Output voltage  - Page 1
Forum section: electronics / beginners

author: hpibmx
author_rank: Regular Contributor
timestamp: August 02, 2022, 02:58:51 pm
content: Hi all,I have made a adjustable power supply, I don’t have the schematic right now,Everything is ok but why do get loweroutput voltage when load is connected?Basically 220v AC goes to my transformer and a bridge rectifier then to this XL4016 step down module then adjusting for output.If schematic is needed I can draw.Thank you.

author: Ian.M
author_rank: Super Contributor
timestamp: August 02, 2022, 03:28:26 pm
content: Draw the schematic! Showing us it prevents much confusion and time wasted playing twenty questions figuring out your circuit.It also helps to give details of your load.What's happening to the unregulated DC (immediately after the bridge rectifier) when you apply a load?If you set a much lower output voltage, lets say20V2.5V, do you have a similar voltage drop problem?If you have a scope, post screenshots of that point with and without a load. If not, measure the voltage there with a DMM on DC V, with and without the load, and to get some idea of the ripple, measure the ripple with the DMM on AC V and a non-electrolytic capacitor of >=1uF in series with the positive test lead. Please also note whether or not your meter is 'True RMS'.Edit: I didn't notice the d.p. in the photos voltage displays, so have revised my suggested 'much lower' output voltage down to 2.5V

author: Terry Bites
author_rank: Super Contributor
timestamp: August 02, 2022, 03:46:34 pm
content: Draw the schematic!

author: radiolistener
author_rank: Super Contributor
timestamp: August 02, 2022, 04:24:24 pm
content: Quote from: hpibmx on August 02, 2022, 02:58:51 pmtoo high internal PSU impedance. It leads to Voltage drop down for a high current.If your PSU without Voltage regulator still has the same Voltage drop down, then it looks like too thin wire on the secondary coil of the transformer.If transformer is ok, then your Voltage regulator cannot handle so high current.Just try to attach the same lamp to the transformer output, before Voltage regulator and measure voltage, if there is no such Voltage drop down, then this is definitely Voltage regulator issue.

author: andy3055
author_rank: Super Contributor
timestamp: August 02, 2022, 07:50:05 pm
content: You are using a tail lamp bulb from a car that has 2 filaments. 1 could be 5 watts and the other could be up to 21 watts for the brake light. Calculate the current draw based on your output voltage and see if your supply can deliver the current for whichever filament you are using.

author: hpibmx
author_rank: Regular Contributor
timestamp: August 03, 2022, 04:05:23 pm
content: Quote from: Ian.M on August 02, 2022, 03:28:26 pmSorry for bad drawing but that’s what I have done.My main use is 3~5v and up to 3A.I have measured with DMM on dc I get roughly 20v dc at bridge rectifier (not sure about the ripple)

author: Ian.M
author_rank: Super Contributor
timestamp: August 03, 2022, 04:14:04 pm
content: So how much is that unregulated 20V dropping when you load the output with a heavy load?

author: hpibmx
author_rank: Regular Contributor
timestamp: August 03, 2022, 04:19:49 pm
content: Quote from: radiolistener link=topic=335260.msg4335247#[/quoteQuoteHow can check on this? Using dmm on resistor mode? What’s the ideal reading?QuoteShould I check before or after bridge rectifier?Quote

author: hpibmx
author_rank: Regular Contributor
timestamp: August 03, 2022, 04:23:13 pm
content: My badI need to check on this again and update.

author: Roehrenonkel
author_rank: Frequent Contributor
timestamp: August 03, 2022, 04:33:12 pm
content: Hi,Quote from: hpibmx on August 03, 2022, 04:05:23 pmadd a smoothing-cap after the bridge.

author: james_s
author_rank: Super Contributor
timestamp: August 03, 2022, 06:47:30 pm
content: Quote from: Roehrenonkel on August 03, 2022, 04:33:12 pmThat is I think the main issue here. There is no way that regulator module can be expected to work properly without a fairly large smoothing cap after the rectifier. This is a hybrid power supply so the input stage is going to be much like that of a linear power supply. Off the top of my head probably several thousand uF.

author: radiolistener
author_rank: Super Contributor
timestamp: August 03, 2022, 08:12:34 pm
content: Quote from: hpibmx on August 03, 2022, 04:05:23 pmTry to measure DC Voltage at bridge rectifier output without lamp and then attach two 12V lamps connected in series (they should work at 24 V) to bridge rectifier output and measure Voltage with a lamp load. How much Voltage drop is?

author: Terry Bites
author_rank: Super Contributor
timestamp: August 04, 2022, 01:41:16 pm
content: How did you make it without a schematic?

author: sleemanj
author_rank: Super Contributor
timestamp: August 04, 2022, 02:30:15 pm
content: What is the shunt resistance of your ammeter, you are measuring voltage after it. Measure voltage directly at the output of your dc-dc converter before the ammeter, no sense chasing geese looking for problems in your converter or transformer if you are losing the volts after it.

author: james_s
author_rank: Super Contributor
timestamp: August 04, 2022, 04:37:26 pm
content: Quote from: Terry Bites on August 04, 2022, 01:41:16 pmThat's easy, I've built loads of stuff without bothering to draw up a schematic. A power supply using premade modules is pretty trivial to wire up on the fly.

author: mariush
author_rank: Super Contributor
timestamp: August 04, 2022, 04:49:29 pm
content: A bridge rectifier converts AC voltage to DC voltage where the DC voltage peaks 2 x AC frequency (ex 120 times a second for 60 Hz AC voltage).The peak DC voltage will be equal to approx Vdc peak = sqrt(2) x Vac - 2 x (voltage drop on diode of bridge rectifier)So for example a transformer that outputs 24v AC, you will have a peak of Vdc peak = 1.414 x 24 - 2 x 0.8v = ~32vWithout a smoothing capacitor after the bridge rectifier, 100-120 times a second the voltage will drop down to 0 and for a short period it will be below the minimum input voltage of your XL4016 step-down module, so for that short period of time, the module may turn off, if it doesn't have an input capacitor inside the module providing some bit of energy.To make sure the output voltage after the rectifier is always above some threshold, you need to add a capacitor there that can charge with energy and then give that energy when the converted AC output goes down.You can estimate the capacitance required with the formula : C = current / [ 2 x AC frequency x (Vdc peak - Vdc minimum desired) ]So for example, if we continue with the example above where the Vdc peak was 32v and let's say I want to always have at least 24v DC at up to 1A of current, then the capacitor should be:C = 1A / [2 x 60 x (32 - 24) ] = 1/(120x8) = 1/960 = 0.001041 Farads or 1041 uF ... so I'd probably use a 1200uF 50v capacitor.

author: mariush
author_rank: Super Contributor
timestamp: August 04, 2022, 04:58:39 pm
content: In addition ...You say in your post that you're aiming for 3-5v at up to 3A on the output - I assume that's what you mean.So your XL4016 converts some input voltage to the configured voltage at up to 94% efficiency.Let's assume a much lower efficiency of around 90%, because that 94% can only be achieved under some specific conditions.If you want the module to output 5v at 3A ... that means output power = 5v x 3a = 15 watts.In order to output 15w, that means this step-down module will need to take in 15 watts x 100 / 90 (the 90%efficiency) = 16.66 watts.So now you can pick the size of that smoothing capacitor depending on the voltage after the bridge rectifier.For example, if you want the minimum voltage at the input of the XL4016 module to be 20v , that means the input current will be 16.66 watts / 20v = 0.83A so that's the minimum current value you'll want to use in that formula to calculate the minimum capacitor value.Keep in mind that the AC voltage can vary, it's not always 230v, so the output of your transformer will also vary, so when you calculate the peak DC voltage be a bit pessimist and calculate the peak voltage when your AC input is much lower ... for example if you have a 230v AC - 24v AC transformer that's almost a 10:1 ratio ... but when your AC input is only 210v AC you're no longer gonna have 24v AC on the output.

author: hpibmx
author_rank: Regular Contributor
timestamp: August 05, 2022, 09:37:41 am
content: Quote from: radiolistener on August 03, 2022, 08:12:34 pmThe output is 18v again my bad connecting the load same lamp reads 13.5v!

author: hpibmx
author_rank: Regular Contributor
timestamp: August 05, 2022, 09:47:19 am
content: Quote from: mariush on August 04, 2022, 04:58:39 pmHow can I know the spec about my transformer as it was taken from another broken psu,How can I know what is it’s power output?And yes my main work let’s say most of time 5v 3A.

author: james_s
author_rank: Super Contributor
timestamp: August 05, 2022, 05:30:14 pm
content: You can guess based on the size of the transformer and you can test it by applying a load and measuring the voltage drop and temperature rise. Don't worry about the transformer yet though, first try adding a large capacitor after the rectifier, that is mandatory and I think it is the cause of your problem.

author: radiolistener
author_rank: Super Contributor
timestamp: August 08, 2022, 12:13:29 am
content: Quote from: hpibmx on August 05, 2022, 09:47:19 amYou can measure it. Just connect it to the mains and measure Voltage of the open secondary coil. Then attach some known powerful load and measure Voltage drop down. With that data you can estimate how much current it can handle.If you have access to the wire of secondary coil, you can measure it's diameter. Then you can estimate max current:D = 1.13 * sqrt( I / j )whereD - wire diameter in mm^2I - current in Ampsj - current density in Amps/mm^2For 50-150W transformer with copper wire you can use:- for single frame construction: j = 2.0 [Amps/mm^2]- for double frame construction: j = 2.4 [Amps/mm^2]In your case (5V 3A), wire diameter should be at least:D = 1.13 * sqrt( 3 / 2 ) = 1.4 mmAlso you're needs to estimate transformer core cross-sectional area, it can be approximated as:S = 1.2 * sqrt( P / η ) [cm^2]whereS - core cross-sectional areaP - expected transformer power output ( P = I * U )η - efficiency, usually 0.75...0.85For 50-150 W transformer you can use η=0.8.In your case (5V 3A, P = 3 * 5 = 15 W) it should be at least:S = 1.2 * sqrt( 15 / 0.8 ) = 5.2 [cm^2]