Also an Olympus something-or-other, and a Pentax P3.
But I always use my Sony Cyber-Shot ;-)
...Jim Thompson
-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |
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| 1962 | I love to cook with wine. Sometimes I even put it in the food.
That's how I did it for our old Minolta SRT-100. I had tried a Schottky but that ended up at 320mV. Maybe a larger one would have worked. The OA91 dropped exactly 200mV at the current the CdS circuit is drawing and it is also very small. Sweet. Then I made an adapter to fit a commonly available Silver cell in there.
This has spared me the painful two-point recalibration. And who knows whether those little potmeters would have survived since they are about
25 years old and that camera has been through a lot.
Oh, BTW, if you also have a newer Minolta X-series and the shutter quits it most likely has lost its electrolytic. Don't procrastinate, that nasty thing started leaking in my case.
There is a surprising amount of space in the SRT-100. Even a larger Ge diode would have fit in.
Are you sure that the circuit cares? Mine uses a bridge circuit, so the sensitivity increases if the excitation voltage increases, but the point of null is unchanged by the excitation voltage. It wouldn't be a very good design if it depended strongly on the voltage.
On a sunny day (Thu, 18 Jan 2007 17:48:41 +0100) it happened Colin Howarth wrote in :
I have been thinking, and your voltage drop is the wrong approach. What you need to do, is measure the resistance of the meter itself. Then put a resistor in parallel so the circuit sees a 15 % lower resistance.
-------- ------- CDS cell -------------------- | | battery O meter [ ] calibration R. | |
------------------------------------------------
This correct? You can calibrate on the spot in artificial light with old and new battery.
Ah that would be one of those new-fangled things with an incy wincy teeny viewfinder, or some screen you can't see in the sunlight. And that exposes around a second after you wanted to take the shot?
:-)
Is this getting off-topic at all?
colin
PS. I do have a DSLR or two, and even a digital ixus.
Nope. You need to RTFM... that only happens in "power-saving" mode.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
They are a bit more complicated than that, usually two potmeters. It's been a while but on the Minolta SRT the voltage dependency wasn't too bad in low light situations while the error was significant in bright light. Anyhow, after adding a Ge diode it now runs just perfect with a
On a sunny day (Thu, 18 Jan 2007 21:43:45 +0100) it happened Colin Howarth wrote in :
OK, CdS has lower resistance if more light. In the old light meter I had (a stand alone one) more light was more meter deflection. So I'd expect the CdS (LDR) to be in series.
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I still have a couple of these nice LDR (Light Depended resistor) type CdS cells in the box.... Incredible ratio in resistance from dark to light, but slow.
Use a digital mpg4 camera now..... Does nothing in low light :-)
My last 35 mm camera was a 'Werra' Russian one IIRC. One eye flip mirror reflex for next to nothing.
Look, before you open the camera, why don't you just pop in a 1.55V battery for giggles and shoot a test roll? You may find out that you don't have to change anything at all.
Overexposed -- so at the higher operating voltage, the light meter read LESS than it should have for that amount of light. That seems backwards from what one might expect, with the simple model the OP posted. I know that mine is a bridge that's balanced to read the light; the voltage makes little difference. I wonder what's going on in yours.
If the germanium diode would not work for you, you might want to try using an LM10 as "voltage regulator".
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With low load current it should be possible to stay close enough to the positive rail (1.55V) allowing the cell to go down in voltage a little when it gets older/used.
You earlier mentioned thinking of a voltage divider with a fet follower. One problem with such a circuit is the unknown error due to Vgs, the gate-to-source operating voltage of the FET.
My circuit is an opamp-style unity-gain voltage follower, buffering the voltage divider. Q1 and Q2 operate at the same current, thanks to the choice of R3 and R4, and therefore have nearly identical Vbe voltages. This helps to insure a low offset voltage, forcing the regulator's output voltage to be nearly the same as the voltage divider.
You can chose a low quiescent current Iq, like 5uA, to save on battery power.
I think there may be a low-power opamp available that can operate from as low as say 1.45 volts, and that can deliver current to its output with as little as 100mV across its pullup transistor. But here on s.e.d. I like to explore discrete circuit designs, instead of simply suggesting some unusual IC or another.
Yes, my circuit has more parts than say a germanium diode, but it has the advantage of being adjustable and predictable. You can select Q1 Q2 in one small 6-pin SMD package, and use a sot-23 package for PNP Q3, and use 0603 resistors, for a really small PCB. Philips' bcm847bs contains two bc847 NPNs matched to 3mV, in a miniature sot363 package.
I was thinking of it as a differential amplifier connected to a pnp common emitter amplifier providing negative feedback. I'm still figuring out the biasing.
In particular, the biasing of Q3. R3 is chosen to give a 0.6 V drop across it (ignoring Q3's base current). Shouldn't this be a bit more to make sure that Q3 is always active? On the other hand, if it's too much that would be applying a potential difference (Vin - Q1's Vc) across Q3's EB - which would destroy Q3?
Arrrgh. It's nearly 3.00 a.m., I'm not an electronics engineer and my knee is full of blood (post operative) and hurts and I can't concentrate.
Is there a sort of negative feedback there too? If Q3's Veb is too big, large Q3 base current would satisfy Q1, thus less current through R3, reducing Q3's Veb???
Ah. Next challenge would be to get this down to nano amps, when there's no load at Vout... :-)
I much appreciate it. I've learnt a lot - and got some way to go... (am playing with Spice at the moment :-)
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