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The capacitor-only bridge rectifier can't remove the ripple under heavy load and for a linear lab power supply I need to have an output without any ripple under load. after simulating the bridge rectifier circuit under 1A load I ended up with this;

Ripple without an inductor:

Ripple with an inductor:

The transformer is 24/1A and diodes are schottky.

Considering it's going to pass trough a linear regulator which will reduce the ripple to some degree and with such low power transformer, do I really need to reduce the ripple to almost nothing by adding an inductor? or is it better to have the inductor there regardless of the transformer power?

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  • \$\begingroup\$ An output without ripple is naive. \$\endgroup\$
    – Andy aka
    Jan 4, 2020 at 10:55
  • \$\begingroup\$ @Andyaka What do you mean by naive? \$\endgroup\$ Jan 4, 2020 at 10:56
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    \$\begingroup\$ For a linear regulator Vin must be >= Vout + min headroom at Imax. Output ripple is >= input ripple - dB attenuation spec of regulator. I say <= as you can't (usually) get better than spec and Murphy will 'help you' get less than spec. So IF the regulator rejection is adequate fpr your need with a cap filter then you can use one. If not you need to improve the filter. || Note: Add some small input resistance to "spread" cap charging peak - or use an L in filter. \$\endgroup\$
    – Russell McMahon
    Jan 4, 2020 at 11:11
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    \$\begingroup\$ If the psu is 1A limited at 24V then with 1 Ohm you will get about 1 Volt in the load. (V = IR = 1 x 1 = 1. Yes? || You need to tell us what you are REALLY doing as that indicates "summat agalae". \$\endgroup\$
    – Russell McMahon
    Jan 4, 2020 at 11:13
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    \$\begingroup\$ Google translate is your friend. You have also the same project (author: blackdog) commented in eevblog : eevblog.com/forum/projects/… Many usefull info even if it dutch. \$\endgroup\$ Jan 5, 2020 at 17:13

1 Answer 1

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If I understand right, Your question whether you should require little to no ripple in the output voltage from your power supply.

Well, all power supplies in the world have some ripple and typically circuits that eventually are the power consumers will tolerate some ripple. In general you would not want your power supply to provide lesser ripple then what your consumer desires because that would add drastically to the cost and complexity of your power supply. As an example you have used 2.2mF capacitor and 100mH inductors. A 2.2mF capacitor of 50V or above would be a gigantic one. A 200mH 2Ampere inductor wouldn't be a small component either.

In your specific case the answer depends on the type of linear regulator you are planning to use. Lets say you want to use a 5V regulator like a 7805 then having fair ripple like around 8 to 10V could easily be tolerated. Typically linear regulators have internal or external feedback loops to compensate for output voltage changes against a reference but you wouldn't have to worry about that considering that the ripple frequency is going to be around 100Hz(the rectifier doubles the frequency as the waveform is just one sided). The max allowed ripple will depend on the final regulated voltage that you desire. So your linear regulator's datasheet will dictate what is the minimum input voltage it can have and still provide a stable constant output.

Now all being said, in short, if you are eventually going yo use a low voltage linear regulator, it seems you dont need such large inductor or capacitors.

If you have further interest in low ripple power converters and that without using very large components, you should study high frequency/switch mode converters. Typically switched converters would feature an IC/microcontroller that would switch a device(mosfet) using a high frequency PWM to achieve target output voltage and that too without a very large size of components.

FYI thats how phone chargers that used to be really bulky suddenly got so compact around 2 decades ago because all of them now use SMPS.

Best of luck learning.

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