Switching-Mode Power Supply Design Tutorial
Simple Dissipative Topologies

A tutorial on switching-mode power supply design by Jerrold Foutz 

Approach

As explained in the introduction, a power supply is a buffer circuit that is placed between an incompatible source and load in order to make them compatible. In this section we explore some simple dissipative circuits that can be placed between a 12 Vdc battery and a 5 Vdc load to make them compatible. The buffer circuits are simple in that we will restrict the parts to one each or less of the following parts: variable resistor, breakdown diode, switch, diode, single winding inductor, and capacitor.

For the dissipative topologies we will only use the variable resistor and the breakdown diode. The other parts will be used in the next section on switching topologies.

We will be concerned only with the power conversion part of the circuit, not the control. We will assume that the circuits are open loop and the output voltage is controlled manually by the value of a variable resistor or the duty cycle of a switch.

Although simple from a parts basis, the circuits we will explore with these parts are not necessarily simple from an analytical viewpoint. Some of the switching topologies will contain right-half-plane zeroes, an interesting topic that will be discussed later in this tutorial.

The first two topologies discussed will not be switching-mode topologies, but dissipative topologies. The motivation for this is to show the importance of efficiency in power conversion -- the major reason switching-mode power supplies are used.

In all of the following examples we will assume a 12 Vdc source and the goal of providing 5 Vdc at a maximum load of 20 A, which equates to a maximum load power of 100 W and a minimum load resistor of 0.25 ohms.

In comparing regulators we will be using efficiency as one of the figures of merit. Efficiency is defined in terms of input power and output power as:

Efficiency = (Pout/Pin)
or in terms of power loss
= 1/(1 + (Ploss/Pout))
where efficiency is a fraction that can be expressed as a percentage.

Dissipative Power Supplies

Dissipative regulators achieve regulation by a purposeful conversion of excessive power to heat, unlike switching-mode power supplies which do not rely on a heat conversion for regulation. Switching-mode power supplies would be 100% efficient if components were ideal. Dissipative regulators convert heat with either a series or a shunt element.

Series Regulators

We start our investigation of simple topologies by selecting the variable resistor from our parts list and connecting it between the power source and load. What we get is a simple open-loop series regulator shown in Figure 2-1. R1 is often a transistor.

Series Regulator Schematic  D
Figure 2-1: Series Regulator

Given:

Solving and listing the parameters of interest:

R1 is varied to obtain 5 Vdc at Vo resulting in 20 A current flowing in the loop and the load resistor R2. Notice that the 5 V output is obtained by dropping voltage across the series resistor R1, hence the name series regulator. To provide 100 W to the load, 240 W is required of the source and 140 W of power is wasted in R1 -- not very efficient. If the load power is reduced, so is the input power and the efficiency remains the same for all loads. Since the current cancels when made explicit in the efficiency equation, the efficiency of this circuit simplifies to Vo/Vin. As we shall see in the next section, this is not true of shunt regulators.

Notice that the topology is that of a voltage divider. Voltage and current dividers are used extensively in power supply circuit design and modeling. A future section will be devoted to them.

Shunt Regulators

From our parts list, we now add a breakdown diode across the load in Figure 2-1 to get the shunt regulator shown if Figure 2-2.

Shunt Regulator Schematic  D
Figure 2-2: Shunt Regulator

This is a very popular topology because it provides voltage regulation with only two parts, a fixed value R1 and the breakdown diode. The circuit also has inherent short-circuit protection as long as the wattage of R1 is selected so that it can operate into a short circuit. The major disadvantage of the circuit is poor efficiency, especially at lighter than full load. If the load power is a small fraction of total system power, this can often be tolerated. It should be noted that the topology is also the same as a series regulator using a breakdown diode as over voltage protection. As we will see in the section on switching-mode power supply topologies, over voltage protection is a very important consideration.

Assume the same conditions as Figure 1-1 and that the breakdown diode breaks an infinitesimal amount above 5 Vdc so it draws no current. Then the efficiency of the circuit is the same as the 42% of the series regulator circuit.

Now reduce the load to one half by increasing the load resistor R2 from 0.25 ohms to 0.5 ohms so only 10 A flow in the load and the load power is 50 W. The voltage would rise, but the breakdown diode constrains it to 5V and the excess current flows in the breakdown diode. The efficiency is now 50/240 W => 21% compared to 42% for the series regulator example. At zero load current, the efficiency is 0%.

In practice, some current always flows in the breakdown diode. This leads to the conclusion that a series dissipative regulator is always more efficient than a shunt dissipative regulator, everything else being equal.

One interesting observation about the shunt regulator is the maximum efficiency occurs at an input voltage that is only a function of the input voltage tolerance and independent of the R1 or R2. By setting up the power loss as a function of the circuit parameters and input voltage tolerance, taking the partial and setting to zero, the following equation is obtained.

Vin = Vo*(1+SQRT(1-a))/a
where a = 1-tolerance.
and
R2 = a*(Vin - Vo)/Imax
where Imax is the maximum load current

For our example and a 20% input voltage tolerance (a = 1-0.2 = 0.8), the input voltage that yields the maximum efficiency and the value of R1 are:

Example:
Vin = 5*(1+SQRT(1-0.8))/0.8 = 5*(1+0.447)/0.8 = 9.05 Vdc
R1 = 0.8*(9.05 - 5.0)/20 = 0.162 ohms
Efficiency = Pout/Pin => 100 W / 181 W => 55%

By selecting the optimum input voltage (instead of 12V) for a 20% tolerance, the maximum efficiency of the shunt regulator has been increased from 42% to 55%. However, note that the efficiency of a series regulator for the same lower input voltage (5/9.05 => 55%) is also increased to the 55% maximum efficiency of the shunt regulator, as would be expected.

This finishes our look at dissipative regulators. In our examples, the power losses were always greater than the load power for a 12 Vdc input and 5 Vdc output -- a terrible waste of power.

What are the system implications of this power loss?

Implications of Power Loss

This section has been expanded and updated and moved to the Power Supply Efficiency page of our companion website PowerSupplyDesignTutorial.com. It has been removed from this website in order to avoid duplicate content. Eventual this tutorial in its totality will be moved to the new location.


Webmaster and editor: Jerrold Foutz
Original: January 27, 1999, revised October 7, 2006