Lesson 21

Power Supplies and Voltage Regulators

If you have an electric circuit, you need electric power to run it. Power supplies come in many forms, from a battery, to a solar panel, to “mains” power in your home. For an amateur radio HF or Mobile rig, many are expecting a 13.8 volt DC power source. One source is a car or LiFePo battery. In your home, it’s AC converted to DC. For handhelds you get a variety of power requirements between 5 and 13.8 volts.

Illustration of DC battery

Some hams go to parks or up on mountains to operate. They end up away from AC power sources. Some will rely on batteries, and others may use solar.  

If you’re operating on batteries, you’ll want to be able to calculate how long you can operate before you go. It’s a simple calculation for battery operating time. It’s battery capacity in amp-hours divided by average current. If you have a 10 amp hour battery, and average 2 amps of current use an hour, you can expect to operate for 5 hours or so.   

If you are using solar, you need to get the right power to your radio and accessories. Let’s say you are running a radio on DC but you want a computer for logging. If your computer runs on AC, use an inverter connected to your solar panel output. That will convert the panel’s output from DC to AC.

Let’s continue on with the topic of conversion. In your home shack you will probably need to convert your 110 volts AC power to 13.8 volts DC. That’s where a power supply comes in. It has a complex job. While your radio is idle, just listening, it has a fairly consistent low voltage draw. When you transmit, it needs a burst of current. That burst can pull down the voltage. Inside your power supply is a voltage regulator.  It helps maintain a constant voltage in your circuit. The two main types of voltage regulators are linear and switching regulators. 

Let’s start with the switchmode voltage regulator. This may have been part of the first power supply in your shack because it’s light and inexpensive. The switch mode voltage regulator works by varying the duty cycle of pulses input to a filter.

What’s a design difference between a switching and linear power supply? On the switching supply, the high frequency inverter design uses much smaller transformers and filter components for an equivalent power output.  

One other high-voltage power supply feature is a step-start circuit. It allows the filter capacitors to charge gradually.

A linear power supply has a linear voltage regulator. How does a linear electronic voltage regulator work? The conduction of a control element is varied to maintain a constant output voltage. 

Linear voltage regulators usually come in two types:

  • A series regulator. This three-terminal voltage regulator usually makes the most efficient use of the primary power source, and
  • A shunt regulator. A shunt regulator works by placing a load on the unregulated voltage source.

Both series and shunt regulators typically use a Zener diode as a stable voltage reference.   

Here’s a formula you need to know, with no math required on the exam. What is the formula to calculate the power dissipated by a series linear voltage regulator? Measure the voltage difference from input to output multiplied by output current. 

There is a level at which a voltage regulator just can’t keep up. That’s called the dropout voltage. What’s the dropout voltage of a linear voltage regulator? It is the minimum input-to-output voltage required to maintain regulation.

You can use resistors to help the performance of capacitors in a power supply. You can connect equal-value resistors across power supply filter capacitors connected in series. This will allow you to:

  • Equalize the voltage across each capacitor
  • Discharge the capacitors when voltage is removed and
  • Provide a minimum load on the supply

All these are correct when asked about resistors and filter capacitors. 

Basic Circuit Design

Since we’re talking about voltage regulators, let’s start a review of three schematics that are on the exam. Figure E7-2 is a linear voltage regulator schematic. Two exam questions ask about linear power supply components in this schematic:

  • Q1 in the circuit shown controls the current to keep the output voltage constant.
  • C2 in the circuit shown bypasses rectifier output ripple around D1.

Next up is figure E7-1. That is the schematic for a common emitter amplifier.

You’ll be asked about several of the resistors in this schematic:

  • The purpose of R1 and R2 is to provide voltage divider bias, 
  • R3 is a resistor to provide self bias.

Finally, figure E7-3 on the exam is an inverting op-amp circuit.

First, let’s talk about a possible modification to the circuit. Add a capacitor across the feedback resistor. This circuit now has the frequency response of a low-pass filter. 

In the E7-3 op-amp circuit, R1 and RF control the gain. The questions for this schematic are based on gain calculations of their ratios. The formula for gain is RF divided by R1.

Gain = RF/R1

  • When RF is 470 ohms and R1 is 10 ohms, the gain is 47. This means the output voltage will be 47 times the input voltage!
  • When R1 is 1800 ohms and RF is 68 kilohms absolute voltage gain, the gain is 68000 ohms divided by 1800 ohms, which is 38.
  • When R1 is 3300 ohms and RF is 47 kilohms, the absolute voltage gain is 47000 ohms divided by 3300 ohms = 14.

There is one more calculation and it has a bit of a twist.

  • What will be the output voltage of the circuit shown in Figure E7-3 if R1 is 1,000 ohms, RF is 10,000 ohms, and 0.23 volts DC is applied to the input?
  • First calculate the gain.  RF is 10,000 ohms and R1 is 1,000 ohms. Divide them to get a gain of 10.
  • Take the 0.23 volts and multiply by 10, to get 2.3 volts.
  • One more thing to remember. This is an inverting op-amp circuit. So 0.23 volts in gives you -2.3 volts out.  

You have both plus and minus 2.3 volts as a possible answer. So, if you can remember this is an inverting op-amp, you’ll choose the right answer.

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