What do you do after rectifying voltage? - Page 1
Forum section: electronics / beginners

author: LoveLaika
author_rank: Frequent Contributor
timestamp: January 30, 2024, 03:15:52 pm
content: I'm playing around with simulations of an AC/DC circuit to understand the idea better. I know the fundamentals of how the AC voltage has to be stepped down through a transformer and passed through a rectifier to be converted to a DC voltage for your circuits. After that, that's where I am lost. After the rectification stage, what do people usually do with the rectified DC voltage?So, my first thought was that with big output capacitors, the DC voltage would be constant (or at least have minimal ripple) to be used as a DC voltage for a load. Unfortunately, that idea didn't pan out in my simulation. I've attached a picture of it below. I used LTspice as my simulator with the built-in diode models with no external SPICE models. Even though I simulated it with a big transformer (14A-56-20, 56 VA, parallel windings allow for 5.6 amps), the output can only handle milliamps before the output voltage oscillates. Why would this happen even though the transformer can handle a higher current?One option is to pass it through a voltage regulator, but why couldn't you use the rectified voltage directly? For that matter, if you had no voltage regulator ICs, what would you do next?

author: tunk
author_rank: Super Contributor
timestamp: January 30, 2024, 03:30:04 pm
content: I think a rule of thumb is to use 2000uF/A and thatthis will give you ~4V ripple. If I understand the figurecorrectly, the transformer has a 10Vac secondary.After rectification and smoothing this will give you13-14Vdc (unloaded). With a 13.8ohm load (~1A)and 1880uF, this should give you ~4V ripple.

author: Terry Bites
author_rank: Super Contributor
timestamp: January 30, 2024, 03:55:19 pm
content: I simulated your circuit in MicroCap and I saw about 2.5V pk-pk of ripple.The following formula is used to calculate the peak-to-peak ripple voltage for a full-wave rectifier.Vpp=I/(2∗f∗C)Where Vpp is the ripple voltage (peak-peak)I is the current in the circuitf is the source frequency of AC Power. f is Line freq for halfwave rectifier or 2*Line freq for fullwave. So, in your circuit f=120HzC is the capacitance in FaradsSee hyperphysics.phy-astr.gsu.edu/hbase/Electronic/rectct.html#c6]http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/rectct.html#c6To get zero ripple you need infinite capacitance. So you have to compromise.An LC pi filter works really well in theory but you need a monster inductor for 120Hz ripple@5A.You can reduce the ripple enormously with a regulator. Typically 65-80dB. The regulator's "dropout voltage", the voltage betwen input and output needed to operate the regulator's internal circuitry can be a problem. Typically a couple of volts for common types eg LM7812. Low dropout regulators (expensive and finicky) can go as low as 100mV. (Vi-Vo)*Iload is power that the regulator has to dissipate. A heatsink is often needed to keep the regulator from overheating.To minimse losses a DC-DC converter fed from the raw dc could be used.

author: Wallace Gasiewicz
author_rank: Super Contributor
timestamp: January 30, 2024, 03:57:16 pm
content: Use a Pi RC FilterRead down to this type of filter in this article:https://ecstudiosystems.com/discover/textbooks/basic-electronics/rectifiers/smoothing-filters/Two caps separated by a high watt maybe 5 Ohm resistorI like the Pi LC filter if you have an extra transformer you can use as a choke. Needs to handle all the power however.

author: bdunham7
author_rank: Super Contributor
timestamp: January 30, 2024, 04:37:17 pm
content: You have to understand something called 'conduction angle', which is the amount of time that the transformer is actually supplying current. (one complete AC cycle is 360 degrees) The transformer and rectifier will only conduct into the capacitors when the voltage from the transformer is a volt or so higher than the voltage already across the output capacitors. If you have no load, those capacitors charge up to very nearly the peak of the AC cycle, meaning that even though your load might be small, the transformer is only supplying current for a very short period of time at the peak of the AC wave. In between those peaks, the capacitor voltage will decrease and thus you see ripple.As for what you use it for, some things don't care about the ripple and some do. In a classic audio amplifier, for example, the power output stages are driven directly by unregulated power because the power amp can reject the ripple, but other circuits in the same amplifier or receiver may need regulated ripple-free power.

author: macboy
author_rank: Super Contributor
timestamp: January 30, 2024, 06:57:01 pm
content: A 470uF capacitor is very small. Four isn't much better. For an unregulated supply, you need a very large capacitance. You can get away with less if using a regulator, as it will reject most of the ripple at its input. Another option is a capacitance multiplier, which differs to a regulator in that the output tracks to the (low peak of the) input voltage rather than providing a pre-determined output voltage. As attached

author: LoveLaika
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:17:49 pm
content: Thanks for the reply. I'll have a look at these filters and learn what I can about them. It's all a learning process for me, and I'm really grateful for the replies.

author: CaptDon
author_rank: Super Contributor
timestamp: January 30, 2024, 08:41:52 pm
content: How about Google up 'Linear power supply design' and read. It will cover every detail from transformer to output. "What do you do after rectifying" is to open ended a question and depends on what you want to do or the task at hand. I will surely do something different "after rectifying" if I am building a D.C. arc welder vs. an ultra low noise BioMedical E.C.G. module. Read up on power supply design, filtering, current and conduction in the diodes, pass transistor series regulators, Shunt regulators (not common due to power loss), three terminal regulators, positive and negative simultaneous supplies from center tapped transformers. Learning through simulations is kind of backwards. Normally you learn about the building blocks first, THEN start putting them together in a well thought out design, THEN test your design through simulation and if it looks reasonable go ahead and build it and hope the SIM and the real world agree. SIMs get a lot of stuff wrong!!!

author: LoveLaika
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:44:29 pm
content: Thanks. It seems that modern power supplies all feed the rectified output to a DC/DC converter. I can see the advantages of doing so. Looking at the inductors available that can handle 5-amps or more, they are monsters. I find these huge transformer-sized inductors, not small ferrite ones that you find in a laptop power supply. This makes me wonder how these power supplies are able to keep their components small (like 100 uF caps and such) while passing high current (like 1~2 amps or so). Sure, they may pass it through a regulator, but don't the components have to be able to handle the high current?

author: LoveLaika
author_rank: Frequent Contributor
timestamp: January 30, 2024, 08:50:02 pm
content: Thanks for the idea. I heard of the capacitance multiplier before, but I didn't know how to apply it. It's tricky. The transformer is rated for a large amount of current, so you would think that you would be able to pass that to the resistive load (or at least, what you need) without the output voltage dropping. But, as shown in my sim before, the circuit could only handle small current, like 50 mA or so. Any more than that, you see ripple. Using higher capacitances is all well and good, but why do some supplies manage to make do with smaller caps and still supply high current?

author: BrokenYugo
author_rank: Super Contributor
timestamp: January 30, 2024, 09:55:19 pm
content: A SMPS uses a much smaller transformer and caps than a linear supply because it runs at a much higher frequency, 20+kHz instead of 50-60Hz, some switch at over 1MHz.Think about the ripple waveform, the closer together the peaks are the less time the capacitor has to hold up the voltage, so it can be smaller for a given targeted ripple value.

author: Ian.M
author_rank: Super Contributor
timestamp: January 30, 2024, 11:21:49 pm
content: There are other considerations: Below the onset of saturation, a conventional (line frequency) transformer's output voltage is proportional to its input voltage. As you are in North America, the usual limits for line voltage are +/-5% on nom. 120V, though if you in a remote area or on private hydro (or a generator) you may see more variation than that. Your regulated power supply design needs to function from the lowest expected mains voltage to the highest, without browning out and letting excessive ripple through to the load as it starts to drop out of regulation at low line voltage, and at high line voltage, without blowing the regulator due to excessive input voltage at no load, or overheating it due to excessive dissipation at full load current.The extra 'headroom' required for regulation during low line conditions kicks you in the pants during high line conditions, especially at low output voltages, as regulators with low dropout voltage generally are quite limited with respect to max. input voltage, so you may well have to settle for a higher dropout HV capable regulator, a higher unregulated voltage and thus significantly greater losses and worse efficiency. The extra heat dissipation has knock-on implications for cooling, which drives a steep increase in cost, size and weight as the load current increases. If you need more than a few hundred mA of load current a linear supply is generally not cost effective unless low HF noise is an absolute requirement.