Secondary current rating of a center tap transformer? - Page 1
Forum section: electronics / beginners

author: techguru
author_rank: Regular Contributor
timestamp: November 05, 2018, 03:44:43 pm
content: Hi to all,I have center tap transformer of secondary rating 24V-0-24V @ 5 ampere.I use bridge rectifier for constructing power supply which will power opamp circuits. what is the current i could draw form each line?my thought is my peak voltage after filtering is 33.94V(24 square root of 2). so i could take 33.94V and 5 ampere, so i could total DC power is nearly 170 W with out regulation circuit. My conclusion is that i could take 170W from positive voltage line and another 170W from negative line with respect to ground.correct me if i am wrong???

author: iMo
author_rank: Super Contributor
timestamp: November 05, 2018, 04:20:06 pm
content: My understanding is you have the bridge wired to 24-24 leads and the 0 lead is the ground.I think your peak voltage will be 68V (measured on the capacitor wired to the Greatz' output with no load), subtract few volts lost on the diodes, based on the filtering capacitor you get a ripple (ie with a 10mF capacitor and 5A loads it will be 10Vpp ripple around 53V mid), the peak diode's current will be about 20A, and you can draw 5A from each line.

author: mariush
author_rank: Super Contributor
timestamp: November 05, 2018, 04:23:48 pm
content: It's 48v at 5A = 240VAIf you convert to DC using a regular bridge rectifier, then the current will be ~ 0.62 x 240 = 150W but naturally, the peak DC voltage would be higher.

author: techguru
author_rank: Regular Contributor
timestamp: November 05, 2018, 04:55:59 pm
content: I want the power that could be taken form +ve line and -ve line with respect to ground?. my transformer is 24V-0-24V @ 5 ampere..

author: iMo
author_rank: Super Contributor
timestamp: November 05, 2018, 05:08:00 pm
content: Yes, as I wrote above.. You may use 1 capacitor or 2 capacitors..You will get aprox 68/2 -> +34V and -34V with no load, and aprox +24 and -24V with 5A load. Mind the ripple on each rail - it depends on the capacitor's value, with (for example) 10000uF it will be about 5Vpp @5A on + and - rail.

author: David Hess
author_rank: Super Contributor
timestamp: November 06, 2018, 04:14:53 am
content: Mariush has it right. The transformer's power rating is limited by heating when the power factor is less than 1 which is the case with simple rectification into a capacitive load. So in practice only about 60% of the rated power is available.In theory power factor correction can be used to get the full rated power but it is usually less expensive to just get a larger transformer.

author: 001
author_rank: Super Contributor
timestamp: November 06, 2018, 03:29:52 pm
content: Quote from: mariush on November 05, 2018, 04:23:48 pmWhat is the book is? Thanx

author: mariush
author_rank: Super Contributor
timestamp: November 06, 2018, 04:00:42 pm
content: It's not a book, it's a page extracted from Hammond Product Catalogue (see page 11) :https://www.hammfg.com/files/literature/5c.pdf

author: ArthurDent
author_rank: Super Contributor
timestamp: November 06, 2018, 04:01:03 pm
content: The chart is a single sheet PDF called "Design Guide for Rectifier Use" by Hammond Mfg., a transformer company.http://www.hammondmfg.com/pdf/5c007.pdf

author: not1xor1
author_rank: Frequent Contributor
timestamp: November 07, 2018, 07:24:48 am
content: Quote from: techguru on November 05, 2018, 03:44:43 pmas other people have already stated you can get no more than about 3A DC from the dual railsif you are adding voltage-current regulators the maximum output voltage would be around 22-26V per rail according to the quality of the transformer (24-24V at full load?) and the regulator design (drop-out, current sense resistor drop, etc)

author: techguru
author_rank: Regular Contributor
timestamp: November 09, 2018, 05:51:26 pm
content: I have to design a regulated power supply with following specification;DC output voltage: 0 to 30 V adjustable.DC current:0 to 3A adjustable.I am a second year under graduate student, I cant understand difference between variable current function, current limiting function and fold back current limiting?Kindly Help me in designing the circuit and Explain the above question.

author: drussell
author_rank: Super Contributor
timestamp: November 09, 2018, 06:04:46 pm
content: Quote from: not1xor1 on November 07, 2018, 07:24:48 amWell you certainly can intermittently. It is all about temperature rise. If you can tolerate extra temperature rise and the transformer is beefy enough, you can intermittently overload it far beyond that orwithappropriate cooling (like forced air) can run over the nameplate rating of most transformers even continuously.Good companies like Hammond and Tirad talk about this in their specifications and make quality, conservatively rated transformers. You need to evaluate the actual unit you have to be sure what you can get away with. If in doubt, monitor the temperature rise over the course ofseveral hoursof continuous loading at your desired level and watch what it does. A transformer rated for intermittent duty won't be able to blast out full power continuously while something designed for continuous maximum-blast will hold up well in most uses.Rectifier topology obviously plays a role in this as it will dictate "how" the power is delivered from the transformer, what part of the cycle is actually conducting, etc.Do you know where the transformer came from originally, what it's intended purpose was?Additionally, what isyourintended use?

author: mvs
author_rank: Frequent Contributor
timestamp: November 09, 2018, 07:22:57 pm
content: Quote from: techguru on November 09, 2018, 05:51:26 pmI would assume that current limiting function is fixed. It is used to protect the power supply from getting damaged through excessive current draw.Variable current limiting can be additionaly adjusted to some lower value, so that power supply would act as a programmable current source.Foldback current limiting acts as a fixed or variable current limiting with additional reduction of current and voltage to some very low value once it get triggered. It was used quite often in old linear power supplies to reduce power disipation and risk of fire during output short circuit condition.

author: Ian.M
author_rank: Super Contributor
timestamp: November 09, 2018, 07:45:28 pm
content: In response to drussell's comments:OTOH if you simulate the rectifier circuit, you'll find that Hammond's derating factors for rectifier use are the ratio of the avg. DC current out of the rectifier to the RMS AC current into it, so are generically applicable to any competently designed line frequency transformer, even cheap ones.Therefore, if your transformer is rated for continuous operation at a particular max. RMS secondary current at a specified max. ambient temperature, Hammond's derating factors *are* applicable to calculate the max continuous DC load current at that max. ambient temperature.

author: not1xor1
author_rank: Frequent Contributor
timestamp: November 10, 2018, 08:06:10 am
content: Quote from: techguru on November 09, 2018, 05:51:26 pmYou can get 5A out of your transformer if you limit the output to half voltage. You can use a switch (acting on a relay or power mosfet, scr, etc) to change between 0-12V 5A | 0-24V 3A as various bench PSU manufacturers do.If you want to get 30V out of your transformer, then you have to use a large capacitor to keep the ripple at minimum and limit the maximum output current to 1.5-2A to reduce the power dissipated by the transformer.Regarding the various kind of current protection/limit/regulation, you can have (even more of this in the same circuit):-1) overcurrent protectiona comparator wich switches-off the PSU output when the output current gets above a set limit (e.g. for battery protection or electric motor PSUs).-2) current limityou can use a sense resistor and the Vbe threashold of a BJT to "steal" current from the base of the pass BJT when the voltage drop through the sense resistor gets above the trhreshold. The regulation is quite coarse as the current increases when the output voltage gets down. This is a problem as a pass BJT has to dissipate a lot of power in case of shortcircuit.-3) foldback current limita foldback circuit uses a trick, adding a 2 resistors voltage divider to the base of the current sense BJT of the previously described circuit, in order to reduce the current as soon as the output voltage decreases.-4) proper current regulationthat is done by using an operational amplifier (or a differential amplifier made by discrete BJTs) and allows the PSU to work within the limits of a set voltage and current, automatically switching between constant voltage and constant current mode according to the load.For more details I suggest you to carefully read the Agilent "DC power supply handbook"http://literature.cdn.keysight.com/litweb/pdf/5952-4020.pdf

author: techguru
author_rank: Regular Contributor
timestamp: November 10, 2018, 04:11:50 pm
content: Thank you all. Please suggest regulated power supply circuit which has both current and voltage control(0-3A and 0-24V). thank you..

author: techguru
author_rank: Regular Contributor
timestamp: November 11, 2018, 07:37:56 am
content: Good After noon to all,Please suggest voltage and curreng control IC for regulated power supply circuit which has both current and voltage control pins (0-3A and 0-24V). thank you all.

author: mariush
author_rank: Super Contributor
timestamp: November 11, 2018, 08:17:02 pm
content: What's the point, if you're not going to understand the circuit anyway, based on the questions... you want someone to give you the solution to the problem directly, no effort on your partA simple start would be a simple google search using terms like "adjustable power supply schematic" and maybe replace adjustable with linear (as opposed to switching)...I'm attaching a bunch of schematics which use simple opamps/comparators/whatever and transistors allowing you to control voltage and current, no single chips ...most can be adapter for higher voltages or whatever (for ex. the 18v one can work with 24v), and the hp one has a lot of useful info.

author: techguru
author_rank: Regular Contributor
timestamp: November 13, 2018, 03:46:28 pm
content: Mariush sir,It will be useful if you give some video which explain current control. Voltage control is possible with normal LM317 Regulator but when we go for current control there is less explanation and literature to explain .thank you sir for schematics but you are correct with out understanding the schematic , it will not be usefull for construction.

author: mariush
author_rank: Super Contributor
timestamp: November 13, 2018, 04:08:47 pm
content: Current control in a very simple explanation : you monitor the current flowing through the output wire and you measure it. If the amount is too high, you restrict the voltage overriding if needed the set voltage.In all the schematics I uploaded, you'll see a resistor somewhere on the output wires, with a very small value, like 0.01 ohm for example (R57 0.1 ohm in the 18v schematic, R38 in the picture schematic) .... that's the "current shunt" ... when current goes through it, there's a voltage drop across the resistor (because formula v = i x r ... so you have a fixed r , and depending on current (i) the voltage changes)This tiny voltage is amplified using one opamp to some level and then its compared with another voltage (for example in the picture attached, see the second opamp where you have the course and (optionally) fine potentiometers for current) and if the voltage is higher than what's set using potentiometers, a signal is sent to reduce the voltage output .... and it keeps going in a loop reading the current comparing to current set until current output is equal to current set

author: techguru
author_rank: Regular Contributor
timestamp: November 14, 2018, 05:11:16 pm
content: hi to all ,Just now i saw LM723 regulator for my design. It is good as variable voltage regulator but how about variable current regulator? Suggest some circuits.with regards.