Rectifier Design Help
AwesomeCronk
Posts: 1,055
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.
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.
Comments
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.
Thank you! I will try that schematic and see what happens.
EDIT: Words of wisdom when soldering diodes: They are EXCELLENT conductors of thermal energy
-Phil
16ac x 1.41421 = 22.62 Volt dc
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)
There is an ST-22 fuse I can desolder from the board to act as overvoltage protection, I think.
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.
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.
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.
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.
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.
Images restored. Thanks for alerting.
Thanks,
-Phil
During the update, rendering of some attachments were disabled. They are now re-enabled.
-Phil
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 ?
> 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.
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.