What does this circuit do ?

[gerry maddenParticipant @gerrymadden53711]

[gerry maddenParticipant @gerrymadden53711]

Hi All,

I pulled this 'black box' out of my microlight which I installed in it about 25 years ago. I'd always assumed it was a small voltage regulator but when I now look inside I don't think it is. I checked this by putting my benchtop PS on the input side and voltmeter and load on the output side. Regardless of whether the output is open-circuit or given a 0.2A drain, the output voltage is the same as the input.

So what does it so ? I imagine the capacitor is there to smooth things. But the diode (I assume thats what it is) …. is it just there to protect for accidental polarity reversal ?

By the way, this 'device' would normally feed into the intercom system.

Gerry.

[EmgeeParticipant @emgee]

Gerry

Does as you describe plus provides a fused connection to your intercom device.

Emgee

[not done it yetParticipant @notdoneityet]

Look, to me to be nothing more than a voltage smoothing device.

[Stuart Smith 5Participant @stuartsmith5]

Gerry,

It is to stop a dip on the supply to the intercom, presumably when the engine is started.

I made a similar circuit for my car when I had a 2 way radio for work. Before I fitted this, the radio would cut out and have to be reset every time I started the engine.

The large capacitor holds the voltage stead for a short time if the supply voltage dips and the diode isolates the capacitor and intercom from the supply.

Stuart

[AdrianRParticipant @adrianr18614]

Another possibility is that the engine may be producing a pulsed supply i.e. does not have an alternator but is using a simpler generator and it needs more smoothing than the battery can give to reduce noise on the intercom.

[AJAXParticipant @ajax]

The circuit is a half-wave rectifier with capacitive smoothing, but in this application (DC supply) the diode protects against reverse bias and the bulk capacitor improves load regulation by reducing ripple.

The open circuit voltage should be around 0.7V greater than the output voltage under load. Consider the IV characteristics of the diode to understand why.

[SillyOldDufferModerator @sillyoldduffer]

10,000uF is a lot of capacity to put on an existing DC supply and is unnecessary if the diode is only there to protect against a reversed connection.

What 10000uF would usefully do is a basic smoothing job on the choppy DC that comes out of a car battery charger and many simple transformer wall-warts. It would even sort out an AC input. I expect that was its purpose; providing some reasonably smooth DC.

The fuse ought to be on the input side. As is, if the 10000uF capacitor shorts out, as electrolytics are prone to do, then the capacitor will go pop, possibly blowing the rectifier and igniting the unfused cable too.

Dave

[not done it yetParticipant @notdoneityet]

All the diode does is prevent the voltage dropping suddenly if the input voltage drops, by back-feeding. No rectification here. Ripple on the DC input, yes, but there ain’t two diodes – as required for a half-wave rectifier..

[Stuart Smith 5Participant @stuartsmith5]

See my earlier post.

It is to prevent the voltage to the intercom dipping when the supply dips.

Stuart

[SillyOldDufferModerator @sillyoldduffer]

Posted by Stuart Smith 5 on 26/05/2022 20:00:50:

See my earlier post.

It is to prevent the voltage to the intercom dipping when the supply dips.

Stuart

Ah, I missed the Microlite!

Dave

[Nicholas FarrParticipant @nicholasfarr14254]

Posted by not done it yet on 26/05/2022 19:16:06:

All the diode does is prevent the voltage dropping suddenly if the input voltage drops, by back-feeding. No rectification here. Ripple on the DC input, yes, but there ain’t two diodes – as required for a half-wave rectifier..

Hi, well according to this, you only need one diode for a half wave rectifier and Gerry's diagram looks a lot like the bottom one.

Regards Nick.

[AJAXParticipant @ajax]

Posted by SillyOldDuffer on 26/05/2022 18:24:55:

10,000uF is a lot of capacity to put on an existing DC supply and is unnecessary if the diode is only there to protect against a reversed connection.

What 10000uF would usefully do is a basic smoothing job on the choppy DC that comes out of a car battery charger and many simple transformer wall-warts. It would even sort out an AC input. I expect that was its purpose; providing some reasonably smooth DC.

The fuse ought to be on the input side. As is, if the 10000uF capacitor shorts out, as electrolytics are prone to do, then the capacitor will go pop, possibly blowing the rectifier and igniting the unfused cable too.

Dave

I can't quite read the capacitor markings, but it may well be 1000uF and not 10000uF (10mF).

I agree with your comment regarding fuse on the supply side, but if the capacitor is large there may be an inrush current sufficiently large to blow the fuse.

[not done it yetParticipant @notdoneityet]

Posted by AJAX on 26/05/2022 22:38:29:

I can't quite read the capacitor markings, but it may well be 1000uF and not 10000uF (10mF).

I agree with your comment regarding fuse on the supply side, but if the capacitor is large there may be an inrush current sufficiently large to blow the fuse.

I can’t either, but I expect Gerry has read it. He may have made a mistake in writing it down, but there is less chance of that than a simple keying-in typo.

[BazyleParticipant @bazyle]

You also mentioned the box has a current drain even when open circuit. This is to be expected with an old and large capacitor. Often with equipment with well under than 25 years on the clock you need to replace the capacitors to restore proper performance or even avoid fatal damage if trying out your old 1980s computer.

[Michael GilliganParticipant @michaelgilligan61133]

This might assist with identification of that capacitor value: **LINK**

https://www.ebay.co.uk/itm/324716002691

MichaelG.

[Jon LawesParticipant @jonlawes51698]

The value looks reasonable given the size, which again gives weight to the reservoir capacitor idea.

[Nicholas FarrParticipant @nicholasfarr14254]

Hi, I'm assuming that the diode is a 1N5408 which I have four of and they measure 5.2mm x 9mm, so scaling Gerry's photo, his capacitor works out to 22.5mm x 40mm approximately. This PDF is a data sheet on CPC's web site and shows some physical measurements of various capacitors. I could though, be totally wrong.

Regards Nick.

[gerry maddenParticipant @gerrymadden53711]

Thanks chaps for the excellent food for thought. This device in fed by rectified power (sorry I should have included this) so it must then be for smoothing and volts dips as you have suggested.

Gerry

(ps. the capacitor is 10,000 uF)

[not done it yetParticipant @notdoneityet]

As a matter of interest, from where/how is the rectified supply derived? Most engines, these days, are fitted with a three phase alternator which rectifies the A/C and feeds that power to a battery which will take out any ripple from the alternator line.

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