Cap and resistor size for full wave rectifier learning circuit - Page 1
Forum section: electronics / beginners

author: DougSpindler
author_rank: Super Contributor
timestamp: April 17, 2021, 02:02:41 am
content: I'm just learning how to read scope patterns by building half wave and full wave rectifier circuits. (See the attacked pic)I'm using a stepdown isolation transformer with an output of 24 vac. (So as not to blow the scope.)My question, what size cap should I use to reduce ripple? Or can I use a range of caps? Can you give me some values to try?Should there be resistor on the positive rail to protect the cap from inrush current? If so what size?Thanks

author: rstofer
author_rank: Super Contributor
timestamp: April 17, 2021, 02:14:02 am
content: It's a little more complicated than just picking a size. It has to do with the load current as well. Furthermore, there is also the complication that if the capacitor is too large, the diodes have a reduced conduction angle (because the capacitor voltage stays high, low ripple) and they have to pass all the power in a short period of time and coast the rest of the time. This leads to overheating.https://electronicbase.net/smoothing-capacitor-calculator/Google for 'sizing rectifier filter capacitor'

author: ledtester
author_rank: Super Contributor
timestamp: April 17, 2021, 03:03:17 am
content: QuoteThe purpose of a smoothing cap is to keep the voltage from the bridge above a certain level. For instance, the 7805 linear regulator needs an input of at least 7 volts to provide a regulated 5V output. A smoothing cap with enough capacitance will ensure that the input to the 7805 is always >= 7V. The required capacitance will depend on transformer's voltage and the current draw through the regulator. The calculator rstofer mentioned will tell you what the value needs to be.The 7805 itself will need small caps on its input and output to stabilize its operation. On the input side you'll see 1 - 10 uF and on the output usually a 0.1 uF cap. Depending on the regulator and application sometimes multiple small caps of varying values are put in parallel on the output to help with transient responses (i.e. short bursts of activity).QuoteI wouldn't worry about inrush current until your smoothing cap starts to get big - like in the tens of millifarads. And then, something that only temporarily introduces resistance is used to counter the inrush current such as:- a timed relay which switches in a resistor when the circuit is powering up but then removes it shortly afterwards- a thermistor which when cold has a high resistance but after current runs through it for a while heats up and then the resistance dropsThese inrush limiting measures usually last for only a short period of time - at most a couple seconds.

author: bdunham7
author_rank: Super Contributor
timestamp: April 17, 2021, 03:44:18 am
content: You don't need a resistor for inrush protection, but it might be instructive to put a 1 ohm resistor between the transformer and the bridge and then connect the scope across it to view the current pattern. Don't worry about the size of the capacitor, just try some with various loads. The 'appropriate size' will vary with the amount of load, as well as the type of the load and the load's sensitivity to ripple.

author: DougSpindler
author_rank: Super Contributor
timestamp: April 17, 2021, 04:58:13 am
content: Would 100 microfarad be about right for 60 hz, .5 a?

author: Ian.M
author_rank: Super Contributor
timestamp: April 17, 2021, 05:16:07 am
content: No. 100uF is an order of magnitude too low. As a rough back of the envelope calculation, consider that the capacitor has to supply the load current between the ripple peaks, which occur at double the line frequency, i.e. every 8.33ms.As a first approximation, 0.5A*8.33ms/100uF=41.7V drop, which is patently ridiculous as its comparable to the peak voltage so the ripple will obviously be grossly excessive. This is the same calculation as the tool at Rstofer's link does.Of course, this doesn't allow for the fact the discharge line is a tangent to the falling side of the peak, so starts later than the peak and ends when it intercepts the rising side of the next peak rather than at the time of the peak itself. Therefore the actual ripple will be smaller. Its a PITA to manually calculate, but is quite easy to solve graphically, if you want a better estimate of the capacitor value required for a particular pk-pk ripple. However that is rarely particularly useful due to the wide tolerance of most large electrolytic capacitors.I'd start with 1000uF if your application can tolerate 4V pk-pk ripple.

author: ledtester
author_rank: Super Contributor
timestamp: April 17, 2021, 05:17:45 am
content: Quote from: DougSpindler on April 17, 2021, 04:58:13 amIt all depends on how much of a voltage drop you can accept.Full bridge rectifier means a "charge time" of 8.3 ms.Plugging in 100 uF and 500 mA into the above mentioned calculator gives a ripple of 41.5 V.1000 uF will give 4.15 V ripple10,000 uF will give 0.415 V ripple

author: Jwillis
author_rank: Super Contributor
timestamp: April 17, 2021, 08:14:07 am
content: Calculators are fine but when your trying to learn it's much more informative to learn the formulas to get what you need.C = I / 2 x f x VppC is the value of capacitor required in Farads multiply this value by 1000000 to get micro FaradsI is the desired load currentf is the the AC frequency of mains supply (60Hz)2 is because the frequency is doubled in a full bridge rectifier (120Hz)Vpp is the exceptable peak to peak ripple when fully loadedfor example you would like .5 amps of fully loaded current and your except-able ripple is 500mA .C= .5A / 2 x 60Hz x .5Vpp = .008333 F = 8333uF (chose the next highest common value of 8400uF )Now theirs something you need to be aware of. Every time you need to half the ripple , the capacitance will need to DoubleFor .25Vpp ripple you will need 16666uF (17,000 µF) , .125Vpp will require 33333uF (34,000 µF) , so onAlso the more capacitance you add , the higher the inrush current . To much in rush current can damage transformers ,rectifiers and even in some cases trip breakers . It's a balancing act between how much current you require , how much ripple noise is exceptable and how much inrush current is exceptable .

author: mariush
author_rank: Super Contributor
timestamp: April 17, 2021, 10:06:36 am
content: Jwillis gave you the right formula to approximate the capacitor you need.Your transformer outputs 24v AC and has a certain VA rating (let's go with 50 VA since you don't mention it) ... with the example 50VA this means the AC current is 50VA / 24v = 2.08AYou can approximate how much DC current you're gonna have with the formula Idc = 0.62 x Iac ... in my example, that would be Idc = 0.62 x 2.08 = 1.29AThe 0.62 is a constant which works well with transformers in the 50..500 VA but it's just an approximation.Anyway, the bridge rectifier (the four diodes) converts your AC to DC, and you get a peak DC voltage that's equal to :Vdc peak : sqrt(2) x Vac - 2 x (voltage drop on diodes inside rectifier)Most bridge rectifier diodes have a voltage drop of around 0.8v...1.0v so let's go with that...So your 24v AC becomes a DC voltage with a peak of Vdc peak = 1.414 x 24 - 2 x 0.8 = 32.3vThe capacitor after the bridge rectifier makes sure the minimum voltage will be at least some value, at a particular current amount.So let's say that with my example transformer (Vac = 24, Iac = 2.08 -> Vdc peak = 32.3 , Idc = 1.29A) we want at least 24v DC at 1.25A of current.You can use that formula to approximate : Capacitance (Farads) = Current (A) / [ 2 x AC frequency x (Vdc peak - Vdc min desired) ]With our numbers , C = 1.25A / 2x60x(32.3-24) = 1.25/996 = 0.0012550 Farads or 1255uF , so you'd probably want to use at least a 1500uF capacitor to guarantee the minimum of 24v at 1.25A.At less than 1.25A of current draw, the minimum voltage will be closer to the peak DC voltage.You can use more and that will raise the minimum voltage like I said but note that if you use too much, the capacitor will draw a lot of current when the circuit is turned on the first time (capacitor sucks in energy like a black hole for a few ms), so you'd have to be careful with fuses (use time delay, maybe use a temperature sensitive resistor to limit inrush current like computer power supplies do)Also worth noting... with no load, all transformers but especially the low VA rating transformers output a higher voltage (10-15% extra is not unusual) but as there's some load applied the voltage sags down to the advertised voltage.So keep that in mind when measuring things.

author: David Hess
author_rank: Super Contributor
timestamp: April 17, 2021, 02:55:30 pm
content: I would start with the load resistance.What is the lowest value of 1/4 watt resistor that we can use with a 24 volt AC source? That ends up being 2304 ohms. Once the AC is rectified to DC and filtered, the voltage will be about 1.414 times higher, so then the 1/4 watt resistor needs to be 4624 ohms. Add some margin so the 1/4 watt resistor is dissipating less than 1/4 watt, and 10 kilohms is a good value.Now 10 kilohms at 34 volts yields 3.4 milliamps and that will be the discharge current from the capacitor between peaks in the AC cycle. The time between peaks is 8.3 milliseconds for full wave rectified 60 Hz power. From this we can calculate that a capacitance of 28 microfarads will allow a voltage drop of 1 volt between peaks at 34 volts DC. Lower values of capacitance will yields more voltage drop to observe.So use a load resistance from 5.6 to 10 kilohms, and a capacitance of 10 to 22 microfarads to get some results that you can observe.

author: floobydust
author_rank: Super Contributor
timestamp: April 17, 2021, 08:05:52 pm
content: +1 to learn about it, use a small valued capacitor.In practical circuits, I use a simple rule of thumb: 500uF-1,000uF per amp of DC load.The math behind sizing the capacitor is very complicated and fails because the power transformer's impedance is not included and it's non-linear. You don't get a pretty sine-wave at its output. Observe with a scope, it has flattened peaks and generally looks terrible nothing like textbook theory.Also, electrolytic capacitors have ripple-current ratings and will have a short lifetime if you operate near limits e.g. 100uF 50V are rated 220mA-600mA at 120Hz. For this reason, you would upsize and not use 100uF for 500mA output.Your 24VAC transformer will have higher output voltage at no load, depending on its size VA rating. So you can see 40VDC or more. If it's a Class 2 transformer like used in HVAC or some doorbells, it will have internal fuse that blows easily. so don't short it out by accident.

author: DougSpindler
author_rank: Super Contributor
timestamp: April 18, 2021, 04:13:18 am
content: Thank you everyone. Again I'm building circuits with learning, not function. For the load, I'm using holiday incandescent lights. I'm using 10 lights out of a string of 100. I'm in the US /120 volts so that's a 1.2 v drop per light. Nice cheap resistive load which works with AC and DC. I found I can drive the lights up to about 6-7 volts before they "blow".

author: TimNJ
author_rank: Super Contributor
timestamp: April 19, 2021, 05:39:36 pm
content: I think this would be a good application of SPICE, at least that's how I would do it. I don't find that the analytical approach is all that useful for my intuition. You can set up an sine wave source, maybe add some series resistance to show the effect of loading on transformer voltage regulation. Then you can try a bunch of different capacitance and load combinations to see how the peak-to-peak ripple changes.I like to run ".step param X" or ".step param X list" in LTPSICE. Give the value {X} to the component value you want to vary, and then give a few values in the step param X list line and you can see a few results plotted at once.https://www.analog.com/en/technical-articles/ltspice-using-the-step-command-to-perform-repeated-analysis.html

author: Jwillis
author_rank: Super Contributor
timestamp: April 20, 2021, 02:13:57 am
content: Quote from: TimNJ on April 19, 2021, 05:39:36 pmInteresting . I've never used a Gauss meter and not sure if they would work to measure the field of a small transformer . Or if there is a mathematical prediction of how a transformer field collapses during load and if that could be incorporated into the simulator if it doesn't already . All I have been able to see is how the sine wave drops when loading the transformer .There is a online calculator that shows how inrush current and ripple change by altering variablesLINEAR POWER SUPPLY DESIGN