bench power supply

Question

hello, I started drawing in eagle this afternoon to make my first home made pbc board for a bench power supply that with an input of 12V/5A will have two output of 12V, 5V, 3.3V and an adjustable voltage (lm317) the switches are these with ground in the middle I hope that the schematic is clear (it is also in the attachments), maybe you could give me some hint or advice about something that i m doing wrong about the .brd i didn't route it since maybe you will say me to modify my schematic thanks in advance 
 Attachments: bps.rar

geralds · Accepted Answer

hi, please look exactly. Gerald ---

COMPACT · Answer

Here is an example of what I mean. 8V in from a regulated power supply being output as 5VDC using an LM7805 Linear Regulator. (The current limit has been set to 50mA on the bench power supply as not to risk melting the breadboard or voltage regulator without a heatsink.) The most important point here is that the circuit works! The test apparatus on the left is a TENMA bench power supply; it is a beauty of a unit and can be remotely controlled by a computer. The test apparatus on the right is a DC Electronic Load; it provides a test load for the regulator output whilst acting voltmeter and ammeter. If you look at the numbers on the apparatus; you'll notice that the input power used is greater than the power being output; the missing power is expelled as heat! "Wherever I lay my cap, that's my home". - Paul Young "You can keep your cap on" - Joe Cocker

mcb1 · Answer

the rated current it is 0.5 at 50v so I could reach 2A at 12v It doesn't work that way. The current specifies how much the switch will pass without damaging the contacts. The voltage specifies how high a voltage you can apply before you have problems with insulation, arcing over, etc. At 12v you still have 0.5A. The voltage/current can be different between AC and DC but it's usually not much. You could change your design. If you switched the ADJ pin with different values, you'll have the regulator doing all the work. Mark

mcb1 · Answer

Good questions. Often the potentiometer/adjustable resistor is shown as above, when the correct drawing should be as per your link. (shown below) Diode D1 is there to protect the regulator from a voltage on the output flowing back through the regulator when the input voltage is removed. D2 would offer protection if a reverse voltage source was connected at the output. My home built supply has a 'LOAD' switch added between the output and the terminals, to allow the voltage to be removed instantly, or set before powering up the connected device. The only issue you'll find is that you are limited to 1.5 A from the regulator. I'd have a look at jw0752 interesting foray into power supplies. Comparing Power Supplies (Links to both articles) Mark

geralds · Answer

Hi, please check your circuit! Best Regards, Gerald ---

jc2048 · Answer

Not sure that the way you have the switch is right. My interpretation of the data would be like this (but I could be wrong):

Do you have a meter you can use to see which terminals are joined for the different switch positions?

The switch is only rated to switch 0.3A.

Calculate the power dissipated by each of the regulators before you do the board (you might find they need a heatsinks).

Connect R2 as shown in a 317 datasheet. The way you've got it, though it will work you won't be able to adjust all the way down to the reference voltage.

mcb1 · Answer

Glad we can help. The datasheet gives some explanation and an example of increasing the current. www.ti.com/lit/ds/symlink/lm317.pdf https://www.google.co.nz/ " My home built supply has a 'LOAD' switch added between the output and the terminals" Could you explain this? I m sorry i didn t I have a switch that interrupts the voltage going to the output terminal.

aster94 · Answer

I will use the suggestion of Mark for the size and position of the caps, but i completely agree with compact that testing would be the best solution I wanted to say that i m not doing it not because I am lazy and i don t want to build and test the circuit on a breadboard but because with my 3 dollars voltmeter i don t think that i will feel any difference between a smooth or a disturbed line

COMPACT · Answer

You'll find that having both +5V and +3V3 simultaneously handy to have. If you're going to have low impedance regulator input capacitor like C2 you should have one for each regulator. Apparently, the shorter the distance from the voltage regulator the better. Lose the slide switch and use an appropriately rated toggle switch.

geralds · Answer

Hi, Attached you'll find the schematic. You can use freely all what you needs; its my pleasure. Then with this you can build your pcb. There is also a heatsink on there you can mount the V-regulators. But use for all an isolation pad between the device and heatsink. Also isolate all screws. After mounting check it with an Ohm (wire finder) -Meter !!!!!!! The isolated DC/DC converter is a pcb mounted type like NME1209, or similar. An ISOLATED DC/DC converter is necessary. At the output of the DC/DC you have a resistor with 270 Ohm. That can be a higher value <1k (9mA, 9V); >270 (30mA, 9V) The thinking about is that the DC/DC needs a load resistance for regulation the output voltage to the nominal value (9V). With that R you makes the minimum burden current for this converter; see in the datasheet which minimum current it needs and then calculate the resistor value. Yes, IC1 is a LM317T. The other regulators are LDOs with attribute S - 78S12.... that have a better dropout voltage. At the input you find a filter - that filters out spikes from the power supply that you use for this regulators. -> You have to see that you use up to > 3A if all regulators are working. -> the filter with min. 6A would be ok. The max. input voltage will be 15V because the DC/DC converter can not more stressed. Here you have a zener diode ZD15V with 1 Watt. This is for security if the input voltage increase over 15V. But stay always under 1 Watt power lost (over voltage and current through the zener). I've build the schematic with Eagle V7.7.0. I wish you the best for your project. Best Regards, Gerald --- 
 Attachments: powertripple1.zip

COMPACT · Answer

I'd definitely put polarity protection at the input. Your original design also makes the assumption that you are feeding the board with 12VDC regulated. Is this the case? With a linear regulator you can't output 12V from an input of 12V. A linear regulator works on the premise that excess power is dissipated as heat but can't absorb heat to boost voltage. I suggest that you start with a prototype first and test it out to see if it meets your needs. 1. with input protection. 2. with a single output. Once this is working then add your bells and whistles. And once this has been done then layout your PCB with sufficiently sized copper traces. Make sure that the regulators are sufficiently heatsunk to dissipate the unwanted power keeping the regulator temperature within operating bounds.

mcb1 · Answer

As COMPACT said the regulators need some voltage above their output. A good rule of thumb is 2v. Rather than go ot the 24v supply, many of the power bricks for laptops are 15v and have more than 2A capability. That assumes you want to have the adjustable voltage got from zero to 12v. Also a minor difference but if you can the voltmeter input should be on the Output of the ammeter. Most ammeters use a small resistance in series and measure the voltage across it to derive the current figure, some are negliglble but others are higher. You will need to be very disciplined when using the fixed output switch. You really don't want to have 5v (or worse) into a 3v3 device Otherwise it looks like it will work. Mark

mcb1 · Answer

Maybe just a 0.1uF ceramic in the input and one identical at both outputs will be ok The 0.1uF is a must, and it should be as close to the output terminals as possible. You can always add another one across the terminals when you build it, and place the other closer to the regulator. The 0.1 removes high frequency ripple and spikes, while the larger capacitors tend to smooth out the voltage fluctuations. You need both to do proper filtering. A good rule of thumb is 1,000uF per amp as the absolute minimum, so if we assume that C2, 4, 6, 8 electrolytics then 2,200 - 4,700 uF is a good size and you can stick with 25v versions to cover any location. Much bigger and you have to leave a little more room (which you can). Over the years the sizing has gotten smaller despite the capacity and voltage rising, so you're winning already. Mark

COMPACT · Answer

Actually all of the capacitors are not mandatory. If you're feeding you circuit with a regulated DC power supply it'll already have its own filtering. A low impedance (ceramic or otherwise) capacitor at the input leads may only be required for situations where longer leads are used. The only way to determine whether capacitors are required is from testing.

geralds · Answer

Hi, please check the detail with the amp-meter - its something wrong as I seen in the blockdiagram of the meter. Best Regards, Gerald ---

COMPACT · Answer

The easiest is just to get a 3 pin Linear Regulator and a heatsink as a starting point. In this day and age almost all Wall warts and power supplies are already regulated so it is likely that the necessary capacitors are already present. The learner will then understand that larger voltage In can be output as an lower voltage with the unwanted power dissipated as heat. This was the technology of the of the late 20th century until recently. If a learner wants to know more then they can attach an oscilloscope and observe the waveforms of their power supply and see how the addition of capacitors may or may not affect it. 21st century Switch Mode Power Supplies (SMPS) are much more efficient at the cost of electrical noise, complexity and material usage. Here the usage of capacitors is critical to obtain the desired performance. To highly excite learners one might deliberately have polarised capacitors attached back to front or use capacitors beyond their rated working voltage to see what happens. If the learner just wants a power supply just get them to use a couple of modern wall warts. USB adapters provide at least 1A @ 5VDC and 12VDC adapters are readily available. To get the 3V3 just use a DC:DC Converter Module.

COMPACT · Answer

NEVER BREADBOARD a power supply design (Unless you have the means to limit the power.) The resistance and power handling characteristics of a breadboard are not necessary suitable Always ensure that you used appropriately sized conductors and insulation.

mcb1 · Answer

In this day and age an Oscilloscope is a mandatory piece of electronics hobbyist equipment You're right, BUT they aren't always cheap unless you aim for an older 2nd hand one with limited bandwidth ... Many newbies struggle to see the value in purchasing good meters, and it takes either good advice or experience to see why they should apy for better quality. I think jw0752 power supply blogs are a good example of an easy to implement supply. The results also show they work well. deliberately have polarised capacitors attached back to front or use capacitors beyond their rated working voltage That used to be a demonstration when we were venturing into electronics. With the newer case nowdays, the result is less than spectacular. Mark

geralds · Answer

Hi, My pleasure. Please wait a bit for etching, because the circuit diagram has some fundamental errors. For example: ) The plug can not be used for 5A, at least only special types of plug. ) There is a better solution for polarity reversal protection, eg with a P-channel MOSFET. ) A filter (C-L-C at the input) should be inserted, especially bypass capacitors should be installed in the right places. ) The supply for the panel meter (V / A) I would solve differently (what kind of panel meter have you?). ) The switch at the output I would make with a rotary switch and relay, because the relay contacts can handle the current. ) The sliding switch described above is more suitable for signal currents. ) Or, even better, direct all outputs directly to the terminals, so you can use all outputs at the same time. ) The part with the LM317 should be modified. There was some hints for checking the schematic. Best Regards, Gerald ---

COMPACT · Answer

Vincenzo, You seem to be focused on ancillary considerations and not the primary considerations as I've tried to lead you. Don't concern yourself about using power plane fills but get the essentials in order first. Here are some observations. 1. PCB traces are too thin; they need to be thick enough to handle the envisaged current and be thick enough firstly to avoid acting as an unwanted fuse and handling the envisaged load. 2.Your low impedance, small capacitance (100n) capacitors for the voltage regulator outputs are far too far away from the voltage regulators; ideally they should be adjacent. Have a look at Steve Ciarcia's Circuit Cellar How to build your own Z80 computer book which is available online which has a great section on power supplies for beginners. https://books.google.com.au/books?id=mVQnFgWzX0AC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

geralds · Answer

Hi, fine, congratulations! A little bit: 1) place the output connectors and poti also the V-regulators a bit to inside of the board, then you have a closed GND plane with this polygon. Then you don't need wire jumpers. 2) turn the input plug backwards to the IC side then you can make a small enclosure around the PCB that have on the front- and backside the mounting parts but not on the l/r sides. The 100n bypass capacitors have mounting so close to the ICs as you can. They are ceramic capacitors, withstanding > 50V. Best Regards, Gerald ---