Spec'ing capacitor and(?) inductor for variable output DC power supply

Thread Starter

George Dennison

Joined Apr 27, 2018
5
I'm just getting started in electronics. I've done some repairs on a number of battery chargers and welders, and misc other projects, but I'm at the stage where I have way more questions about most things, than anything else. I don't have the money to drop on a bench full of equipment, so I'm having to scrounge and build, as I can. I looking for a deal on a scope, but one hasn't come way way, yet.

I came across the guts of a 'Variac' being used to control a bank of ceiling fans. It is a Luxtrol WBD750, (0-125V, 750W, 6. 25A). I was looking at buying a Variac when I was given the Luxtrol, so I changed directions and am, instead, building one.

I have it in a prototype case with both analog and digital gauges, and a variety of connections for output voltage. I'm currently looking for a suitable 1:1 transformer to place between it and the wall voltage, to create an isolation transformer.

I'm a bit cramped for bench space, so I got to thinking about also using this as a DC power supply. I couldn't come up with a reason not to combine the two power supplies into a single unit, but if there is, please let me know what I missed.

I have a rectifier, (1600V, 100A), I didn't use for a welder repair. I figured that would certainly handle anything I put through it, but I'm not sure how to approach calculating the capacitor size for smoothing the rectifier output.

I have done a lot of research on spec'ing a cap for a fixed output, but haven't actually done one, yet. From what I learned, it comes down to formula math based on several data points derived from the application. Seems straight forward.

Something I discovered in my research, was a number of different formulas so, one challenge may be choosing the 'right' formula! (I figured I do calcs with several, and see how close their relative outcomes turn out.)

Regarding the variable power supply, do I use what would be maximum output from the power supply to calc the cap size?

(I wouldn't ever drive anything at a full 6.35A @ 125VAC, but I assume designing at max would give me 'headroom', and adequate function.)

The other question I have is about inductors. Some of the devices I have repaired, (battery chargers, welders), have had induction coils, in addition to capacitors, for smoothing the rectifier output. Do I need to plan on including an inductor?

Any help, or clarifications, (if I'm heading in the wrong direction with some aspect), will be greatly apppreciated.

Thanks, in advance.

GeoD
 

Hymie

Joined Mar 30, 2018
711
I'm just getting started in electronics. I've done some repairs on a number of battery chargers and welders, and misc other projects, but I'm at the stage where I have way more questions about most things, than anything else. I don't have the money to drop on a bench full of equipment, so I'm having to scrounge and build, as I can. I looking for a deal on a scope, but one hasn't come way way, yet.

I came across the guts of a 'Variac' being used to control a bank of ceiling fans. It is a Luxtrol WBD750, (0-125V, 750W, 6. 25A). I was looking at buying a Variac when I was given the Luxtrol, so I changed directions and am, instead, building one.

I have it in a prototype case with both analog and digital gauges, and a variety of connections for output voltage. I'm currently looking for a suitable 1:1 transformer to place between it and the wall voltage, to create an isolation transformer.

I'm a bit cramped for bench space, so I got to thinking about also using this as a DC power supply. I couldn't come up with a reason not to combine the two power supplies into a single unit, but if there is, please let me know what I missed.

I have a rectifier, (1600V, 100A), I didn't use for a welder repair. I figured that would certainly handle anything I put through it, but I'm not sure how to approach calculating the capacitor size for smoothing the rectifier output.

I have done a lot of research on spec'ing a cap for a fixed output, but haven't actually done one, yet. From what I learned, it comes down to formula math based on several data points derived from the application. Seems straight forward.

Something I discovered in my research, was a number of different formulas so, one challenge may be choosing the 'right' formula! (I figured I do calcs with several, and see how close their relative outcomes turn out.)

Regarding the variable power supply, do I use what would be maximum output from the power supply to calc the cap size?

(I wouldn't ever drive anything at a full 6.35A @ 125VAC, but I assume designing at max would give me 'headroom', and adequate function.)

The other question I have is about inductors. Some of the devices I have repaired, (battery chargers, welders), have had induction coils, in addition to capacitors, for smoothing the rectifier output. Do I need to plan on including an inductor?

Any help, or clarifications, (if I'm heading in the wrong direction with some aspect), will be greatly apppreciated.

Thanks, in advance.

GeoD
If you are to use a variac as the source for a dc output, then an isolating transformer is a must, otherwise your dc supply will be mains referenced – and it would be dangerous to contact any part of the dc supply output.

In relation to the value of the smoothing capacitor placed on the output of a bridge rectifier, you can use the following formula to give the resulting ripple voltage on the dc output:-

V = (I x t)/C, where V is the peak-peak ripple voltage, I is the current in amps, t is the time in seconds (half the period of the mains frequency for full wave rectification), and C is the capacitor value in farads.

As an example, if you draw 3 amps and have a 4,700 uF capacitor (based on 50Hz mains) we get

V = (3 x 0.01)/0.0047 = 6.4V

One of my concerns with your proposed set up is that you might accidentally turn the variac fully clockwise, resulting in 340V at your dc output.

Therefore I would recommend that your variac is driving the primary of a transformer with a 48V secondary, which will allow a dc voltage of up to around 70V.
 

crutschow

Joined Mar 14, 2008
23,521
I couldn't come up with a reason not to combine the two power supplies into a single unit, but if there is, please let me know what I missed.
If you want to use them with a common ground, then the DC circuit could only use a half-wave rectifier, which has a high ripple value.
If you don't need a common ground between the two, then you an use a full-wave bridge rectifier, which has lower ripple.
I have a rectifier, (1600V, 100A)
If you only use one as a half-wave rectifier, the ripple will be high.
Do I need to plan on including an inductor?
An inductor can be used to reduce the ripple, but they big and expensive for use in this application.
One of my concerns with your proposed set up is that you might accidentally turn the variac fully clockwise, resulting in 340V at your dc output.
You must be thinking 240Vac, but his Variac is 120Vac so the maximum DC would be about 170V.
 

crutschow

Joined Mar 14, 2008
23,521
Notice that the ripple voltage can be significant for larger currents, even with a large value of filter capacitance.
If that's a problem in your application you might consider some type of capacitance multiplier circuit to act as an electronic filter and reduce the ripple voltage.
 

MrChips

Joined Oct 2, 2009
19,404
The size of the capacitor is dictated simply by how much current your load is taking and how much ripple you are willing to accept. There is a simple formula for this.

Going whole hog with a "one size fits all" philosophy is not a good idea for safety and cost reasons.
It is better to design the power supply to fit the need of your load requirements.
 

-live wire-

Joined Dec 22, 2017
897
Keep in mind, there would be NO REGULATION with that setup. I recommend making a proper SMPS. This is one approach: Get a very stable HVDC rail. I mean 80-150V. Isolate it and make sure it can supply 10-15 amps. Then PWM it at a medium-high frequency and use an L/C low pass filter. Use a current shunt with a good op-amp circuit, and a voltage divider on the output to measure voltage and current. Feed that into your oscilator, add some more control chips, maybe use a microcontroller for a display, and you should be good. You may also want to include a BJT in series with it with a bypass MOSFET. That would allow accurate constant current, in addition to the constant voltage with a current limit. Also add a zener on the output to prevent momentary overvoltages above the maximum voltage for it. Just in case there is some big surge or something.
 

Thread Starter

George Dennison

Joined Apr 27, 2018
5
Hi Hymie,

Thanks for the quick reply and info. I have a couple of questions, though, about your comments.

You wrote the following three statements. My questions follow each statement.

1. =======

"As an example, if you draw 3 amps and have a 4,700 uF capacitor (based on 50Hz mains) we get

V = (3 x 0.01)/0.0047 = 6.4V"

Thanks for the info, but what I'm looking for is the value I should use in the formula for current. Do I use the max amperage rating for the Luxtrol, 6.35 A? I don't have a good enough understanding of capacitors to know if using a capacitor much larger than needed will cause problem(s). IE: If I use 6.35 A, (based on the max capacity of the auto-transformer being 6.35 A @ 125 VAC), will there be problems if I set the auto-transformer to, say, 43 V, and put a 1.25 A load on it.

I guess the question comes down to, is it possible to OVER 'capacitate' the output of a full wave rectifier?

2. =======

"One of my concerns with your proposed set up is that you might accidentally turn the variac fully clockwise, resulting in 340V at your dc output."

When the Luxtrol auto-transformer, (finally remembered the name!), is turned fully clockwise, the output voltage is 120-125v. If the output was 340v, it would be violating a basic law of physics, wouldn't it? It would making something from nothing, isn't that the Law of Conservation; the reason there are no perpetual motion machines, (in spite of SOME YouTube video claims! LOL)

3. =======

"Therefore I would recommend that your variac is driving the primary of a transformer with a 48V secondary, which will allow a dc voltage of up to around 70V."

You have me a bit confused, (more than a bit, actually), with this statement. If a transformer with a 48v secondary output is used, how do get the 70v? That's equivalent to over driving the transformer by almost 50%. (assuming it is a 120v primary, 48v secondary transfrormer, that would mean having to put about 170v into the primary. I was reading about transformers just this morning, and they talked about risk of 'saturation', (I think that was the term used), of the transformer when they are overdriven. There wasn't much in the way of specific effect of saturation, but the indication was it was negative, and should be avoided.)

Where do you come up with the 70v value?


Thanks,

GeoD
 

Thread Starter

George Dennison

Joined Apr 27, 2018
5
Hi Crutschow,

Thanks for the quick reply.

I have questions about the two statements you wrote. My question(s) follow each quote.

1.======

"If you don't need a common ground"

I don't know if I need/want a common ground. Why would a person want or need either a common ground, or to NOT have a common ground? How would having one or the other impact use of the power supply in general electronics work/repair?

2.======

"An inductor can be used to reduce the ripple"

Something I have NOT found in my reading & research is if EITHER a capacitor or an inductor can be used, individually, to smooth rectified AC waves, or if it is better to use BOTH, for better results.

---------

It's a challenge to do mess around with the effect of caps & inductors on rectifier output without a scope.

If I had one, I could experiment with capacitors & inductors, individually, and in combo, to see just how much better the wave gets with different setups.

Thanks for you input.

GeoD
 

Thread Starter

George Dennison

Joined Apr 27, 2018
5
Hi Live Wire,

Thanks for your reply, and comments.

I have either questions or comments about your statements, they follow each quote.

1.=====
"Keep in mind, there would be NO REGULATION with that setup."

I was about to answer this with, 'except for the voltage regulator I plan on installing', but it looks like the early draft of my original post DID mention adding a voltage regulator to the DC output, but it looks like I accidnetally deleted the statement.

I DO plan on adding a DC regulator for fine-tuning the DC output. I figured it would be essential for any kind of low voltage DC work I may do on the bench. (My thinking is the AC output should be fine straight out of the auto-transformer, and higher voltage DC should be OK unregulated, (for bench testing DC motors, for example), but it seemed to me if I was working on a 3V, or 5V DC circuit, I need as much accuracy as I can get. If my thinking is haywire, please let me know.)

2.=====
"proper SMPS", "HVDC rail", "PWM it at a medium-high frequency", "L/C low pass filter", "current shunt", "op-amp circuit", "voltage divider", "BJT"...

When I have the time to look up the meaning of the acronyms, figure out what some of these are, and how they work, I'll get back to you about your suggestion. Maybe you skipped over the first sentence in my original post: "I am just getting started in Electronics." LOL...

I appreciate your suggestion and know it was well intended, but when it is so far above someone's knowledge and experience-wise, it may as well be in a foreign language. (For over 30 years, I was in the academic world. I wrote and edited research papers, technical articles and instruction and how-to manuals. A habit drilled into me, in my early days, was to always define an acronym the first time it is used. It helps your intended readers comprehension, tremendously, especially if they are new to a field.)

At the stage of learning electronics I'm at, I simply need an AC power supply safer than a power strip plugged into a wall outlet, and capable of delivering less than full line voltage. I also need a DC power supply I can connect a DC motor to, for a bench test, and also for supplying various lower DC voltages for learning more about electronics. I need something I can turn on and adjust, instead of gathering enough batteries together, (and making sure they are charged), and connecting them through a regulator, or using a battery charger, DC welder or several car batteries in series, in an attempt to hit the DC voltage I need.

While the NiMH/Lithium battery or car battery or battery charger/welder options can be highly exciting, they are very time consuming; reducing 'learning' time drastically, AND increasing the amount of time to get anything done, tremendously.

Something we seldom, if ever, think about, or experience when we are young is a sense of 'time running out'. After you pass 60, it becomes much more a part of one's life. I only have so much time left to learn electronic, and utilize the knowledge.

A big challenge to learning electronics, (especially if you are self-teaching), is having the equipment to learn WITH. I'm 65, and disabled with numerous physical problems, as well as cognitive problems related to a 2003 brain injury.

Combine those with very much being a 'hands-on' learner, and the need for an electronics bench becomes imperative.

Thanks for your suggestions and thoughts.

GeoD
 

Thread Starter

George Dennison

Joined Apr 27, 2018
5
Hi MrChips,

Thanks for your reply, and suggestions.

My comments or questions follow each of your quoted statements.

1.=====

"There is a simple formula for this."

Yes, as I said in the original post, calculating cap capacity seems like a straight forward calculation.

What I am unsure of is if there are any peculiarities to calculating cap capacity for a variable output transformer, like an infinitely adjustable auto-transformer. Do I just use the maximum amperage value of the unit?

As I asked, in a reply to someone else, can problems arise from 'over-capacitancc' the output of a rectifier? In other words, if I put a big enough cap to handle the max output of 6.35 A @ 125 VAC going into the rectifier, will I have problems when I dial the transformer down to 5 V?

2.=====

"Going whole hog with a "one size fits all" philosophy is not a good idea for safety and cost reasons."

What would be examples of safety reasons for NOT adding DC output to an AC output bench power supply?

3.=====

"It is better to design the power supply to fit the need of your load requirements."

When related to a specific use, such as a power supply for a DC motor on a lathe, or a power supply for a bank of LED lights and cooling fans, I would agree completely with your statement, but I just don't understand how it applies when the load requirement is variable AND the voltage requirement is anywhere from a fraction of a DC volt, to 80-90VDC AND AC voltage needed could range from 10-125 V, as in an electronic bench work environment.

When I read articles/watch vids about setting up an electronics bench, the basics generally include a Variac and a DC power supply. If I can safely combine the two pieces of equipment and get good results, why not? I haven't looked, but I imagine if I had a grand or three to spend, I could probably buy a very nice AC-DC variable output power supply. They are made, aren't they?

I already plan on an isolation transformer to increase safety, (though the biggest safety factor is NOT touching the hot 'bits'), and though I accidentally left it out of the final version of my post, I do plan on a voltage regulator on the DC output for better voltage control, especially at low values.

I do appreciate your input, as I do all the suggestions and comments. I learn something from each one. If there is a something I'm missing here, due to my knowledge or experience, I certainly want to hear what it is.

With that said, the reason(s) do need to be more than 'don't do it', or 'it's not a good idea'; otherwise NO ONE learns anything from the post.

Thanks, again.

GeoD
 

crutschow

Joined Mar 14, 2008
23,521
I don't know if I need/want a common ground. Why would a person want or need either a common ground, or to NOT have a common ground?
What I meant was, with a bridge rectifier you can't use both the AC and DC outputs to a circuit with the same ground.
You can only use one or the other.
Something I have NOT found in my reading & research is if EITHER a capacitor or an inductor can be used, individually, to smooth rectified AC waves, or if it is better to use BOTH, for better results.
You can use either just a capacitor (which is the most common approach) or an inductor followed by a capacitor, which can give low ripple voltages with smaller value capacitors. You never use just an inductor by itself.
But inductors are seldom used because of their high cost, size, and weight.
It's a challenge to do mess around with the effect of caps & inductors on rectifier output without a scope.
You might try using the free LTspice simulator from Linear Technology/Analog Devices.
That way you can readily experiment with different filters and loads to see their effect on the ripple voltage.
It has a somewhat steep learning curve but there are some good tutorials and examples to help you get started. I think you will find it's well worth the effort.
 

Hymie

Joined Mar 30, 2018
711
Hi Hymie,

Thanks for the quick reply and info. I have a couple of questions, though, about your comments.

You wrote the following three statements. My questions follow each statement.

1. =======

"As an example, if you draw 3 amps and have a 4,700 uF capacitor (based on 50Hz mains) we get

V = (3 x 0.01)/0.0047 = 6.4V"

Thanks for the info, but what I'm looking for is the value I should use in the formula for current. Do I use the max amperage rating for the Luxtrol, 6.35 A? I don't have a good enough understanding of capacitors to know if using a capacitor much larger than needed will cause problem(s). IE: If I use 6.35 A, (based on the max capacity of the auto-transformer being 6.35 A @ 125 VAC), will there be problems if I set the auto-transformer to, say, 43 V, and put a 1.25 A load on it.

I guess the question comes down to, is it possible to OVER 'capacitate' the output of a full wave rectifier?

2. =======

"One of my concerns with your proposed set up is that you might accidentally turn the variac fully clockwise, resulting in 340V at your dc output."

When the Luxtrol auto-transformer, (finally remembered the name!), is turned fully clockwise, the output voltage is 120-125v. If the output was 340v, it would be violating a basic law of physics, wouldn't it? It would making something from nothing, isn't that the Law of Conservation; the reason there are no perpetual motion machines, (in spite of SOME YouTube video claims! LOL)

3. =======

"Therefore I would recommend that your variac is driving the primary of a transformer with a 48V secondary, which will allow a dc voltage of up to around 70V."

You have me a bit confused, (more than a bit, actually), with this statement. If a transformer with a 48v secondary output is used, how do get the 70v? That's equivalent to over driving the transformer by almost 50%. (assuming it is a 120v primary, 48v secondary transfrormer, that would mean having to put about 170v into the primary. I was reading about transformers just this morning, and they talked about risk of 'saturation', (I think that was the term used), of the transformer when they are overdriven. There wasn't much in the way of specific effect of saturation, but the indication was it was negative, and should be avoided.)

Where do you come up with the 70v value?


Thanks,

GeoD
In answer to your specific questions, using the formula provided, if your circuit draws 1.25A, with the 4,700uF capacitor, then the ripple voltage will be V = (1.25 x 0.01)/0.0047 = 2.7V.

Since once you have built your circuit you cannot easily change the capacitor value, you need to consider the maximum current that you want to draw from the supply, and then decide what the maximum ripple voltage you are willing to have on the dc voltage.

e.g. assuming you will limit the output to 2A and want a maximum ripple voltage of 1V, then by transposing the equation we get C = (2 x 0.01)/1 = 20,000uF.

The only practical limit on the capacitor value is the peak current that the supply can deliver; most power supply circuits should be OK with a 100,000uF capacitor.

My 340V figure was based on the use of 240V mains, if you have an input of 120Vac to your variac, then the 120Vac smoothed rectified voltage will be around 170Vdc.

So to avoid having 170Vdc on your supply output due to turning the variac fully clock-wise; if the variac output feeds a 120Vac primary transformer with a 48Vac secondary – 48Vac smoothed rectified voltage will be around 70Vdc (with the variac fully clock-wise).

This is not an idea circuit since the primary of the transformer will be supplied at a voltage varying between 0 – 120Vac (depending on the variac setting), whereas it is designed for 120Vac.
 
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