18V

Transient results of max load

Next, the 33% load of calculated value of 36.2 Ohms was set up as shown in the schematic below

Simulation setup for 33% load

The stable output voltage for 33% load was at 9.1902V as shown in the output waveform.

Transient results of 33% load

Conclusion

The design met all specifications and produced a measured result which was very close to the simulated behavior. Some design choices were made to the physical circuit due to component constraints, but even with those differences, the circuit performed well within spec. It was a very fun project but also one to be very cautious with since dealing with the mains voltage can be dangerous. The simulation matched closely to our results which verified that the design process was correct.