Switch-Mode Power Supplies: Optimizing Energy Use in Electronics

This article examines the considerations involved in selecting switch-mode power supplies.

This article is published by EEPower as part of an exclusive digital content partnership with Bodo’s Power Systems.

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Voltage converters are the backbone of many technical systems. Depending on the application, the required power supply unit is realized by a transformer, rectifier AC/DC converter. When high-performance switching power supplies were not yet available, 50 Hz transformer solutions were used almost exclusively.

Power Supply Considerations

Electrical energy is provided almost exclusively as a three-phase current with a system voltage of 10 ...30 kV~ in the power supply plants and transported over long distances at voltages of 38?kV~. Electronic DC voltages like 5 V, 12 V, and 24 V were transformed down from high transmission voltage via transformers to 220 V~/230 V~ voltage by means of a transformation ratio, rectified, smoothed, and stabilized.?Features for selecting a power supply are price, reliability, availability, whether using standard components or special solutions, the required power, needless of wide-range input yes/no 110 V~ ... 230 V~ with/without power factor, operating temperature, EMC, efficiency, MTBF, form factors such as size, weight, plug and the type of installation.

In the 1970s,?solutions?often consisted of the components 50 Hz transformer, bridge rectifier, and smoothing, possibly with downstream U-stabilization. The question is, to what extent should the acquisition costs be the sole deciding factor in the purchase decision? Operating, maintenance, and replacement costs may be a factor that should not be neglected. It is worthwhile using a more energy-efficient solution after just 24 - 36 months of operation time. For industrial goods, depreciation periods of ten years are often applied. However, the machines often run twice as long. It is now necessary to consider whether and to what extent the use of an energy-saving solution can reduce the operating system costs for the energy supply company.

Many electronic devices require a stable DC voltage. Energy is transmitted through AC voltage plants' supply and networks. This means that a DC voltage must first be generated from an AC voltage at the point of use. The method of voltage generation affects both operational safety and operating costs.

Before the introduction of high-frequency power supply solutions, which have now assumed a dominant position with the advent of fast-switching, low-impedance semiconductors, solutions with 50 Hz transformers, bridge rectifiers, and smoothing capacitors were predominantly used to convert the 220 V~/230 V~ mains voltage to 24 V and other frequently required DC voltages. The transformer solutions enable galvanic isolation between primary (230 V~) and secondary circuits (e.g., 12 V, 24 V).

Due to the special way of loading the transformer with the bridge rectifier and smoothing electrolytic capacitor, a pulsed current flow results in the 230V~ primary circuit.

The electricity meter is calibrated in effective values. The pulsed current flow results in a different RMS value for the current, which is calculated despite the same charge carrier flow.

i = C * dU / dt?? ? ? ? ?(1)

i = dq / dt? ? ? ? ? ? ? ? (2)

From equation (1) follows by conversion:

i dt = C dU? ? ? ? (1a)

On the DC side, a certain power or energy is required to perform the work.

dW = u i dt? ? ? ? (3)

Substituting the expression from equation (1a) into equation (3) results in

dW = u C dU? ? (3a)

If both sides are integrated, the result is

W (work, labour, energy) = ∫dw = C ∫UdU = ? CU2 W = ? CU2? ? ?(4)

From the square of the voltage in the formula, you can see straight away that the energy or work does not change linearly with the voltage. The higher the crest factor ξr = ? / Ieff, the higher the electricity costs.

Energy, Work

W = P * t? ? ?(5)

A capacitor, an accumulator can now be filled in various ways, i.e., ‘filled’ with electrical charge carriers (q). Of particular interest in this context is the energy required for this on the AC side.

Example: On the DC side, an electrical power of P = 50W with a voltage U = 24V is required. The energy in the storage unit (battery) should be sufficient for an operating time of t = 24h.

E = 50W * 24h = 1’200Wh? ? ? ? ? (6)

The current I is calculated as follows:

I = P/U = 50VA / 24V = 2.08A? ? ?(7)

The battery has a storage capacity of C = 2.1A * 24h / 24V? ?(8)

The charge in the battery: Q = I t = 2.08A 24h = 50 Ah? ? ? (9)

This charge must be supplied by the AC voltage source. In addition, the AC voltage source must provide the losses in the circuit components involved. These losses will be omitted for the sake of clarity.

The amount of charge supplied must be the same when charging with direct current or alternating sine current (rectified value) and with pulsed current (red), otherwise the battery, the accumulator and the capacitor would be charged at different rates. The surface areas for the charge must be the same size.

Q = I * t direct current

Q = 2 * ? / π Sine after bridge rectifier

Q = iRE * tp tp = T/12 with a typ. Current flow angle of Φ = 30° for bridge rectifiers with smoothing electrolytic capacitor.

However, to transfer the same amount of charge to the DC voltage side 24 V, a different amount of work (energy) must be applied to the AC voltage side.

To compare only the influence of the waveforms, the time (t) of charging must be set to the same value.

Energy for 1 complete charge

a) Direct current E = 24 V 2.08 A 24 h = 1,200 Wh

b) Alternating current sine waveform

P = UEeff * IEeff = 50 W (without losses in the transformer, rectifier)

The following applies to the secondary side:

Q = 50 Wh ? ? = 50 Ah/24h = 2.083 A * π / 2 = 3.25 A

This results in IAeff = ? / √2 = 2.31 A

UE,eff = 230 V~ ? IEeff = 50 VA/230 Veff = 0.217 A

c) If the energy is transferred in pulses to the secondary DC voltage side, the following ratios result:

required electrical charge Q = 50 Ah

iRE,puls =??? I = 1/T ∫ iRE,puls * dt

iRE,puls = 6 * I as the charge carrier transport takes place per sine half-wave in T/12, i.e., 2 x per complete sine input oscillation. Consequently, the current must be greater by a factor of 6 so that the same amount of charge is transferred to the secondary side in T/6 of the period duration T.

The effective value of the input current, which is also billed, is calculated as follows:

TEeff2 = 1/T ∫(6*I)2dt = 1/T 36 T/6

This results in Ieff,puls = √6 = 2.45 times greater.

The crest factor is therefore: ξr = ? / Ieff = 6 I / √& I = √6 = 2.45

In many mains applications in the 10 W to 75 W power range, high energy costs often occur unnoticed when consumer devices that use non-linear components in their inputs, such as bridge rectifier circuits, are connected to the 230 V~ mains. The so-called power factor λ for such appliances is often only between 0.3 and 0.6. As the energy meters of the energy supply companies are calibrated in effective values, it is important to ensure efficient energy consumption of the consumer appliances. Otherwise, it will be expensive. As long as electrical energy costs 5 - 6 cents, this consideration did not play such an important role. But with today’s costs of 20 cents /kWh upwards, this is wasted money.

As the integration takes place inside bottom and top boundaries (the current flow in the example takes place between 60° and 90° corresponding to T/12 in relation to 2 sine half-waves), this means twice within one complete cycle T.

Therefore iRE,puls will 6 x higher than the I for the eual charge transfer towards the secondary path.

Assumption energy costs: 20 cents/kWh

Calculated under the following assumptions:

ξr = 2,22 o. Power Factor PF = 0.45 and 365 * 24h operation time

η = 0.92

ξr o. PF = Power Factor; λ = real power/apparent power

η = efficiency

Differences in Switch-Mode Power Supplies

Switch-mode power supplies with a sinusoidal current consumption can help here. However, there are differences in the way they work. Classic solutions are a step-up converter consisting of a storage choke, MOSFET trs. rectifier diode, smoothing capacitor, and IC circuit with various additional components, which initially generates an intermediate circuit voltage of approximately?380 V ... 400 V DC from the rectified mains voltage. A downstream DC/DC converter, also in high-frequency operation f ≥ 50 kHz, then generates the required secondary voltage of 5 V, 12 V 24 V, galvanically isolated from the mains side, short-circuit and open-circuit proof.

However, Grau Elektronik power supply units use a PFC transformer solution that follows a sinusoidal input current curve and, therefore, has a low crest factor and high power factor λ with a maximum reduction in circuit components. By using transient-resistant components, a mains transient of 1.6 ... 2.3 * UE,nenn for t <= 0.1msec can also be handled.

This article originally appeared in?Bodo’s Power Systems?[PDF]?magazine.

This text is taken from https://eepower.com/technical-articles/switch-mode-power-supplies-optimizing-energy-use-in-electronics/