Rectifier Design Help

AwesomeCronkAwesomeCronk Posts: 899
edited 2020-01-04 - 22:48:20 in General Discussion
I am trying to build a rectifier for my prop-driven DCC remote. I know this is possible with the components I have, because I pulled them off the old dcc remote and verified that they work. I have four 1N5401 diodes and two 35v 2.2mF capacitors.

I am trying to rectify a 16VAC supply to at least 15VDC. This supply is wierd, though. It is rated for 16VAC, but my meter reads 21.8VAC. I suppose that I need to rectify 21VAC in reality. Should be no issue for my circuitry.

I ohmed out the board of the old remote, trying to find out how they placed their filter caps, but what I got clearly was not right. I have a simple schematic of what I think would work underneath the schematic I got from the old remote attached. What I need help with is getting the DC supply smoothed out enough that I can power a model train from it (through an L298N in order to generate DCC signals) and power a prop from a linear regulator.

It should be known that I only have a multimeter, no scope.
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Comments

  • I think this is what you want. (Pardon the chicken scratch. Did this in the McDonalds drive thru. Hehehe)
  • I am trying to build a rectifier for my prop-driven DCC remote. I know this is possible with the components I have, because I pulled them off the old dcc remote and verified that they work. I have four 1N5401 diodes and two 35v 2.2F capacitors.

    I am trying to rectify a 16VAC supply to at least 15VDC. This supply is wierd, though. It is rated for 16VAC, but my meter reads 21.8VAC. I suppose that I need to rectify 21VAC in reality. Should be no issue for my circuitry.

    1 - The schematic JRoark posted is basically what you need to wire up.

    2 - With no load on the transformer 21.8V is a reasonable output voltage. That will drop a few volts when the train is moving.

    3 - Do you really have two 2.2Farad 35V capacitors or was that a typo? Typically a power supply like this would have a capacitor in the 2,200uF to 10,000uF range depending on the current draw and max permitted ripple.

    4 - No scope required. When the circuit is wired up you can can check the DC level and AC ripple at no load and with the train moving. Measure the DC on the appropriate DC range (probably 20 or 50V), and the AC on a lower (10 or 20V) AC range. At no load the AC reading should be very low, and at or near full load should be around 10% of the DC voltage or less.
    I ohmed out the board of the old remote, trying to find out how they placed their filter caps, but what I got clearly was not right. I have a simple schematic of what I think would work underneath the schematic I got from the old remote attached. What I need help with is getting the DC supply smoothed out enough that I can power a model train from it (through an L298N in order to generate DCC signals) and power a prop from a linear regulator.

    It should be known that I only have a multimeter, no scope.

  • AwesomeCronkAwesomeCronk Posts: 899
    edited 2020-01-04 - 16:55:35
    Here's the caps, diodes, and the board.
    2048 x 1536 - 1M
  • @JRoark
    Thank you! I will try that schematic and see what happens.
  • Those look like typical 1N400x diodes and 2200uF capacitors used in power supplies like that so you should be fine reusing them.
  • AwesomeCronkAwesomeCronk Posts: 899
    edited 2020-01-04 - 19:02:13
    OK, good. I will try and wire up a rectifier following JRoark's schematic by the end of today.

    EDIT: Words of wisdom when soldering diodes: They are EXCELLENT conductors of thermal energy
  • I put together a GreatScott! style midair soldered jury-rig. I was really scared when I plugged it in, I'm not gonna lie, but I got 30vDC measured at the caps, under no load, with no explosions!
  • That's correct. A full-bridge rectifier circuit will give you the DC equivalent of the peak-to-peak AC voltage -- at least until you apply a load.

    -Phil
  • U x √2

    16ac x 1.41421 = 22.62 Volt dc
  • AwesomeCronkAwesomeCronk Posts: 899
    edited 2020-01-04 - 22:48:57
    I am currently testing this jury-rig and I am concerned about something. it seems that the caps charge up over time. I know they charge during the peaks of the cycles, but they also appear to be charging more as time goes by. when I measure the rectifier now, I get 29.9VDC. In order to test this theory, I unplugged the wall pack, then got a piece of 16AWG stranded and touched the two leads on the caps. I got a BRIGHT blue spark and a wonderful popping sound that ironically sounded like it was saying "CAP!". Lesson learned. Don't discharge capacitors, especially 2.2mF caps, with a piece of wire. I managed to spot-weld the wire to the solder that I had on the terminal.

    Then I read the DC voltage a minute later. 1.88VDC and climbing at about 0.005V per second. I suspect the caps are gathering charge from the nearby devices (EDIT: I think it was just the transformer coils deenergizing. The caps appear to stop recharging about 10-20 minutes later if left unplugged.) using power (PC, power strip, ect.) I am concerned because this voltage may climb excessively if my brothers or I leave the system on without the trains running. it would soon reach 22v, which is the peak leagal voltage supply for DCC power stations. (upper right graph on page 3 of this document)
  • Have you got a discharge resistor on those caps? That might solve that problem.
  • AwesomeCronkAwesomeCronk Posts: 899
    edited 2020-01-04 - 22:18:33
    No discharge resistor. How would I wire one up? Is 10kOhm big enough? That is the biggest one I have.

    There is an ST-22 fuse I can desolder from the board to act as overvoltage protection, I think.
  • Just take a rather big resistor (is 100KΩ a good value?) and wire it in parallel to the capacitor
  • kwinnkwinn Posts: 8,440
    edited 2020-01-04 - 22:41:13
    The circuit you have built with a transformer, full wave rectifier, and filter capacitor is a circuit that has been used in thousands, possibly millions of pieces of equipment over the years. The voltage on the capacitors will vary with the amount of current drawn (ie the load) and the voltage in from the power grid. That grid voltage will vary quite a bit over a 24 hour period. I have seen the nominal 120V from a receptacle vary from as little as 95V to as much as 135V over a 12 hour period at some sites. Typically it is less than that at most locations, but this is not unusual, and normally not a problem. I would suggest you measure and record both the voltage input to and output from the transformer every half hour over an evening to see how they vary from 17:00 to 21:00.

    PS, 2200uF is 2.2milliFarads (mF), 2.2F is 2.2 million uF.
  • AwesomeCronkAwesomeCronk Posts: 899
    edited 2020-01-04 - 22:50:55
    kwinn wrote: »
    PS, 2200uF is 2.2milliFarads (mF), 2.2F is 2.2 million uF.

    Oops. Thanks for catching that. I'll edit my posts.

    I am not in a position at all to measure my grid's supply. I do know, however, that my PC runs fine 24/7, so there can't be too much variation.
  • ercoerco Posts: 19,587
    AwesomeCronk: You're in very good hands all the Forum professionals weighing in with comments & suggestions. I just had to mention Electroboom's channel for comic relief. He loves full bridge rectifiers and often wears a FBR T-shirt. His comic videos are informative, funny and mostly clean. He self-bleeps when he constantly blows things up and shocks himself. A pleasant diversion anytime you need a break. Here's a typical video with his FBR shirt:

  • Hi AwesomeCronk

    Please, pay attention to your computer's power supply line input rating from mains!

    As an example, the one where I'm typing that post has a power supply whose input is rated from 100VAC to 240VAC, delivering steady 19 VDC at its output, in order to feed the internal regulators and battery charge circuits of the notebook.

    So, if one equipment works well, there is no garantee other devices will do, even if they are fed by the same mains AC voltage. It's only a matter of knowing the right specs for each one.
  • @erco
    He is funny, indeed!

    @Yanomani
    Thank you for the warning. I think I will take that fuse that was on the board and add it into the circuit.
  • Your observation that the caps are “self-charging” is actually correct. However it has nothing to do with the transformer coils. Some big caps (usually electrolytics) display this behavior. What happens is the electrolyte inside is releasing a tiny bit of voltage due to the reaction of the electrolyte with the plates. Over time this builds up. Essentially it works like a very (very!) weak battery.

    You can prove this to yourself easily: Take a big low-voltage cap, charge it up for a few hours near the rated voltage, then discharge it flat. Hook it up to your meter intermittently over the next few hours. You should see this effect. It wont be a lot of voltage usually, but it is there. At some magic value the internal leakage balances the charge being created, and the rise stops. The rate of rise, final voltage attained, etc, are determined by the cap materials (plates, electrolyte) and the temperature.

    This effect is why really high voltage caps always have a bleeder resistor across them. It isnt just to bleed down the stored charge when the equipment is turned off. It is also to prevent the cap from building a charge back up when the equipment is off. Some really big high voltage caps can only ship with the electrodes shorted. Why? Because if a 20,000 volt (rated) cap were to charge to just 0.5% of its rated voltage, that is 100 volts... a potentially lethal hit to anyone touching it.
  • @AwesomeCronk, the second circuit of your initial post shows your two caps in series. This will result in 1/2 of the value of the capacitors (opposite of resistors which would show double the value). Good explanation via link:

    https://eecs.tufts.edu/~dsculley/tutorial/diodes/diodes3.html

    It talks about time constants and ripple, but may be easiser to think of the cap as the device that tries to maintain the voltage level when it drops below what the cap was charged to until the next charge cycle. How fast it discharges depends on the load.

    Some circuits that have really large caps will use a series resistor and bypass relay contacts to limit the current when the supply is first turned on (discharged cap looks like a short circuit with no charge on it). The current will drop as the capicitor charges. Once a certain time has passed the relay will bypass the series resistors so the caps are directly connected to the source.

    As to measuring the ripple, I have only ever done it with a scope. Have not tried to see if a meter would give a reasonable value since the waveform is not likely sinusoidal and not at 50/60Hz in frequency.

  • What happened to the images in this thread?
  • What happened to the images in this thread?

    Images restored. Thanks for alerting.
  • It wasn't just this thread, and the thread I started regarding the image issue got deleted. Can you explain what's happening?

    Thanks,
    -Phil
  • VonSzarvasVonSzarvas Posts: 1,808
    edited 2020-01-05 - 19:06:17
    There was some improvement work on the code tags output today.
    During the update, rendering of some attachments were disabled. They are now re-enabled.
  • Got it. Thanks.

    -Phil
  • So, I have noticed that the rectifier seems to crank up the voltage over time, likely due to the caps charging up and not discharging. If I leave a train on the tracks, not running for 10 minutes, will the load of a propeller and lcd display (powered by linear reg.), will this charge up continue to occur?
  • jmgjmg Posts: 14,177
    edited 2020-01-06 - 18:34:32
    ... It is rated for 16VAC, but my meter reads 21.8VAC. I suppose that I need to rectify 21VAC in reality....
    .. when I measure the rectifier now, I get 29.9VDC..
    I am concerned because this voltage may climb excessively if my brothers or I leave the system on without the trains running. it would soon reach 22v, which is the peak leagal voltage supply for DCC power stations. (upper right graph on page 3 of this document)

    AC packs are usually spec'd at full load, and the smaller transformers have significant droop effects, so 16VAC at full load can easily be 21.8VAC at light loads. ie sounds ok.
    If that 22V-DC is a true MAX, you may need to add a regulator, as your measured 29.9VDC out, is well above the 22V limit.
    What current do you expect to supply ? A common LM317 may be good enough ?
  • > @AwesomeCronk said:
    > So, I have noticed that the rectifier seems to crank up the voltage over time, likely due to the caps charging up and not discharging. If I leave a train on the tracks, not running for 10 minutes, will the load of a propeller and lcd display (powered by linear reg.), will this charge up continue to occur?

    No. Basically, the caps are done charging after a few dozen cycles of the AC line. Certainly after a few hundred cycles, they’re as full as they can get.

    Visualize this; you have an empty cup buried even with the sand surface on a beach right at the very highest point where the waves run out. Each time a wave runs up on the sand, the cup is filled. Once the cup is filled, you cant fill it any more.

    That cup is a metaphor for your capacitor. You cant fill it beyond the height of the tide regardless how you may try.

    Putting a train on the tracks or adding a small load will have a small, measurable effect on the peak voltage, but if that transformer is rated properly for the job, the effect should not be very pronounced.
  • The rectified AC wave will charge the cap on the rising side of the wave and as it falls to where it will equal the voltage on the cap at which point no more charge occurs until next rise. As noted above, with no load to discharge the cap it will equal the peak of the rectified DC value after a few cycles since discharge only possible with a load (or leakage). The capacitor voltage will never exceed the input voltage.

    Once a load is applied, the charge being removed by the load will cause the capacitor to discharge into the load (rc time constant, load being the r part of the equation) so the voltage falls, until the rectified voltage coming in exceeds the cap voltage again, charging the cap and raising the voltage to the peak value and then dropping due to discharge through the load. That is why the output looks like DC with a ripple riding on it. The heavier the load, the farther down the the capicitor voltage will fall, the more the ripple will be present in the output. This is offset by the amount of current and therefore charge into the capacitor while providing the required load current. The bigger the cap, the more current can be sourced when it is not being charged, but then a higher current capacity to charge the cap while supplying the load current will be required.

    Not to sound like a UDEMY shill, but one of the long time forum members, Andre Lamothe, has a course called "Crash Course in Electronics". It covers a lot of the things including this topic.
  • The max DC voltage on the caps will be 1.414 x the ac volts sine you have full wave rectification
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