Linear regulated power supplies
With a 7.5 V secondary, a 7805 loses regulation a hundred times per second and an LDO regulates with almost 1 V of margin. What decides it is the bottom of the ripple, not the average voltage.
01From the mains to a stable DC voltage
A linear power supply has four blocks: a transformer that steps down the mains voltage and isolates, a rectifier, a filter capacitor that converts the pulses into a DC voltage with ripple, and a regulator that eliminates the ripple and the variations of the mains and of the load. The regulator is a feedback amplifier: it compares a fraction of the output with a reference and drives a series transistor that absorbs the voltage difference.
Rectifier diodes were presented in semiconductors and diodes, in the technician program, and there is a publication with the complete design of a regulated power supply with an LM317.
02Rectifier and filter capacitor
With a bridge, the capacitor charges at the peaks of each half-cycle and discharges through the load the rest of the time. The peak is that of the secondary voltage minus two diode drops, and the ripple depends on how much charge is lost between peaks:
Recharging occurs over a short angle, so the current through the diodes and the secondary arrives in pulses of several times the load current. That is why the transformer is sized with an apparent power greater than the DC power it delivers, and the diodes and the capacitor for their peak current and their RMS ripple current.
03The series regulator
The classic linear regulator has a voltage reference, a compensated zener or a 1.25 V bandgap reference, an error amplifier and a pass transistor. Feedback makes the output follow the reference multiplied by the divider, and reduces by the loop gain the effect of the mains and of the load:
To regulate, the pass transistor needs a minimum voltage between input and output, the dropout voltage (dropout). In regulators with an NPN emitter output, such as the 7805, it is about 2 V; in LDOs, with a PNP or a PMOS pass device, it falls to a few tenths of a volt. If at the bottom of the ripple the input falls below \( V_o + V_{DO} \), the ripple passes to the output.
04Dissipation and efficiency
The design is a compromise: the secondary voltage has to be enough for the bottom of the ripple to exceed the regulator drop with the mains low, but every extra volt is heat. LDOs allow the secondary and the dissipation to be lowered; switching power supplies solve the problem at the root, in exchange for switching noise.
Time-domain simulation of a diode bridge (0.8 V each) fed from a secondary with 0.5 Ω of winding resistance, at 50 Hz, with a filter capacitor and a constant-current load. The gray trace is the rectified voltage; the blue one, that of the capacitor; the red line, the minimum the regulator needs, and the green one, the output.
Integrated regulators add protections: current limiting, sometimes with foldback (foldback) to reduce the dissipation in a short circuit, thermal shutdown and safe-operating-area protection of the pass transistor.
05In the lab
Measure the ripple across the capacitor with the oscilloscope for several load currents and compare it with \( I_L/(2fC) \). Also measure the peak current through the diodes with a small series resistor.
Lower the mains voltage with a variable autotransformer and observe at what point the ripple appears at the output of a 7805 and of an LDO.
Measure the line and load regulation of an adjustable regulator and compare them with the datasheet.
06Common mistakes
- Looking at the average voltage and not at the bottom of the ripple. The regulator stops regulating at the minimum, not at the average.
- Designing with the nominal mains. The mains can be 10 % low and the supply must keep regulating.
- Forgetting the transformer resistance. Under load, the real peak ends up well below \( \sqrt 2 V_s \).
- Raising the secondary “just in case”. Every extra volt turns into heat in the regulator.
07Self-assessment
What ripple does the formula estimate for 0.5 A with 2200 μF at 50 Hz?
\( 0.5/(2 \cdot 50 \cdot 2.2\cdot10^{-3}) \approx 2.3 \) V.
What output does an LM317 give with R1 = 240 Ω and R2 = 720 Ω?
\( 1.25 \cdot (1 + 3) = 5 \) V.
A 12 V regulator delivers 0.8 A with 17 V average input. How much does it dissipate?
4 W.
Why does an LDO allow a lower-voltage transformer to be used?
Because it needs only a few tenths of a volt between input and output, instead of about 2 V.
08Further reading
- Paul Horowitz and Winfield Hill. The Art of Electronics. 3rd ed., Cambridge University Press, 2015. The chapter on linear regulators, with the design of complete power supplies.
- Adel S. Sedra and Kenneth C. Smith. Microelectronic Circuits. 8th ed., Oxford University Press, 2020. The rectifier with a capacitor and the conduction angle.
- Texas Instruments. Datasheets of the LM7800 and LM317 regulators and of the LDO family. Available on the manufacturer's website. The specifications for regulation, ripple rejection, dropout and protections.