From: John on
I have an electric shutter. The circuit works the shutter when testing
with a simple switch, but because I know nothing about electrics, it
puzzles me.

Here is a drawing of it: http://www.digoliardi.net/electrics2.gif

I am dangerously ignorant of electrics so I'm asking: How does it work
that one AC pole goes to one pole of the orange thing (capacitor?) and
then also go around it? It's there a short there?

The electro magnet works just fine. Nothing gets hot even with a long
'on' time (for focusing). Do you think it is safe to hook up to the
timer (F-Stop timer by Darkroom Automation)?

Thank you,

PS: The shutter is on the back of the lens board because if it were on
the front it would make changing the F-stop almost impossible. I've also
made an extension of the F-stop ring to permit adjustment from below the
lens. (Shutter is from a long-rollfilm camera and it fits in a well
fitting enclosure around the rear of a Red Dot Artar.) Can provide
pictures or a build-article if you like.)
From: Nicholas O. Lindan on
<nohj(a)droffats.ten>

> http://www.digoliardi.net/electrics2.gif

> I am dangerously ignorant of electrics so I'm asking: How does it work
> that one AC pole goes to one pole of the orange thing (capacitor?) and
> then also go around it? It's there a short there?

No, the capacitor is charge from the AC line with a resistor and a
rectifier diode.

You have some problems here that will cause trouble down the line:
1) I am assuming that "TR R210" is a rectifier - the number comes up with
zilch when googled. It should be rated 700V (~2x Vp-p) or greater. The
standard popcorn part would be a 1N4007.

2) The capacitor needs to be rated for a minimum of 200VDC continuous - it
will charge to 170V in this circuit.

3) You should have a fuse in the circuit. It is hard to say what size
without knowing the solenoid rating. If the Lectrohm is rated for 5W then a
good guess is the circuit needs about a 1/8 amp with the shutter closed. A
1/4 amp slow-blow fuse should be adequate. If you don't have a fuse, and
there is a short in the shutter solenoid or capacitor, then there will be 33
watts dissipated in the Lectrohm. This is a very unfortunate number if the
Lectrohm is rated for 5W - the Lectrohm will get yellow hot but not fail and
you have yourself a fire starter. You should test the fuse, making sure it
will blow, by shorting the solenoid and closing the switch: make sure the
fuse blows before smoke starts to rise from the Lectrohm. The fuse gets a
bit of a workout in this circuit and may fail from fatigue every few years.
In lieu of a fuse you can use a 50W Lectrohm.

I am assuming the circuit works as follows, though I have no idea of the
shutter solenoid's ratings, so some of it is guestimation:

The capacitor is normally charged to 170V and it sits there at this voltage
until the shutter is switched on. No current is drawn from the 120VAC once
the capacitor charges.

When the switch closes the capacitor provides a shot of 170V power to close
the shutter solenoid.

Solenoids need a lower voltage to hold closed than they need to activate. By
reducing the holding voltage the solenoid coil can be made smaller - the
solenoid doesn't have to be sized to take full voltage all the time.
Continuous full voltage will overheat the coil in a solenoid designed to be
used in this fashion.

When the shutter is closed, and the capacitor has partially discharged, the
circuit provides a much lower current to hold the solenoid closed. The 200
ohm resistor drops some of the voltage going to the solenoid and capacitor.

When the switch is opened the shutter solenoid opens and the capacitor is
charged back to 170V for the next shutter closing. The recharging takes less
than half a second.

Connecting the circuit to a timer:

First, a problem: a timer is wired to supply 120V when exposing. You can't
wire the timer's socket across the switch without major pyrotechnics.

The "standard" way to connect this circuit to an enlarging timer would be to
leave the switch closed and plug the 120V leads into the timer. This may not
work very well.

If the timer controls the 120V then the capacitor doesn't get fully charged
and the shutter solenoid doesn't get a high voltage kick to slam it closed.
Without the higher voltage the solenoid may not work or work sluggishly and
erratically.

You need to wire a relay where the switch is located - the coil of the relay
goes to the timer and the relay contacts are wired across the switch. You
need a SPST relay with a 120VAC 60Hz coil. You can get one at Radio Shack.


--
Nicholas O. Lindan, Cleveland, Ohio
Darkroom Automation: F-Stop Timers, Enlarging Meters
http://www.darkroomautomation.com/index2.htm
n o lindan at ix dot netcom dot com


From: John on
ThaNicholas O. Lindan wrote:
> [... snip excellent article ...]

Thanks for the help, Nicholas. Clearly I'd better leave this to a
professional to build. I don't want to make an expensive mistake after
wasting my time screwing up the build.

I'll stick to analog photography, nuts, bolts, machines.

Best,
John
From: jch on
John wrote:

> Here is a drawing of it: http://www.digoliardi.net/electrics2.gif
>
> I am dangerously ignorant of electrics so I'm asking: How does it work
> that one AC pole goes to one pole of the orange thing (capacitor?) and
> then also go around it? It's there a short there?
>
> The electro magnet works just fine. Nothing gets hot even with a long
> 'on' time (for focusing). Do you think it is safe to hook up to the
> timer (F-Stop timer by Darkroom Automation)?
_____
The circuit diagram looks correct. It is a half wave rectifier with an
80 microFarad storage capacitor. The capacitor is always kept charged
through the 200 Ohm resistor by the silicon diode (TR210). When the
switch to the solenoid is open, the DC voltage across the capacitor will
be about 169V (square root of 2 times 120V). When the switch to the
solenoid is closed, DC current will flow through the magnetising coil,
thereby opening the shutter. The actual voltage seen by the solenoid
coil will be lower than 169V because of a small drop across the 200 Ohm
resistor. The drop depends on how much current the coil requires. The
capacitor could be of a higher voltage rating to be safer. If the
solenoid "hums" due to the half wave nature of the circuit, you could
increase the size of the capacitor to, say, 150 microFarad, or use a
full wave rectifier. You should put a master power switch on the hot
side of the the AC line (black wire in USA/Canada). It would be no
problem plugging this device into a timer since the current draw will be
in the order of 250 mA (my guess).
--
Regards / JCH
From: Jean-David Beyer on
jch wrote:
> John wrote:
>
>> Here is a drawing of it: http://www.digoliardi.net/electrics2.gif
>>
>> I am dangerously ignorant of electrics so I'm asking: How does it work
>> that one AC pole goes to one pole of the orange thing (capacitor?) and
>> then also go around it? It's there a short there?
>>
>> The electro magnet works just fine. Nothing gets hot even with a long
>> 'on' time (for focusing). Do you think it is safe to hook up to the
>> timer (F-Stop timer by Darkroom Automation)?
> _____
> The circuit diagram looks correct. It is a half wave rectifier with an
> 80 microFarad storage capacitor. The capacitor is always kept charged
> through the 200 Ohm resistor by the silicon diode (TR210). When the
> switch to the solenoid is open, the DC voltage across the capacitor will
> be about 169V (square root of 2 times 120V). When the switch to the
> solenoid is closed, DC current will flow through the magnetising coil,
> thereby opening the shutter. The actual voltage seen by the solenoid
> coil will be lower than 169V because of a small drop across the 200 Ohm
> resistor. The drop depends on how much current the coil requires. The
> capacitor could be of a higher voltage rating to be safer. If the
> solenoid "hums" due to the half wave nature of the circuit, you could
> increase the size of the capacitor to, say, 150 microFarad, or use a
> full wave rectifier. You should put a master power switch on the hot
> side of the the AC line (black wire in USA/Canada). It would be no
> problem plugging this device into a timer since the current draw will be
> in the order of 250 mA (my guess).

I used to design electronic circuits, much more complicated than this one.
N.B.: that was a long time ago.

IMO, the AC line slowly charges the capacitor and when the switch is
pressed, the energy stored in the capacitor is what fires the shutter (like
electronic flash circuits). The current through the resistor does not
directly fire the shutter. In fact, with the circuit as shown, the resistor
should have high enough resistance so the shutter solenoid will release
after tripping the shutter so it can try again.

--
.~. Jean-David Beyer Registered Linux User 85642.
/V\ PGP-Key: 9A2FC99A Registered Machine 241939.
/( )\ Shrewsbury, New Jersey http://counter.li.org
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