Sciencemadness Discussion Board - Inductor and capacitor for rectifier output filter

Inductor and capacitor for rectifier output filter

yobbo II - 16-2-2024 at 03:40

Can anyone point me in the direction of making filter circuits for the output of a 2 or four diode rectifier.
I am having no luck on obtaining information regarding the size of inductor and capacitor required. The inductor needs to be air gapped to deal with the DC.
I will be using the core of an MOT for the inductor.

The usual advice is that the whole thing will end up being larger and heavier than just simply using a larger transformer at the start but I want to go the inductor/capacitor(s) route.

Thanks,
Yob

Sulaiman - 16-2-2024 at 04:14

Look for PSUD2 by Duncan, windows only I think.
Much easier than a full simulator.

An inductor can really reduce ripple and/or improve efficiency,
best suited to loads that draw a fairly constant current,
remember,
the flyback voltages can be quite high,
when you try to open circuit an inductor that has significant current passing through it,
I've given myself quite a few surprises this way

PS try C-L-C, eg 1,000uf high ripple - inductor - 10,000uF

Often electrolytic capacitors are sold with a minimum order quantity, eg 5
You can simulate 1cap - inductor - 4caps
Or 2caps - inductor - 3caps etc.

Also, your choice of rectifier can reduce psu heating a lot.

[Edited on 16-2-2024 by Sulaiman]

Rainwater - 16-2-2024 at 12:04

Depends alot on which type of rectifier you use.
3phase is best, next up is full bridge, and the worst is single diode rectifier.
Basicly your building a low pass filter.
Need a few numbers to provide useful feedback.

Code:


Input voltage (ACrms):
Input frequency:
Output voltage DC:
Output current:
Maxium desired ripple:

hyperphysics online calculator

[Edited on 16-2-2024 by Rainwater]

semiconductive - 16-2-2024 at 15:06

It usually is larger to have both a transformer and an inductor; because there is wasted space in each. But, a larger transformer doesn't necessarily cause less ripple in the output; so I'm not sure why people are suggesting a bigger transformer is better.

You're using a two diode rectifier, which is split phase. Therefore, your mains transformer is likely center tapped. You will get pulsing at twice the frequency of the mains; eg: either 100 or 120 pulses a second for 50 to 60Hz mains.

Sauliman's suggestion of using two capacitors on either side of the inductor is recommended.
There ought to be a capacitor from the center tap of the transformer to both the diodes rectified outputs.

A typical motor winding (MOT?) has a few millihenries of inductance in my experience.
Since you've already chosen the inductor, you should give more information about it since it basically limits what can and can't be done with the filter circuit.
Do you know the inductance of the motor? Or do you know it's DC resistance, and can you estimate how many winds of copper it has?

Without this information; we'd have to walk you through an experimental way of figuing out what you have; and what you can do with it. I'm not sure what you are expecting.
Rainwater's questions would be very helpful if you are able to answer them.

Twospoons - 16-2-2024 at 15:41

Before we can answer your question we need to know how much current you want to supply, and how much voltage ripple is acceptable.
Basic rule of thumb for a full bridge rectifier is to use roughly 1000uF per amp of output current. Use of an inductor is usually not necessary - there are better ways of producing smooth DC if thats what you need.

Supplying some background info to these questions would be enormously helpful: e.g. what is it you want to power?

yobbo II - 16-2-2024 at 16:56

Firstly an MOT is a Microwave Oven Transformer!
I will be dismantling the transformer, removing all windings and putting a winding onto the core and gapping the core.
These transformers have a current shunt in them (which will be removed). They are also a gapped core as the E and the I pieces are not interleaved like you have in a 'more normal' transformer. Found this out when I cut the two weldes that hold on the I lump (all the I's are in one lump, and all the E's are in another single lump) and seperated the I from the E.

I don't have a detailed specification of a particular power supply that I want to build. I want to know how you would calculate or estimate the capacitance and inductance needed for this particular type of filter.
I would be using a four diode rectifier. Single phase 50Hz. Output voltage in the region of 5 or six volts DC. The transformer that I intend to use is another MOT with a output winding wound onto it to supply from a few amps to perhaps 100 amps.
I would probably use two MOT's with the primaries in series on a 240v supply (the mains) as these transformers have a very skimpy number of primary turns (they run a very high magnetizing current when not connect to a load). This means that you have a center tap sitting there when you wind an output winding on each transformer so I might use a two diode rectifier (to contradict what I said above!). I would probably have a number of taps on the secondary's so that I can have different voltages at the output but not that far away from 6 volts DC.
Ripply, I don't really know. As little as possible without breaking the bank on caps. The MOT inductor will not break the bank.

The caps/inductor filter is very simple and rugged (I think).

I have a toroid from a fairly large motor (few horse power single phase) with all the motor windings removed. Perhaps I could try winding some turns of wire onto it and see what inductance value I get. It is not gapped but is so large it may not saturate?

I went to the PSUD2 by Duncan page but the software does not seem to be available at the moment. I will have another look tomorrow.

A reason to use a bigger transformer that what you might think you need is that the transformer needs to be able to supply the large current pulses needed to keep the (single capacitor no inductor filter) capacitor topped up. If you use a larger and larger capacitor to get a smoother output, the current pulses get larger and larger in magnitude and smaller and smaller in duration.
This also leads to lots of harmonics on the input line.

Thanks for your time
Yob.

[Edited on 17-2-2024 by yobbo II]

Twospoons - 16-2-2024 at 21:47

http://sim.okawa-denshi.jp/en/RLCtool.php
https://www.analog.com/en/resources/design-tools-and-calcula...

Two tools for you. The filter calculator will help get you in the ball park with the L -C filter, and the Spice simulator will let you try it out without releasing the magic smoke. Modelling the MOT properly will be all but impossible, but you should be able to estimate ripple voltage and the currents and power loss in the rectifiers.

If you really want to do 100amps I'd start with C in the region of 47,000uf up to 100,000uf. 16V electrolytics of that size are 8 to 20 USD from Digikey. You wont need a particularly big value inductor, maybe 10uH to 100uH. And for 100amps you dont want a high value or the thing would have to be enormous in order not to saturate.

You could also look at some passive PFC techniques, such as the passive valley fill PFC.

Are you using MOTs because you have some and are trying to save money? Its a sub-optimal solution. They are not really designed for continuous use - they use the absolute minimum copper and iron, so losses get pretty high compared to a better built and rated transformer.

Rainwater - 17-2-2024 at 02:23

If your starting with microwave oven and want to finish with a low voltage dc power supply, your going to have a lot of steps. You have all of the needed parts, but, they have been configured in a way that is not perfered for your application. The transformer core will be in a flux shunted configuration. This limites the maxium flux available for use without saturation of the core. Really fancy talk for putting that weld line across the core plates. Its partly responsible for limiting the power available at the output. And the main reason you'll get a very non sinusoidal output voltage and input current. Cut the cores plates apart, fix the rough edges and bers, apply a thin film resistive coating( more fancy talk for rust) then reassemble.
They core will have an air gapped bridge, this to needs to be removed, usually by grinding they other side for a proper fit, as it acts as a flux limitor under all but the smallest loads
Once you have an isolated, solid core, you have a power transformer.
Your primary current will be greatly increased and the output power waveform stabilized. This will require rewinding the primary coil to match he increased permeability of the core.
After all that you can focus on rectification.

It should be noted that large industrial scale electrolysis setups do not use filtered DC, but pulsating DC due to the high cost of the filters. It makes calculating the used current more difficult

[Edited on 17-2-2024 by Rainwater]

semiconductive - 17-2-2024 at 17:24

I see.

Yes, the smoother the supply, the less time a typical transformer actually outputs energy into the capacitor. This does cause harmonics issues in analog supplies with no switching circuitry.
Switching circuitry tends to cause high frequency noise, and is tricky in a different way.

At the high current end of your range (100 Amps), you're also going to have huge inrush currents into the first capacitor of your filter. The diodes will take a beating every time the power supply is first plugged in and the capacitors initially charged. I've had several designs fail because the diodes were not rugged enough to survive a high inrush; and that's something you'll want to consider.

The other (typical) choice is to use a positive coefiicient thermistor (PTC) device to limit current inrush. Calculation will be required, but you're not close enough yet to work that out. 100Amps is a pretty ambitious goal for low ripple output!

Note:
I'm writing my own simulator, and have been studying transformers for a while; so all the ideas are fresh on my mind for what you are doing:

https://physicsdiscussionforum.org/ohm-s-law-states-relation...

In the thread I linked, I posted some links to core material information at colleges that might be helpful; and am developing formulas after reviewing online literature (which has mistakes...!).

Note: Spice simulator is *horrible* at simulating inductors, accurately. It will likely crash.
As you develop the design, since it's simple, I don't mind coding up a simulation for you the same as I'm doing for my friend on the other thread. I'm not fast at this, but I am thorough. I expect to be running more simulations for my friend within a week, after doing a test on a ferrite core transformer. (I'm FAR better at electronics, than I am at chemistry! )

Do you have an inductance meter?
Are you familiar with/do you have an Oscilloscope?
How about digital multimeters?

Sulaiman - 17-2-2024 at 19:54

I think that an MOT core is an excellent base for a dc choke,
no secrets, fill the winding area with as much copper as you can,
high current single-strand copper is difficult to wind,
you can use as many strands of thinner wire as required to carry the required current.
(2 or 3 can be wound neatly in layers,
more than that it's easier to twist them into a cable/rope)
I use 2.5/5/10 A(rms)/mm² for cool/hot/short-term
Adjust inductance by varying the air gap.
(more inductance = lower saturation current)

Use massively overrated rectifiers, preferably on a fan-cooled heatsink,
(cheap compared to the transformer, capacitors etc)
for a high current psu every millivolt helps so
Simulate, determine Peak rectifier current and use rectifiers
that you have or can get, that have a low (<1V) forward voltage at that peak current.
Saves a lot on heatsink requirements.

I think that PSUD2 could help design, simulate and 'get a feel for' various variables.

Spoiler alert : my summary
I've not found a situation, at 50/60 Hz, where using a
large inductor is the best choice.
(my personal diy supplies are mostly 10A or less)
I normally use power supplies that give a near constant voltage under varying loads.
Smaller inductors are commonly used in simple high current (3-phase) rectifier systems to reduce the peak, and rate of change of, currents.

Just like in chemistry,
an hour of research (simulation) can be worth 10 in the lab.
But I learn more by blowing stuff up!

PS winding is a chore, and copper is not cheap,
but you can make multiple windings almost as easily as a single winding, allowing flexibility.
eg two similar windings (bi-filar or separated)
allows parallel or series connection for higher (2x) current or higher (4x) inductance, etc.

Often multiple lower capacitance units in parallel have lower cost and esr,
and higher life and current rating
than a single larger capacitor.

PPS in place of a normal transformer,
you could use an existing MOT core, shunt and primary winding as-is,
just replace the high voltage secondary (and filament winding) with a diy high current secondary.
Electrically equivalent to a transformer with output inductance.
Almost bomb-proof, especially if two transformers have their primaries in series.
Reduce the shunt to reduce the output inductance/impedance, increasing regulation, and short-circuit current.

As mentioned by others, a switch mode power supply would be cheaper
(and a lot less heavy - more important with an ageing body

[Edited on 18-2-2024 by Sulaiman]

Twospoons - 18-2-2024 at 14:45

Quote: Originally posted by semiconductive

Note: Spice simulator is *horrible* at simulating inductors, accurately. It will likely crash.

Have you tried Simetrix? They have a saturable/hysteretic inductor in their arsenal that uses the Jiles-Artheron model specifically modified to avoid convergence issues at the extremes of the B-H loop.

Simetrix is my preferred Spice simulator - I don't like the UI of LTSpice.

Refs:
Theory of Ferromagnetic Hysteresis, DC.Jiles, D.L. Atherton, Journal of Magnetism and Magnetic Materials, 1986 p48-60.

On the Parameter Identification and Application of the Jiles-Atherton Hysteresis Model for Numerical Modelling of Measured Characteristics, D Lederer, H Igarashi, A Kost and T Honma, IEEE Transactions on Magnetics, Vol. 35, No. 3, May 1999

As has been said, switching supplies are not that expensive. You can buy a Meanwell 500W 5V 100A supply on Aliexpress for around 70-80USD. They are surprisingly robust - they will shutdown without damage on overload or overtemp.

[Edited on 18-2-2024 by Twospoons]

[Edited on 18-2-2024 by Twospoons]

Rainwater - 18-2-2024 at 17:57

Modeling non ideal transformers is next to impossible without a phd.
Experimentation would be quicker.

Just remimber that electricity will hurt like hell while it kills you.

You will not be able to let go once you grab it.
Use chassis grounds
Core bonds
gfci
And low current fault protection.
A varable transformer like the ones tomholm has been auctioning off are a must.
Your transformer will work at 1v just as good as 250v, youll just have to change the coefficients in your equations

yobbo II - 22-2-2024 at 05:01

"PPS in place of a normal transformer, "
What is a pps?

I have a multi meter, oscilloscope, large variac etc.

When taking an MOT with all windings (except the primary) and magnetic shunt removed I get the following values for inductance.

0.41 H using mains 240v 50Hz
0.98 H using 22 Volts 50Hz
and 0.126H using the Inductance facility on a multimeter
The DC resistance is 2.4 Ohms.

Why is there such variation in the value of H.
Perhpas the core is saturating with the 240 Volts.

I simply connect up the 240volts and the 22 volts TO the primary and measure the current (using a multimeter on AC) to get a value of H
from the formula XL = 2 pie f H

I found this page gives transfer functioins of lcr combinations.
http://sim.okawa-denshi.jp/en/RLClowkeisan.htm

Yob

[Edited on 22-2-2024 by yobbo II]

Sulaiman - 22-2-2024 at 09:25

PPS, comes after PS.
So the text is just
'in place of a normal transformer'
..........
At 240Vac 50Hz the core will almost certainly be beginning to saturate, (minimum materials cost)
you will hear it buzzing more than humming.
and the multimeter probably works at kHz where the steel core will be very lossy.

Twospoons - 22-2-2024 at 13:22

Quote: Originally posted by yobbo II

Why is there such variation in the value of H.
Perhpas the core is saturating with the 240 Volts.

[Edited on 22-2-2024 by yobbo II]

Simple: MOTs are built with the absolute minimum iron they can get away with, purely for cost reasons. So they are operating well up the BH loop, where the inductance drops away and iron losses are much higher. If you are running one more than a minute or two it will need forced air cooling.

This is why MOTs are a poor choice for the primary step-down transformer for building a DC supply - you really need a transformer built for continuous use, where the primary magnetizing current isn't pushing the core into saturation.

There's a good reason for the proliferation of switching supplies - better regulation, better efficiency, better protection features, smaller size. Its worth noting that microwave ovens are now available using switching converters instead of chunks of iron - allowing direct regulation of the magnetron power, instead of the crude on/off control of older generation microwaves.

Rainwater - 22-2-2024 at 14:14

Another possibility will be your current meter, calculating the impedance of an inductor at mains ac, you will have to use a true rms meter. A sample rate greater than 2 to 10 times your input frequency is desirable.
If using averaged or apparent readings for voltage or current, your SOL.
For field work I use the ideal 61-765 but there are cheaper options to do what your doing.

wg48temp9 - 25-2-2024 at 05:13

Below is a set of B-H curves. H is equivalent to voltage and B is equivalent to current density.

A typical MOT from a 800W 204V oven has primary winding that draws about one ampere with the secondaries open circuit or removed. Typically the primary has about 250 turns with a magnetic path length of about 0.25m . So that is 1,000 At/m (250/0.25). You can see from the curve for silicon steel, that at 1,000 At/m the core is very saturated (the curve is almost flat). I don't know if that curve corresponds to the core material of the above MOT but it is probably similar. The inductance of a particular primary winding is approximately proportional the ratio of H to B. So at Higher primary currents the inductance can be drastically reduced.

The curve is also useful for understanding why an air gap is required in a choke. Note that in an MOT with the magnetic shuts removed there s no deliberate air gap between the E and I stacks of laminations. The weld line between the E and I stacks and on other outer faces are there to hold the stacks together and to reduce noise.

[Edited on 2/25/2024 by wg48temp9]

wg48temp9 - 2-3-2024 at 15:50

So below is the B H curve from my previous post with an added curve of an air gap shown in green.

Sorry, I could not add the reluctance of the air gap to the reluctance curve for silicon steel. You have to add the two curves horizontally in your head. The reluctance of the air is equal to the reluctance of the silicon steel at G2. This significantly reduces then flux and therefore the saturation, but it also halves the inductance proportional to the flux decease. The inductance can then be restored by increased the number of turns.

So now can I design a 100A air gapped choke using an MOT with the following characteristics:

Original primary
250 turns, 3.2 ohms, inductance 05 H
No load rms current, 1 A
Full load rms current 5 A

Given that the core losses will be greatly reduced because of the reduction in amplitude alternating flux, so I will assume the usually high copper losses can be the same as the original windings. To keep the same primary copper losses the same as the original winding (5A), with a 100A current the number of turns must be reduced while keeping the amount of copper unchanged. As current has increased to x 20, so the winding resistance must be decreased by 400. For a fixed volume of copper, the resistance of a winding is proportional to the square of the number of turns. Therefore, 12.5 turns at 100 A will dissipate the same power as the 5 A did in the original primary winding of 250 turns. The secondary space can also be used, so the total turns are 25 assuming the original secondary dissipated the same power as the primary. That will have an inductance of 100 time less than the original primary 0.005 H and halved again to account for the air gap, so the final inductance 0.0025 H.

For full wave rectification of 10 volts...

Sorry, I will have to finish and check this later

[Edited on 3/2/2024 by wg48temp9]

Sulaiman - 3-3-2024 at 09:41

Quote: Originally posted by yobbo II

. I want to know how you would calculate or estimate the capacitance and inductance needed for this particular type of filter.

Ripply, I don't really know. As little as possible without breaking the bank on caps. The MOT inductor will not break the bank.

as a starting point, my quick rule-of-thumb is,
For 1 volt pk-pk ripple using full-wave rectification of a 50Hz supply,
use 4,700 uF per amp dc output.
eg a 100A dc output requires about 0.47 Farad (470,000 uF)
Probably super-caps, or a lot of electrolytics.
An inductor does reduce the value of capacitance required for a given ripple voltage,
and the ripple current requirement of the capacitors.

wg48temp9 - 4-3-2024 at 01:37

Quote: Originally posted by Sulaiman

Quote: Originally posted by yobbo II

That looks very optimistic. 100A for 10ms is 1C. 1C into 1 Farad is a change in voltage of 1V ie pk to pk.

Sulaiman - 4-3-2024 at 08:33

Quote: Originally posted by wg48temp9

That looks very optimistic. 100A for 10ms is 1C. 1C into 1 Farad is a change in voltage of 1V ie pk to pk.

you are correct,
I don't remember where I got my rule-of-thumb from but mathematically it is wrong

Sorry all

yobbo II - 7-3-2024 at 15:14

What would be the incremental permeability (IP) of the steel used in microwave transformers.
The IP is the figure you use (I think) as there is DC present. The usual permeability is not revelant.
Also what would be the saturation flux density of the core?

I am going to use the 'E,s' from two transformer cores. The transformers are cut open, the 'I,s' are discarded and the two E parts are put together which gives double the space for winding. I will have to put somethig to space the two E parts apart (the gap) put on the windings and then weld the two E parts together.
Doing calculations I am getting about 1.9mH. Using a gap of 1mm in all three legs, 24 turns, and an (incremental) permeability (a guess) of 200
With a gap of 0.5mm I get 2.63mH.

What a miserable amount of Henry's!

Thanks,

Yob

[Edited on 7-3-2024 by yobbo II]

Twospoons - 7-3-2024 at 17:41

Quote: Originally posted by yobbo II

What would be the incremental permeability (IP) of the steel used in microwave transformers.
The IP is the figure you use (I think) as there is DC present. The usual permeability is not revelant.

Since IP is dependent on DC bias there is no single answer to that question.

I suggest you download Femm 4.2 and model your core assuming the use of grain oriented silicon steel.
Thats going to give you better answers than anyone here can give you.

[Edited on 8-3-2024 by Twospoons]

yobbo II - 7-3-2024 at 17:48

Thanks Twospoons,

Femm may not be too easy to learn? I don't know a whole pile about magnetics.

Yob

Twospoons - 7-3-2024 at 18:14

Its well documented and there are plenty of examples. It would be well worth your time to learn to use it.

yobbo II - 16-7-2024 at 16:46

Finally got arount to this.

If using just one capacitor and an inductor for the filter. Where is the sensible place to put the inductor?
Before the capacitor or after the capacitor.
I have done both and took some scope pictures.

Also i think i need to get a proper true rms reading meter.
I have a cheap amazon meter.

Yob

Rainwater - 16-7-2024 at 17:22

basic filters

Depends on your configuration, got a schematic?
I may have missed it, but what is your final application and specifications?

If no ripple is the goal, active regulation is 3-20 times better than passive filtering

yobbo II - 17-7-2024 at 14:46

This is just a simple filter to reduce ripple (no actual pacific target ripple) on a full wave rectified signal. 50 amps supply approx.
Is there a (more) correct place to put the inductor.
I will try a pie filter too.
The caps i have are 68000 micro F. The inductor is 18 turns on an MOT core.

Yob

Twospoons - 17-7-2024 at 19:04

The answer is rather complicated. If you put the L first then your unloaded output voltage will be higher than your transformer output voltage because the system operates a bit like a boost converter. With a load you get an initial output voltage boost on power up, which then drops over time (load dependent) to a voltage lower than your transformer output (with some ripple).
In the attached sim you can see the initial inrush current, which over-charges the capacitor, then every settles down to a lower voltage with ripple.
For this sim I used 1mH inductance, 10mF capacitance and a 10 ohm load.
Current is green, input V is blue, output V is red

[Edited on 18-7-2024 by Twospoons]

Twospoons - 17-7-2024 at 19:07

If you put the inductor after the capacitor then you dont get the initial overshoot, the average output voltage is a bit higher, ripple is similar, but peak diode current is considerably higher.
Input V is green, current is blue, output V is red

[Edited on 18-7-2024 by Twospoons]

Twospoons - 17-7-2024 at 19:13

In short, L first is easier on the diodes and transformer due to lower peak currents, but DC load regulation suffers.
L second means no overshoot, but is much harder on the diodes and transformer. It also means any load transients are going to seriously affect your output voltage.

So - choose your poison

See, this is why it is far more usual to use a switchmode PSU if you want 100A at 5V. You get stable output, usually with overcurrent protection, and a power factor corrected input which is friendly to your utilities company.
It will also weigh a lot less.

[Edited on 18-7-2024 by Twospoons]

yobbo II - 18-7-2024 at 01:31

Thanks for replies.

I will post some scope pictures.
The inductor is around 1.3mH when measured with an inductance function on a multimeter. No DC through inductor and no gap in core.
Cap(s) are 68000mF. Yob

Rainwater - 18-7-2024 at 18:53

68F total? Imagine the bass you could pump with that
Supercaps or a big array?
Supercaps are not designed for ripple filtering, most employ thin film technology and many are a hybrid between a cap and a batterys. This makes them easily damaged.
Inrush current will be an issue,... supose an C_esr + Vcc_z = 1 ohm.
68_F × 1_ohm × 5 = 340_S of charge time

Twospoons - 18-7-2024 at 20:13

I guess he's just confused the prefixs for milli (m) and micro (u). To be fair, both words start with m.

yobbo II - 19-7-2024 at 01:35

Yes indeed

Its micro F

Yob

Sulaiman - 19-7-2024 at 04:38

Quote: Originally posted by yobbo II

Also i think i need to get a proper true rms reading meter.
I have a cheap amazon meter.

I have a cheap true rms meter and a nice one,
- no significant benefit to me so far, compared to a normal dmm.

If budget allows, I really recommend an oscilloscope.
Measurements are fine, but waveforms are so much more instructive.
Excellent models are available for a few hundred dollars, but..
I recently bought a small Chinese oscilloscope (SCO_2) and I can recommend it for dc to >1MHz work.
You could go for a single channel 'scope and it will be almost as useful, with more MHz/$
A great benefit of these mini-'scopes is that there is no earth connection so you can connect almost anywhere - but be careful.
........................
I once had access to scrapped mainframe computer parts,
one thing that I built was an audio amplifier (to match my diy speakers)
(8x 30W + 2x 120W rms ILP modules), 4x transformer and 6x giant electrolytics.
At normal listening level the sound would continue undistorted for over a minute after power was disconnected...
slightly over-engineered

(and HEAVY)

[Edited on 19-7-2024 by Sulaiman]

yobbo II - 19-7-2024 at 05:36

I have a scope which can be verified with all my lovely pictures.
The earth can be infuriating.

Attached is diagram + scope pictures of PIE filter PSU using MOT's and an inductor made from MOT's.
Not a whole pile of science but that's compensated for by the fact that there is a whole lot of pretty pictures.

Don't purchase an Amazon 80CM810 true RMS clamp meter. They don't read correctly.
You should get the same value of RMS current (meter set to AC and clamp used) going into the rectifier as you get DC going into the load, after the filter, (meter set to DC and clamp used)? There was little ripple on load voltage/current. For example, I get 49 Amps AC (true RMS??) going into rectifier and 36 Amps DC (true RMS) going into load.
If I place a 0.001 Ohm resistor between the transformers and rectifier and use the AC voltage facility (the two leads) on the meter, it still reads a wrong value of 49 Amps
(it shown an AC Voltage (true RMS??) of 0.049V across the resistor).
There IS 36 Amps going to load according to scope and load resistor (0.11 Ohm). The meter seems to work OK on the DC (clamp) true RMS readings.
Yiz gets what ya pay for I suppose.

Anyhow the supply shown uses a pie filter made from two capacitors and an inductor made from two MOT's. The ripple decreases when spacers are placed into the inductor. I guess it is saturating without them.

Yob

Attachment: mot_circuit(1).zip (7.6MB)
This file has been downloaded 104 times

wg48temp9 - 20-7-2024 at 02:39

Quote: Originally posted by Sulaiman

Quote: Originally posted by wg48temp9

That looks very optimistic. 100A for 10ms is 1C. 1C into 1 Farad is a change in voltage of 1V ie pk to pk.

you are correct,
I don't remember where I got my rule-of-thumb from but mathematically it is wrong

Sorry all

My calculation for the pk to pk ripple voltage shown above is worst case (max pk to pk) for ideal components and instant recharge of the capacitors on each cycle)

I wondered what the minium ripple voltage would be. I think the minimum is the capacitors are charged for half cycle and then discharged for half cycle. That makes the capacitor current a triangular waveform as opposed to a saw tooth waveform. So for the first half of the cycle, the rectifier supplies 200A (100A charge current and 100A load current). For the second half of the cycle, the 100A load current is only supplied by the capacitor. This would make the ripple voltage half of my previous calculation and equal to the rule-of-thumb.

So perhaps the rule of thumb should be : the pk to pk ripple is between the min and max values shown above.

It's interesting to note that switch mode power supple filters with square wave inputs could operate near the lower ripple limit, while filters for rectified AC will operate near the upper limit, assuming the first filter component is a capacitor
Perhaps that's were the rule-of-thumb originated.

yobbo II - 12-8-2024 at 10:17

Sulaiman said:

Also, your choice of rectifier can reduce psu heating a lot.

What choices do i have for a rectifier?

Do you mean to split the output of the transformer and use a two diode rectifier.
My output is too.low for that.

Is there rectifiers that have less V drop than the usual silicon diode ones.

Yob

[Edited on 12-8-2024 by yobbo II]

Sulaiman - 12-8-2024 at 10:39

For low voltage high current full wave rectification,
At any particular current, a higher current rated diode (eg 250A) will have less voltage drop than a lower rated (eg 50A) diode of the same type.
Schottky diodes have lower voltage drop than normal silicon rectifiers. (I've not seen a 250A Schottky diodes)
Switched MOSFETs can have very low voltage drops.

Currents on the transformer side of the capacitor are peaky/pulses so the rms currents will be higher than the (average or rms) DC current.
Your measurements are consistent with reality

Twospoons - 12-8-2024 at 13:12

Your clamp meter is reading correctly - on the ac side it is higher than your DC load current because it is also seeing the reactive current flowing in your filter capacitors, especially the capacitor just after the rectifier.

Attached is a spice simulation with L and C values similar to your setup, showing the current waveforms around the circuit and their RMS values.
Note the AC RMS current is double the DC load current!
Also note the magnitude of the diode peak currents, as compared to your
output current.
I would reduce the first capacitor by a factor of 10 to 6800uF and increase the second capacitor to 120000uF. That will halve the diode peak current while still keeping the output ripple low.
Also the Pi filter does not have an 'e' - its named after the Greek letter π, not the pastry.

[Edited on 12-8-2024 by Twospoons]

[Edited on 12-8-2024 by Twospoons]

Rainwater - 12-8-2024 at 18:33

Quote: Originally posted by yobbo II

What choices do i have for a rectifier?

many.

Quote:

Is there rectifiers that have less V drop than the usual silicon diode ones.

Yes, your forgetting, we live in the future.
its been around for a while in welders.
Its call an active rectifier. But basically it is a full bridge with the diodes replaced by mosfet transistors.
forward resistance can be in the order of 10^-4 if the $$$ is right(requires LN cooling) but typical values are 10^-1
A lot less than the normal 0.6V drop of a diode.
wiki
Youtube
And not as many limitations.

If you can force your input A/C frequency up to around 300KHz, you can reduce the size of that transformer

[Edited on 13-8-2024 by Rainwater]

[Edited on 13-8-2024 by Rainwater]

yobbo II - 13-8-2024 at 11:45

Can i use one of there as a rectifier?

https://www.ebay.co.uk/itm/196553357725?itmmeta=01J56KXM2JTC...

2AAAOSwU9hmt3kb&itmprp=enc%3AAQAJAAAA4HoV3kP08IDx%2BKZ9MfhVJKn%2FEmx06xzAZI2LxwOzULg2uREWKc%2BvgrHTodwb3NanWj%2Fu7RgFyx69JEoaHDeMEwhBGpbBbq4RxWa JhTLVZXijreLx0b7cPbsQPnbPaIaxSg01qBgRQ2SYKj%2BCBXsimjG39idJ6JZH95Tu97Urg--lPY7ssoEgV72Y00UrrUd8Tq2YhjSXUnH0Hu02YvMcrdd7F6fUt6OFbW0523GFTnjrxLY408hU0cR 6wMpQh3a4NVqvlKW%2BZ1I3TQVSQgQWsL8GVlQk5Xvr32rmr8Hv2ZzF%7Ctkp%3ABk9SR6rB9tOpZA

Descrilfion below
Feature:

1. This is a MOSFET high current (50A) H‑bridge driver module
2. With the microcontroller PWM signal isolation, effectively protect microcontroller
3. To achieve the motor forward and reverse rotation, two PWM input maximum 200kHz frequency
4. 3.3V to 12V power on average use, fully compatible
5. The power supply voltage 5V to 15V

Specification:

Description:
The drive uses a full H-bridge driver module full-bridge MOSFET driver chip and composed with very low internal resistance and high current. MOSFET H-bridge driver circuit, with strong drive and braking effect, signal isolation chip isolates motor drive effectively. 3.3V to 12V, PWM level is fully compatible. High current 50A.

Features:

This is a MOSFET high current (50A) H-bridge driver module.
With the microcontroller PWM signal isolation, effectively protect microcontroller.
To achieve the motor forward and reverse rotation, two PWM input maximum 200kHz frequency.
3.3V to 12V power on average use, fully compatible.
Power indicator is schematically clear.

Specifications:

Max. Cu(rrent: 50A

Circuit Structure: H Bridge

Size: 4.3*4.8*2cm/1.7*1.9*0.8in
P

Package Includes:

1 x Motor Driver

Rainwater - 13-8-2024 at 13:05

So long as the maxium reverse drain source voltage (VDS) is not exceded.
But you will need a driver circuit.
Something like the TEA2209T. Ive worked with it before