## Background Information

On more than one occasion, the author has experienced mains powered switch mode power supplies being destroyed when run from a petrol generator during field day type events. The component that seems to be the root of the failures is the electrolytic smoothing capacitor that carries the rectified mains voltage. The failures occur when the power supplies have only a small load or occur a few minutes after being powered by the generator. It is believed that voltage transients from the petrol generator (that aren't present in the domestic mains supply) are rectified by the bridge rectifier in the switch mode supply to voltages much higher than normally encountered.

For these reasons the author advises extreme caution when powering sensitive expensive electronic equipment containing switch mode power supplies from petrol generators.

One such occasion involved a switch mode power supply that formed the supply for a logging laptop used during a 2 metre (144MHz) QRP contest. The supply failed and generated lots of smoke a few minutes after power-up. The author has now decided to banish switch mode power supplies from field day events and replace them with linear supplies that don't suffer the same problem.

One solution is to use a 12V lead acid battery or a standard 13.8V linear power supply and use this to supply a 12 to 19.5V switch mode psu. The linear supply will effectively filter out the voltage spikes, and the smps will then transform the 13.8V to the 19.5V required for the laptop.

Converters for 12V to 19V \ 19.5V voltage step-up are commercially available, but it was decided to design a switch mode supply as a "paper exercise". This page presents that design.

For the purposes of the design, 12V wil be assumed as the supply voltage.

## Design Specification

The specification is to run a small laptop requiring 19.5V at 3.5A maximum from a 12V car battery. The design will use a step-up switch mode power supply.

The reader may wish to familiarise themselves with the formulas for Energy Stored in the Magnetic Field Associated With a Current-Carrying Inductor (links to http://www.mw0llo.com/EnergyStoredByInductorPage.aspx) and Rate of Current Increase in an Ideal Inductor for a Fixed Applied Voltage (links to http://www.mw0llo.com/CurrentRiseInInductor.aspx).

## Design

The proposed circuit is a boost smps, the simplified circuit of which is shown below:

The switch in the above diagram is usually a transistor, and this "switch" opens and closes repeatedly at high frequency. When the switch closes, a current will build up over time through the inductor coil and will store energy in the magnetic field. During this "transistor on" time, the diode will be reverse biased and will act as an open circuit. When the transistor switch opens, the current will continue to flow through the inductor and the forward biased diode into the load. The load requires a steady voltage, so a capacitor is used at the output to accept the pulses of energy from the inductor. The result is a "smoothed out" voltage at the output. If a load demands more current, the relative "on" to "off" time of the transistor switch is increased so that the inductor stores more energy on each cycle. A practical circuit is somewhat more complicated and there many integrated circuits available to simplify the design. For this design a Texas Instruments TL494 (links to http://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=slva001&fileType=pdf) will be used. This has an in-built oscillator, the frequency of operation of which is determined by two components: a resistor R_{T} connected between pin 6 of the device and ground; a capacitor C_{T} connected between pin 5 of the device and ground. The frequency of operation of the oscillator is given by:

MathML formula | Straight text formula |
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$f=\frac{1}{{R}_{T}{C}_{T}}$ | f = 1 / R_{T}C_{T} |

Where:

f = frequency of operation, Hertz

R_{T} = timing resistor value, Ohms

C_{T} = timing capacitor value, Farads

For a target frequency of operation of 20kHz a resistor value of 10kΩ gives a timing capacitor of 5nF.

For a 20kHz oscillator, the time of each cycle is 1/20000 = 50µs

The output power of the circuit is 19.5 x 3.5 = 68.25W = 68.25 J/s

So in a single cycle, the amount of energy to be transferred from input to output is 68.25 / 20000 = 3.413 mJ

### Duty Cycle

The ratio of the "on" time of the switch to the "off" time is referred to as the duty cycle. Assuming that the transistor works well as a switch, it will drop a certain voltage, say 0.2V, when switched fully on. Note that an NPN bipolar transistor when switched fully on (referred to as "saturated") will actually have a *lower* voltage between the collector and the emitter than the voltage between the base and emitter (which will be somewehere around the 0.6 to 0.7 volts range). So the voltage applied across the inductor during "on" time is 12 - 0.2 = 11.8 Volts. During the "off" period of the switch, the magnetic field stored by the inductor will start to collapse and generate an e.m.f. that will act to keep the current flowing at the same magnitude. It is known that the output voltage will be a reasonably stable 19.5V, the supply voltage is 12V, and the forward biased diode will drop about 0.6V. So the voltage that the inductor will generate to keep the current flowing is: (19.5 + 0.6) - 12 = 8.1V.

The current rise or fall in an inductor is proportional to the voltage across it. So in this design, the current will rise quicker during the on time (when the voltage applied to the inductor is 11.8 Volts) than it will fall during the off time (when the inductor will be "generating" 8.1V). The maths is as follows:

MathML formula | Straight text formula |
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$\frac{{di}_{\mathrm{on}}}{{dt}_{\mathrm{on}}}=\frac{{V}_{\mathrm{on}}}{L}$ | di_{on}/dt_{on} = V_{on} / L |

$\frac{{di}_{\mathrm{off}}}{{dt}_{\mathrm{off}}}=\frac{{V}_{\mathrm{off}}}{L}$ | di_{off}/dt_{off} = V_{off} / L |

${di}_{\mathrm{on}}=\left(\frac{{V}_{\mathrm{on}}}{L}\right){dt}_{\mathrm{on}}$ | di_{on} = (V_{on} / L) dt_{on} |

${di}_{\mathrm{off}}=\left(\frac{{V}_{\mathrm{off}}}{L}\right){dt}_{\mathrm{off}}$ | di_{off} = (V_{off} / L) dt_{off} |

At steady state operation with a fixed load, the increase in current through the inductor during transistor switch "on" time is balanced by the decrease in current during transistor switch "off" time. So:

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$\left(\frac{{V}_{\mathrm{on}}}{L}\right){dt}_{\mathrm{on}}=\left(\frac{{V}_{\mathrm{off}}}{L}\right){dt}_{\mathrm{off}}$ | (V_{on} / L) dt_{on} = (V_{off} / L) dt_{off} |

Multiplying both sides by L gives:

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${V}_{\mathrm{on}}{dt}_{\mathrm{on}}={V}_{\mathrm{off}}{dt}_{\mathrm{off}}$ | V_{on} dt_{on} = V_{off} dt_{off} |

$\left({V}_{\mathrm{supply}}-{V}_{\mathrm{switch\; loss}}\right){dt}_{\mathrm{on}}=\left(\left({V}_{\mathrm{load}}+{V}_{\mathrm{diode\; fwd\; bias}}\right)-{V}_{\mathrm{supply}}\right){dt}_{\mathrm{off}}$ | (V_{supply} - V_{switch loss}) dt_{on} = ((V_{load} + V_{diode fwd bias}) - V_{supply}) dt_{off} |

${dt}_{\mathrm{on}}=\frac{\left(\left({V}_{\mathrm{load}}+{V}_{\mathrm{diode\; fwd\; bias}}\right)-{V}_{\mathrm{supply}}\right){dt}_{\mathrm{off}}}{\left({V}_{\mathrm{supply}}-{V}_{\mathrm{switch\; loss}}\right)}$ | dt_{on} = (((V_{load} + V_{diode fwd bias}) - V_{supply}) dt_{off}) / (V_{supply} - V_{switch loss}) |

$\frac{{dt}_{\mathrm{on}}}{{dt}_{\mathrm{off}}}=\frac{\left(\left({V}_{\mathrm{load}}+{V}_{\mathrm{diode\; fwd\; bias}}\right)-{V}_{\mathrm{supply}}\right)}{\left({V}_{\mathrm{supply}}-{V}_{\mathrm{switch\; loss}}\right)}$ | dt_{on} / dt_{off} = (((V_{load} + V_{diode fwd bias}) - V_{supply})) / (V_{supply} - V_{switch loss}) |

$\frac{{dt}_{\mathrm{on}}}{{dt}_{\mathrm{off}}}=\frac{\left(\left(19.5+0.6\right)-12\right)}{\left(12-0.2\right)}$ | dt_{on} / dt_{off} = (((19.5 + 0.6) - 12)) / (12 - 0.2) |

$\frac{{dt}_{\mathrm{on}}}{{dt}_{\mathrm{off}}}=\frac{8.1}{11.8}=0.686$ | dt_{on} / dt_{off} = 8.1 / 11.8 = 0.686 |

So if the switch is "on" for 0.686 units of time, it is "off" for 1 unit of time. This is an interesting result, because it shows that the duty cycle is related to the input and output voltages (plus the voltage losses across the transistor switch and diode), but is *independant of the value of the inductor*.

For a frequency of 20 kHz the time of one cycle is 50 µs corresponding to 1 + 0.686 = 1.686 units of time.

So the switch is on for (0.686 / 1.686) x 50 = 20.34 µs

and the switch is off for (1 / 1.686) x 50 = 29.66 µs

All of the energy supplied to the load is supplied in pulses to the output smoothing capacitor during the "off" times of the switch.

We know that 3.413mJ must be supplied per cycle, so 3.413mJ must be supplied in the 29.66 µs of the off time.

The combination of the supply and the inductor in series during the "off" time "generates" a voltage of 20.1V and during the off time the power supplied is:

P = 3.413 mJ / 29.66 µs = 3.413 x 10^-3 / 29.66 x 10^-6 = 3.413 x 10^3 / 29.66 = 3413 / 29.66 = 115W

So the "average" current during off time is I = P / V = 115 / 20.1 = 5.72A.

For the purposes of calculation, we should assume that the input and output capacitors are very large.

The value of the inductor will relate to the amount of current rise and fall during the on and off periods of time.

If a large value of inductance is chosen, then the current rise and fall will be small. The disadvantage is that it will take a long time for the circuit to stabilise once first powered up. A further disadvantage is that the inductor will need to be physically large.

If a small value of inductance is chosen, then the current rise and fall will be large. The disadvantage is that the transistor switch and diode will need to be able to handle large peak currents. A further disadvantage is that the losses in the transistor and diode will be large due to the higher currents involved.

In practice a reasonable value of a few amps rise and fall of current should be chosen. If the current in the inductor rises from 4.72A to 6.72A during the "on" part of the cycle, we can calculate L from the "Current Rise for Applied Voltage" formula given above:

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$L=\frac{V}{\left(\frac{di}{dt}\right)}=\frac{11.8}{\left(\frac{6.72-4.72}{20.34x{10}^{-6}}\right)}=\mathrm{120\mu H}$ | L = V / (di/dt) = 11.8 / ((6.72 - 4.72)/(20.34 x 10^-6)) = 120.006 x 10^-6 = 120 µH |