Galvanic isolation without transformers ??

Could some electronics guru please help ? Maybe this is a very silly question, so please pardon me. A voltage transformer is used to provide galvanic isolation, in a power supply, amongst other features. Is there another way to achieve galvanic isolation ? Thanks in advance.

Reply to
dakupoto
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Don't know if I could be called a guru, but consider what "galvanic isola tion" means. It means there's no direct current path between two points.

There aren't a lot of *feasible* ways to do this that don't involve a tra nsformer, especially if power transference is the goal. You could go with a Great Big LED on the supply side and a Sufficiently Efficient PV cell on t he load side, but that makes feedback difficult. You could go the motor-gen erator (AKA dynamotor) route, or use a big speaker-dynamic microphone combo .

The common thing is to convert electric power (embodied as current flow) to some intermediate form (light, torque, sound, etc.) that can be reconver ted. Transformers convert AC current to changing magnetic fields and back, and transformers are literally the most efficient machines ever devised. Wh en transferring power, efficiency matters a lot, which is why you don't oft en see speakers and mikes (and the other examples I mentioned) in power sup ply component catalogs.

Motor-generators can approach transformers in efficiency in certain circu mstances but are primarily used to convert DC to AC or to do phase/frequenc y conversion, harmonic rejection, and other tricks rather than specifically just provide isolation.

Do you have a good reason not to want to use a transformer, or were you j ust curious?

Mark L. Fergerson

Reply to
Alien8752

** I guess you could say 93 million miles of space provides us with galvanic isolation from the sun - though allowing plenty of EM radiation to arrive and do good things.

A electric motor /generator set isolates the original AC or DC supply from the output too.

One could use air pressure too, by bolting two speakers face to face.

.... Phil

Reply to
Phil Allison

When your purpose is to convey a signal of some type, it is not uncommon to use capacitors to block DC and low frequency signals while chopping the signal of interest to turn it into a higher frequency signal which will pass through the capacitors easily. This will require circuitry on both sides of the interface and so is not as simple as a transformer. But transformers can be bulky, heavy and cost more than a capacitor based approach. Capacitive isolation is usually accomplished by using a pair of chips designed for this purpose, sometimes both on a module in a single package.

--

Rick
Reply to
rickman

Well, or AC,one might use a capacitor..and if a DC level needs to be "restored", then modulation may be useful..

Reply to
Robert Baer

Photo coupling?

Jamie

Reply to
Maynard A. Philbrook Jr.

Is this for power transfer or just for sensing?

Reply to
John S

Actually both the Sun and the Earth are immersed in a conducting medium (the solar wind) so their relative potential is very nearly zero. (Feynman has a lecture on that.)

The atmosphere is usually a pretty decent insulator, of course, but it's the last few miles that do the isolating, not the first 93 million.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net
Reply to
Phil Hobbs

Galvanic isolation in what? If you just want to achieve galvanic isolation between two parts, then don't let them touch. If you want to transmit signals from one to the other while maintaining galvanic isolation, that's a different problem. If you want to conduct energy from one point to another while maintaining galvanic isolation, that's yet another problem.

So, tell us what your question is, and maybe you'll get a sensible answer or two.

--
www.wescottdesign.com
Reply to
Tim Wescott

On a sunny day (Thu, 27 Nov 2014 12:15:53 -0500) it happened Phil Hobbs wrote in :

Has anybody mentioned piezo electric transformers yet?

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can be made very small.

Reply to
Jan Panteltje

--
It's entirely possible to isolate a load from the mains using the 
dielectric of a couple of capacitors as the isolating medium, but 
whether it's practical is another matter. 

Do you have an application in mind or is your interest merely 
academic?
Reply to
John Fields

Powered lamp, air gap, photovoltaic "solar panel" Motor, belt/hydraulic pump/nonconductive shaft, generator Buzzer, ceramic rod, contact-microphone-and-rectifiers Hydrolysis apparatus, low-pressure gas tubing, fuel cell Heat pump, circulating air or fluid, thermopile Battery factory...

Reply to
whit3rd

What, No Aether ?

Jamie

Reply to
Maynard A. Philbrook Jr.

Yep; one of many ways.

Reply to
Robert Baer

Level of power unstated.

Reply to
Robert Baer

Thanks to each of you for your responses. My interest is mostly application/work related and a little academic. We build, amongst other things, power supplies of all varieties -- linear, switched mode( full galvanic isolation) and a few very low current output(20 - 30 mA) reactive ones(zero galvanic isolation). Recently, we were discussing the general issue of galvanic isolation, and one of the items discussed was capacitive isolation. Indeed, it is very much possible to provide galvanic isolation with capacitors for AC, and in fact, capacitive reactance may be used to reduce high mains side current(5A) on manageable isolated load side, as well. May be this scheme could be used in some low power power supplies. Surely this would reduce the time needed to wind transformer coils.

Reply to
dakupoto

** You are talking complete bollocks.

Capacitive *galvanic* isolation from the AC supply only works for signals - not power.

The largest value cap you can safely and legally connect to the AC supply and have a human handle the other end is about 2nF. Even with an earthed metal enclosure, the largest allowed value is about 5nF.

There is a very simple reason PSUs all use magnetics to achieve isolation.

... Phil

Reply to
Phil Allison

Sure you can use capacitive coupling, especially with balanced feed and using protocols, not containing a DC component (Such as Manchester coding).

If you are going to measure live line currents, what is wrong with a traditional current transformer ?

To do current measurements, you would need a shunt resistor and then couple the difference voltage and drop down to some "safe" levels by capacitors will cause a lot more headache compared to a current transformer. With a current transformer, you get all this for "free" even without a significant voltage drop.

Reply to
upsidedown

--- While capacitors are attractive for use as series voltage droppers because of their contribution to the circuit's impedance being lossless, the problem which quickly rears its ugly head is that for substantial current into a load the capacitive reactance must fall to a low value.

That means a large capacitance must be used to feed a moderately high-powered load, which quickly gets to be a more expensive proposition than using a transformer.

For example, consider 120 volt 60Hz mains being used to drive a

12V/12mA load through two capacitors for "perfect" isolation from both sides of the mains:

C1

120L>-----[CAP]----+ |R1 [1000R] C2 | 120N>-----[CAP]----+

FIG 1.

In order to make life easier, though, we can combine C1 and C2 -for now - and have our circuit look like this:

E1 E2 / C1 /

120L>-----[CAP]----+ |R1 [1000R] | 120N>--------------+

FIG 2.

Now, since the current in a series circuit is everywhere the same in that circuit and we want the 12 mA through R1 to drop 12 volts out of the 120 available from the mains, we can find the value of the capacitor required to drop the remaining 108 volts, with 12 mA through it, by first considering the impedance, Z, of the circuit, where:

E1 120V Z = ---- = -------- = 10000 ohms I1 0.012A

and, since E1 and E2 aren't in phase and are driven by sinusoidal mains by considering that:

= 100e6 - 1e6

= 99e6 ohms

Then, since Xc is the square root of that,

Xc = sqrt 99e6R ~ 9950 ohms.

and, since:

1 Xc = --------- 2pi f C

we can rearrange to solve for the capacitance:

1 C = ---------- 2pi f Xc 1 = --------------------- 6.28 * 60Hz * 9950R

~267nF

Finally, to return to the configuration in figure 1, the capacitors are equal-valued and in series, so the value of each must be twice the value of the single cap shown in figure 2, or 534nF each.

That's about half a microfarad each for a 12mA load, so if the load

each for a 1200mA load.

Because of that and because of the safety issues surrounding using caps for off-line voltage droppers, transformers start looking better than caps somewhere between three and ten VA, I'd guess, with anything notmuch higher than that being no contest.

John Fields

Reply to
John Fields

--
What might that be? 

John Fields
Reply to
John Fields

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