Simple problem: 0.20 V drop

Well, it's a Leica! I'll probably turn the adapter on a lathe. Probably in copper. That would need gold-plating then, wouldn't it?

:-)

colin

Yes, of course, but opamp followers are easy to understand, and hopefully the circuit has the critical opamp properties we need, e.g., a high-Z input, sufficient gain and low input offsets.

. Micropower low-voltage LDO regulator: . drop 13% = 200mV at 1.55V (if k=0.87) . . Vi___________________pnp Q3 . | | | E C ----+-- OUT --, . R1 R3 | B | Vo | . | | | | | | . | +---- | -----' R5 RL . | Q1 | | Q2 npn | | . | C C | gnd . +---B B -----------' . | E--+--E k = Vo/Vi = R2/(R1+R2) . | | R4 = (kVi-0.55)/Iq . R2 R4 R3 = 2*0.60/Iq . | | choose Iq >= Iout/25 . gnd gnd choose R1+R2 = 10 Vi/Iq

That's not a very useful way to analyze the circuit. Instead, realizing that the feedback loop is closed, we know the correct biasing voltage for Q3 will in fact be present, whatever it is. What does happen though, is that the current through R3 isn't known exactly. We know it's IcQ1 = R3/VbeQ3 + IbQ3, and that both VbeQ3 and IbQ3 = Iout/betaQ3 are functions of the unknown load current. So load variations will cause changes in IcQ1, but we hope that Q1's collector current will be close to half the current through R4, so Q1 and Q2 will have equal currents. Realizing that Q2 takes up whatever R4 current Q1 doesn't want, we see it's the Ic1/Ic2 current ratio that's adversely affected.

Ouch! Get well soon!

OK, you've got it now. Returning to the issue of equal currents in Q1 and Q2: to the extent they aren't equal, we'll suffer an offset-voltage degradation. Voffset = kT/q ln (Ia/Ib), assuming matched transistors. But this isn't serious, see AoE page 80 and the graph in figure 2.53 page 91. We see that a severe 2:1 current imbalance causes only a modest 17mV offset voltage, or less than a 10% error in your specified 200mV voltage drop.

Hmm, actually, what can you tell us about the current load? Is it only the CdS light-measuring cell, or is it the entire camera electronics? If the latter, AWWKK!, my circuit will probably need some beefing up. I had assumed the former, and if so, the circuit's quiescent can be starved to be much less than the cell's maximum current. Say 50 to 100nA? No problem.

Have fun. But I do recommend vetting your transistor's low- current, low-voltage spice model with some bench measurements.

Sorry, oops! Of course I = V/R so that's IcQ1 = VbeQ3 / R3 ...

What is the load current value?

...Jim Thompson

How about something simple-minded like....

...Jim Thompson

On a sunny day (Sun, 21 Jan 2007 11:46:22 -0700) it happened Jim Thompson wrote in :

That sucks about 700uA

So? Don't you press a button to read the meter?

Scale resistors up by 10X and the output varies from 1.42V at 100uA load down to 1.32V at 1mA load

I posted an inquiry as to load current but saw no reply.

...Jim Thompson

On a sunny day (Sun, 21 Jan 2007 12:24:56 -0700) it happened Jim Thompson wrote in :

Well if it has a 100uA meter (sort of standard) then that soloution will allow you pres the button 8 x less 800uA versus 100ua), _at_full_light_. At average light 16 x less.

Correct, I already thought of that solution actually, but then the Ge diode is simpler.

The resistor parallel to the meter, the solution I proposed is the most accurate I think.

hehe

Probably.

I may be old, but I don't live in the past, or waste time (which is the same as money)...

I'd just buy a new camera... or more likely, find a shop that does a rework.

Likewise with all this wasted time trying to make an ancient track-ball work ;-)

I contribute to the economy... I hire lawn maintenance, pool maintenance, vehicle maintenance and house cleaning, carpet and tile cleaning, and I still come out way positive on the dollars per hour.

And I buy (wireless) mice three at a time, for spares ;-)

...Jim Thompson

Ok, that warrants leather seats, caviar and champagne. A OA91 or similar Ge diode might do but it may not be dignified enough to serve here. Although, it does have a German heritage. But AFAIR it was conceived in Hamburg while the Leica was most likely engineered in Wetzlar so it might not like such northern influence ;-)

The one to accommodate a new 1.55V coin cell? I picked one that had about the same height as the old mercury cell but, of course, it was of smaller diameter. So I took a piece of rubbery plastic and cut and filed that to size. This way nothing rattles and the battery has a nice and stable position.

Not on most vintage cameras. For example, my Minolta SRT100 has a turn switch under the housing and you usually leave it one for the whole hike if you want to do wildlife photography. Usually all day. The circuit hardly consumes any power while the lens cover is on, which it should be. Sometimes I left the CdS circuit on for the whole vacation.

I have yet to find a set of lenses that matches the quality of that of my film camera. Some have cost more than a ritzy digital camera. A piece. Also, even at 8Mpixels the quality of the results is just not up to par. Says even my sister and she is an expert on this stuff, she calculates lenses for a living.

:-)))

But you had recently confessed that you've laid tile. Ok, so did I.

I might just do that as well. Will try out the new Trackman when I get to town.

Thanks, that is indeed elegant.

Yes. But I'd need to open the camera to get at the button.

I'm not old (39), but I do waste time (and thanks for taking the time to answer this).

This whole topic is just for fun. I'm learning some useful stuff, which I don't need in my day-to-day life. It's a secondary hobby. Hobbies are supposed to use up time.

I have a new camera. The old ones are more fun. In particular, the viewfinders are gorgeous. And yes, the lenses were well made in the old days.

I missed this, but there you go.

colin

Sorry, I'm still trying to find that out.

colin

Amusing myself....

...Jim Thompson

You'll recall I said I did it for therapy... plus I'm a bit of a perfectionist when it comes to tile work ;-)

I've had too many disasters happen mid-project to risk down-time due to a failed mouse, or monitor, or PC for that matter.

...Jim Thompson

now what if Wins regulator uses a 1.23V reference diode instead of the divider?

Oh yeah, you might quite enjoy this article from the Guardian Jim - how the left lost its way:

and part 2:

Cheers Terry

Cheers Terry

Joerg's "regulator":

Lone diode. Not as scientific as Jim's regulator and certainly not like Win's regulator. But works. Lacquered glass housing matches vintage of camera :-)

Same here. That's the main reason I did it. But I am also a perfectionist when it comes to car maintenance, especially after my father was almost killed but a "pro" who failed to tighten the steering columns properly. The pool is way cleaner after the local pool guy just quit coming one day and we took over ourselves. And my wife stressed out over the work of a gardener, maybe she's also too much of a perfectionist.

I've got two backup PCs :-)

I think she's right. The weak spot of all those new-fangled cameras (cheap compact models as well as epxensive "pro" stuff) are the lenses. I guess there just isn't any appreciable market for high-quality amateur level photoghraphy any more. Actually the lens has always been the bottleneck in photography, and with the rise of a much inferior recording technique (especially in the consumer segment) the industry has happily grabbed the chance of greatly reducing cost by letting lens quality plummet accordingly.

robert