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'[EE] How does this isolated supply voltage circuit'
2011\09\16@195349 by V G

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I want to be able to generate a low current isolated supply (for driving my
solid state relay MOSFETs) gate-to-source. I googled and found this:
http://www.maxim-ic.com/app-notes/index.mvp/id/1932

1. How does this circuit work? I'm guessing the 74HC14s give a sharp signal
to U1 whose outputs oscillate signals into high pass filters C1 and C2. At
that point, I'm guessing the signal looks more like a sine wave which passes
through the full wave rectifier comped of those 4 diodes and the capacitor.
Tell me if I'm wrong.

2. Is there a simpler way to generate an isolated supply

2011\09\16@202516 by peter green

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V G wrote:
> I want to be able to generate a low current isolated supply (for driving my
> solid state relay MOSFETs) gate-to-source.
When designing an isolated power supply your first task is to define the isolation parameters.

How much leakage is acceptable?
What frequencies do you need to isolate at? DC? 50Hz? higher? (in particular remember that a sudden step change is like a high freqency signal)
What are the consequences of an isolation failure


{Quote hidden}

Now the downside of this circuit is that while it provides good isolation against steady DC voltages it's not good at providing isolation against step changes or AC voltages.

> 2. Is there a simpler way to generate an isolated supply?
If you want good isolation you really need to look at transformer based soloutions and since transformers don't pass DC a transformer based soloution is likely to be roughly similar in complexity

2011\09\16@212050 by Dwayne Reid

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At 05:53 PM 9/16/2011, V G wrote:
>I want to be able to generate a low current isolated supply (for driving my
>solid state relay MOSFETs) gate-to-source. I googled and found this:
>http://www.maxim-ic.com/app-notes/index.mvp/id/1932
>
>1. How does this circuit work? I'm guessing the 74HC14s give a sharp signal
>to U1 whose outputs oscillate signals into high pass filters C1 and C2. At
>that point, I'm guessing the signal looks more like a sine wave which passes
>through the full wave rectifier comped of those 4 diodes and the capacitor..
>Tell me if I'm wrong.

Nope - you are pretty much bang-on.  However, the signal coupled through the capacitors is a square wave, not sine.

You can think of the coupling capacitors as high-pass filters but that is NOT their intent and might cause confusion at a later date.  Rather, they present a low impedance at the high switching frequency used, and therefore couple most of the energy to the load.

The coupling capacitors present a fairly high impedance (~13k) at line frequency (60 Hz).

There are several down-sides with the circuit as presented.

1) the rectifier diodes drop the available voltage significantly.  Note that the app note states that this circuit will supply 3v3 from 5V input.  You would have to use a voltage doubler-type circuit if you needed to get an output somewhere near 5V.  Or: run the left side of the circuit from a supply voltage significantly higher than 5V.  You would need to change the 74HC parts to 74C or 4000 family parts if you do that.  The MAX628 is most likely rated to run from 12V or higher.

2) the coupling capacitors will couple unwanted high-frequency energy across the isolation barrier.

3) the amount of energy coupled across the isolation boundary at line frequency is actually fairly high.  If you were relying upon this to provide isolation from a line-operated supply, the leakage current could be as high as 1mA from a 120V supply.  That is well above the 'ouch' level.  (assumes 10% accurate capacitors)


>2. Is there a simpler way to generate an isolated supply?

Several.

1) Use a transformer.  I routinely use really inexpensive Common Mode chokes as isolation transformers.  Advantage: really high isolation voltage, really low coupling capacitance.  Disadvantage: drive circuit can be complex.  However, google for "Build Your Own Ultra-Low-Cost Isolated DC-DC Converter"  for a really low cost implementation: <http://electronicdesign.com/article/power/build_your_own_ultra_low_cost_isolated_dc_dc_converter.aspx>

Disclosure: its my design - and it works REALLY well.

2) Use photo-voltaic opto-isolators.  Feed current into the input LED, get up to 10 or 12Vdc out at several uA.  Advantage: really high isolation voltage, really low coupling capacitance, really simple.  Disadvantage: only capable of supplying uA current levels.

dwayne

-- Dwayne Reid   <spam_OUTdwaynerTakeThisOuTspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

2011\09\16@213837 by V G

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On Fri, Sep 16, 2011 at 9:20 PM, Dwayne Reid <.....dwaynerKILLspamspam@spam@planet.eon.net> wrote:

{Quote hidden}

1. How do you calculate the impedance of a capacitor at a particular
fequency?

2. I thought the definition of a filter is to present low/high impedance at
low/high/high/low frequencies.

There are several down-sides with the circuit as presented.
{Quote hidden}

Wow, that photo voltaic isolator method was right under my nose all along.
Perfect for driving my MOSFET SSR. One chip should provide the voltage
necessary to drive a few SSRs (I'm guessing at least 10) at slow speeds

2011\09\17@021135 by Oli Glaser

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On 17/09/2011 02:38, V G wrote:
> 1. How do you calculate the impedance of a capacitor at a particular
> fequency?

1/(2pi * C * f)

e.g for a 100nF capacitor at 60Hz:

1/(2pi * 100e-9 * 60) = 26.525k

2011\09\17@023100 by cdb

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On Fri, 16 Sep 2011 21:38:22 -0400, V G wrote:
:: 1. How do you calculate the impedance of a capacitor at a
:: particular fequency?

Xc = 1/( 2 * pi  * f * C )

I'll let you transpose.

Colin
--
cdb, colinspamKILLspambtech-online.co.uk on 17/09/2011
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2011\09\17@023829 by cdb

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And just think, bung an Arduino or similar on to that and you'd be able to measure frequency, display a fancy graph and pump the info out to mass storage or a PC.

Easy for some some researcher to knock up and use, providing they had the " intelligence" (humour) to realise they could do this without bothering the EE or the Bio/Pre Med types at University.

Colin
--
cdb, .....colinKILLspamspam.....btech-online.co.uk on 17/09/2011
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2011\09\17@150640 by V G

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On Sat, Sep 17, 2011 at 2:38 AM, cdb <EraseMEcolinspam_OUTspamTakeThisOuTbtech-online.co.uk> wrote:

> And just think, bung an Arduino or similar on to that and you'd be able to
> measure frequency, display a fancy graph and pump the info out to mass
> storage or a PC.
>

$ define bung
*bung*/bəNG/
Noun: A stopper for closing a hole in a container.
Verb: Close with a stopper: "the casks are *bunged* before delivery".

I don't understand.


> Easy for some some researcher to knock up and use, providing they had the "
> intelligence"


Nothing to do with intelligence.


> (humour) to realise they could do this without bothering the
> EE or the Bio/Pre Med types at University.
>

I don't get it :(

2011\09\17@175000 by IVP

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> $ define bung

UK - put in, shov

2011\09\18@072322 by Electron

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At 03.38 2011.09.17, you wrote:
{Quote hidden}

It's called capacitive reactance (XC), and it's obviously measured in ohms:

XC = 1 / ( F*Hz*Pi2 )

Where F is Farad of the capacitor, Hz is the frequency for which you wanna
know the impedance, and Pi2 is roughly 6.28

Cheers,
Mario

2011\09\19@161941 by Dwayne Reid

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At 12:11 AM 9/17/2011, Oli Glaser wrote:
>On 17/09/2011 02:38, V G wrote:
> > 1. How do you calculate the impedance of a capacitor at a particular
> > fequency?
>
>1/(2pi * C * f)
>
>e.g for a 100nF capacitor at 60Hz:
>
>1/(2pi * 100e-9 * 60) = 26.525k

Yep.  Then note that there are two capacitors that bridge the isolation boundary - treat them as being in parallel.  Thus my statement of ~13k at 60Hz.

dwayne

-- Dwayne Reid   <@spam@dwaynerKILLspamspamplanet.eon.net>
Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
(780) 489-3199 voice          (780) 487-6397 fax
http://www.trinity-electronics.com
Custom Electronics Design and Manufacturing

2011\09\20@035055 by RussellMc

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> Yep.  Then note that there are two capacitors that bridge the
> isolation boundary - treat them as being in parallel.

In series, I think, in this context.

_____________

Added factor here is ratio of pump capacitor to floating store cap.
Affects delta change which output can achieve per cycle. (Obvious but
can cause issues in some applications). Here where a power supply is
being created there is no problem. For using this as a FET gate drive
the response time will be affected. (Again, obvious, but ...)
________________

The higher the pump frequency the lower that the capacitor sizes can
be and the higher the achievable mains rejection.

__________

This circuit is useful for eg powering the simple digital panel meter
IC's which demand that the IC float at the measuring voltage
potential. Many such panel meters are offered for sale without enough
information provided on how to power them.





                Russell

2011\09\20@061330 by peter green

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RussellMc wrote:
>> Yep.  Then note that there are two capacitors that bridge the
>> isolation boundary - treat them as being in parallel.
>>    
>
> In series, I think, in this context.
>   They are in series from the perspective of the power being transmitted by the circuit but they are in paralell from the perspective of leakage current across the isolation barier.

The biggest problem I see with this circuit though is it's transient response. A rapid change in the potential difference between the isolated sides (say a spike on the mains if one side is tied to mains live and the other to mains earth) will cause a spike of current through the capacitors.

Remember the real equation for a capacitor is Q=CV differentiating this with respect to time and knowing that current it is the rate of change of charge gives us I=C(dV/dt) (for those who stopped studying maths before taking calculus this means that the current though a capacitor is proportional to the rate of change of voltage across the capacitor

2011\09\20@062945 by Michael Watterson

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On 20/09/2011 11:13, peter green wrote:
> Remember the real equation for a capacitor is Q=CV differentiating this
> with respect to time and knowing that current it is the rate of change
> of charge gives us I=C(dV/dt) (for those who stopped studying maths
> before taking calculus this means that the current though a capacitor is
> proportional to the rate of change of voltage across the capacitor)

i.e. for a really sharp spike the capacitor looks like a short circuit no matter what the voltage is.

Also capacitors are more prone to short circuit isolation failure than a transformer or optocoupler.  Inherently an unhappy solution for mains isolation

2011\09\20@091659 by Electron

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One problem I see with non isolated capacitor-coupled PSU's is that if
a transient hits the line, then the capacitor will happily pass it..


At 09.50 2011.09.20, you wrote:
{Quote hidden}

>

2011\09\20@160840 by Richard Prosser

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On 21 September 2011 01:16, Electron <KILLspamelectron2k4KILLspamspaminfinito.it> wrote:
>
> One problem I see with non isolated capacitor-coupled PSU's is that if
> a transient hits the line, then the capacitor will happily pass it..
>
>

This can be mitigated by adding a shunt capacitor  - effectively
making the input a capacitive voltage divider. Or, more commonly by
using a zener or TVS diode in the same position.

R

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