Regulated Low Voltage Power Supply Problem - Page 1
Forum section: electronics / beginners

author: Iseeker
author_rank: Newbie
timestamp: July 27, 2015, 02:48:23 am
content: Hi Good day everyone.I have a problem on my RLV-Power Supply.Im not sure if where is the problem and what is the problem of this schematic diagram,.The problem is when I used at least 4 of the outputs, the voltage rating of each will go down and going back to its rating.Example: 8v output, when i read it on the VOM the pointer points to 8v then it will move to almost zero volts then back to 8v, and then back to almost zero volts then back to 8v. (Assuming that 4 of its output is being used) =x It's voltage behavior is fluctuating.I hope I make my problem clear to you guys..I suspect the value of the capacitance, i think its not enough.. or the Transformer??PS: the original schematic of that has only 2 outputs, and I edited the schematic diagram to make it 6 outputs and change the value of the capacitance into higher in the smoothing area.

author: mariush
author_rank: Super Contributor
timestamp: July 27, 2015, 04:06:59 am
content: You have several problems here.Your transformer is 14VAC 750mA or 0.75A.When you rectify this using the bridge rectifier, you get a dc voltage with a peak equal to 14x1.414 = ~ 20v , and then you need to substract 2 x Voltage Drop on diode inside rectifier, so 20v - 1.5v = ~ 18.5vThe maximum current after the rectifier will be about 0.62 x Iac = 0.62 x 0.75 = ~ 0.45 AYour schematic shows c1 to c4 equal to 4400uF 16v ... if you have up to 18.5v after the bridge rectifier, these four capacitors will be damaged because 18-19v is more than 16v.In addition, the amount of capacitance is ridiculous and unnecessary, it just stresses the transformer when you plug your power supply into the mains socket (unless it's a time delay fuse - and even then - too much capacitance may blow it).The formula to approximate the minimum capacitance required is Capacitance (Farads) = Max Current / [ 2 x AC Frequency x (Peak DC voltage - Minimum Voltage you want) ]So for your peak of about 0.45A the transformer can give you and a minimum voltage of 14v (because otherwise 7812 won't work right) then you have C = 0.45 / 2x50x(18-14) = 0.45 / 400 = 0.001125 Farads or 1125 uF, so a single 1500uF would be enough for your design.People in general design their linear power supplies around the value of 2200-3300uF for each 1A, it's a compromise between the amount of copper in the transformer and the capacitanceIf I were to make the design below, I'd also add a small electrolytic capacitor after each linear regulator, something in the range of 10-100uF 25-35v would be perfectly fine.Other than this, I have no clue why your meter behaves this way, I'd suggest checking if your meter is on AC or DC mode, and check for bad connections and so on.

author: codeboy2k
author_rank: Super Contributor
timestamp: July 27, 2015, 06:28:05 am
content: It would seem that you duplicated the 5 V and 6V designs, and just added 8, 9, 10, 12 volt outputs.You'll notice that the resistor for the 5V and 6V outputs is 1K ohm. This is because the 78xx series regulators need a minimum 5 mA load current. So the 5V and 6V outputs have a 1K ohm resistor which gives 5 mA and 6 mA minimal load. But you duplicated this all the way up to 12V, which is not necessary, because then you have 12 mA load at 12V. You just want to have the 5 mA minimum load at each voltage, so for 8V, the load resistor can be 1500 ohm (8/1500 = 5.3mA) ,for 9 V you can use 1800 (5 mA), for 10V you can use 2k ohm (5mA) and for 12V you can use 2.2k ohm (5.5 mA)So that meansR31.5k ohmsR41.8k ohmsR52k ohmsR62.2k ohmsQuote from: mariush on July 27, 2015, 04:06:59 amThis is a good idea, but you will also need to add a diode from output to input, because adding output capacitance risks having the output stay at a higher voltage than the input voltage when you power off (especially if you also lower C1, C2, C3, C4 to a single 1500 uF capacitor as suggested).Suppose you have high loads on each output except the 8V output, which is unloaded. Then the 10-100uF cap on the output of the 8V can hold-up the output voltage when the input voltage falls off(due to the other loads, it will fall off fast). This will guarantee damage to the 78xx voltage regulator. So to counter this, each regulator should have a protection diode from output to input. This allows the higher voltage at the output to bleed back and bypass the regulator's internals.EDIT: I'd probably keep the output capacitance at 0.1uF as you already have, and just put the 10uF at the input. Keep the bulk 1500uF capacitance at the supply end near the bridge rectifier. I've drawn this below for a few voltages. Also, I put an RC snubber across each diode of the bridge, this is a good addition if you plan on using your PSU for any audio work. It removes the switching transients on the diodes. If you don't care about audio work, then you can leave the snubbers off.

author: Iseeker
author_rank: Newbie
timestamp: July 27, 2015, 12:05:24 pm
content: Quote from: mariush on July 27, 2015, 04:06:59 amThank you so much for that Mr. Mariush.. I'll noted all your suggestions.I have a question, where did you get the 0.62?

author: Iseeker
author_rank: Newbie
timestamp: July 27, 2015, 12:07:29 pm
content: Quote from: codeboy2k on July 27, 2015, 06:28:05 amYeah, I just edited it by myself. Thanks for the information Mr. Codeboy2kIm going to use your suggested schematic diagram. Thanks a lot!

author: Kleinstein
author_rank: Super Contributor
timestamp: July 27, 2015, 03:30:49 pm
content: The factor from AC (RMS) current to DC current after rectifier and caps is some rule of thumb or from approximate tables. The smaller the transformer and caps the more current you can expect. Usually the factor starts at something like 0.75 for very small transformers and goes down to about 0.5 for large (e.g. 100 VA) ones with large caps. The factor comes from the current not being in the form of pulses and thus has a large RMS value compared to the average.With data on the transformer one can use something like a simulation to get an more accurate number.

author: mij59
author_rank: Frequent Contributor
timestamp: July 27, 2015, 04:30:52 pm
content: In the schematic there is connection between the "earth" symbol and the GND missing.

author: jeroen79
author_rank: Frequent Contributor
timestamp: July 27, 2015, 05:38:21 pm
content: Why are you using a bunch of 78xx regulators to get the different voltages?With one LM317 and a potentiometer you can get all these voltages as well as any in between.

author: Iseeker
author_rank: Newbie
timestamp: July 28, 2015, 12:06:12 am
content: Quote from: jeroen79 on July 27, 2015, 05:38:21 pmI'd like to use all of them for a single time, or at least 4 of them.

author: codeboy2k
author_rank: Super Contributor
timestamp: July 28, 2015, 03:43:54 am
content: Quote from: mij59 on July 27, 2015, 04:30:52 pmI presume you mean my schematic example. In my opinion, for a benchtop PSU, it's better to float the PSU GND, and bring it to the front panel on the traditional black jack, and then also bring the Protective Earth Ground (P.E. GND) to the front panel on the green jack. This way, the user can strap P.E. GND to PSU GND as needed, and when needed. Otherwise it remains floating and will be more useful that way, because it can be placed in series with other PSU's for creating a higher output voltage.Also, if this PSU is put into a metal project case, then the metal case should be P.E. grounded for safety.

author: mij59
author_rank: Frequent Contributor
timestamp: July 28, 2015, 05:54:52 am
content: Quote from: codeboy2k on July 28, 2015, 03:43:54 amSorry, no I am referring to the original schematic, looks like the ground pins of the 8, 9, 10, and 12V regulators are floating.