0

I'm attempting to design my own rectifier circuit to output a positive and negative voltage as shown in the circuit below. I'm running into some concerns regarding the smoothing capacitor choice and how it affects the surge current and the voltage regulator output.

schematic

simulate this circuit – Schematic created using CircuitLab

I have my primary side wired to the same voltage input, and the secondary side gets rectified down to \$20 V_{RMS}\$. Smoothing capacitors C1 and C2 along with resistors R1 and R2 maintain the rectified voltages which go into voltage regulators to get the desired output voltage. According to the datasheet, the regulators are rated for at least 1 A, so I picked my test loads to see if each supply can deliver 1 A each. The issue I'm facing is with my choice of capacitors for C1 and C2. Larger capacitors seem to guarantee a stable voltage at the test loads, but doing so results in a larger initial surge current (which in my simulation with the values below is 2.11 A), and I'm concerned that the surge current might be more than what the transformer is rated for. Smaller values result in smaller surge currents, but the voltage is not maintained at the outputs. What is an appropriate trade off between capacitance value with regards to surge current and voltage regulator outputs given heavy loads?

EDIT: Perhaps my understanding of in-rush currents is missing something. From my textbook, I'm calculating my surge current using the following formula, assuming no initial charge on the capacitor:

\$\ I_{SC} = \omega \cdot V_{Pk-Secondary} \cdot C\$

where \$\ \omega=2\cdot \pi\cdot f \$ and f = 60 Hz

Taking the peak of the secondary to be \$ 20 \cdot \sqrt{2}\$ volts, the resulting current results in a value of 2.13 A. Since the transformer is rated for 2.4 A in parallel, this seems okay in order to not burn it out. Is my understanding correct?

BestQualityVacuum
  • 456
  • 7
  • 32
  • Why are you using two transformers when one LP-40-1200 would likely do the job? – Andy aka Mar 31 '20 at 18:27
  • @Andyaka The LP-40-1200's datasheet shows two primary inductors and two secondary inductors. I couldn't find a component that had that configuration, so I put two transformer components to showcase the connections, but yes, both them are part of the same transformer, so there's only 1 transformer in the entire circuit. – BestQualityVacuum Mar 31 '20 at 19:14
  • Adding a resistor in series with the transformer secondary can greatly reduce surge current -- and, counterintuitively, also improve efficiency. See https://electronics.stackexchange.com/questions/27881/ripple-current-in-a-linear-power-supply-transformer for details. – davidcary Mar 31 '20 at 23:11
  • @davidcary Thanks. I'm looking into the idea of resistors (and/or inductors) on both sides of the secondary. Resistors reduce the inrush current, but too high causes oscillation at C1. Inductors seem to work better, but simulation is slower. Someone suggested NTC Thermistors, but that's a little tricky to simulate in SPICE. – BestQualityVacuum Apr 01 '20 at 14:44
  • Would you ming telling me what is your textbook? I need a cite for such formula. Thanks anyway. – Gabriel Santos Apr 12 '22 at 12:44

1 Answers1

1

Using Q = C * V, differentiate with respect to T (time) and you get

qQ/dT = C * dV/dT + V * dC/dT [ uses d(U*V) = U * dV/dT + V * dU/dT ]

We assume dC/dT = 0 (that C is constant) and set dQ/dT = I current. Find

I = C * dV/dT

Now you have full-wave-rectifiers probucing double the power line frequency as ripple.

Let this ripple be 120 Hz. Let the load current (out of the cap) be 1 amp, assume your regulator will significantly attenuate the 120 Hz ripple.

The math is I = C * dV/dT which we re-arrange to be C = I * dT/dV, and we substitute

C = 1amp * (1/120 second) / 1 volt = 1/120 farad = 8,333 microfarad=ad

However, the rectifier PEAK current is also one of your concerns.

If you have huge ripple (the entire half cycle), the recharge time is 1/4 cycle. And your rectifier current will be about 2x the load current (out of regulator).

If you have small ripple, the recharge time is reduced. Unfortunately the peak of the sinewave is not a triangle wave, or we could predict the recharge time based on the fraction of ripple/peak, where 1v ripple/20v transformer output == 5% and you'd have some confidence the peak was 1/0.05 = 20X the average.

If you have very small rectifier on-time (less than 1 millisecond), the regulator will be less successful in reducing the ripple because of the very fast edge of the re-charge time. And your very fast current transient thru the diodes will cause Magnetic Interference all around.

Thus small inductors in series with the diodes are possibly useful.

Or try 1 ohm resistors. Remember the peak currents are LARGE, even if the duty cycle is low (5%).

analogsystemsrf
  • 33,703
  • 2
  • 18
  • 46
  • 1
    Thanks for your reply. I tried what you said, putting inductors in series with the rectifier diodes, and it seems to work to an extent. Causes the simulation to run slowly. Still getting a large inrush current. Since the output is going to a voltage regulator with a cap at the input, is there some flexibility with the ripple voltage/recharge time as long as the regulator's output provides a steady voltage and outputs the correct amount of current? – BestQualityVacuum Apr 01 '20 at 15:00