Request electrical design help with hot-rodding a Howtek.

[ah693973]

I have been modifying my Howtek 4500 and have run into a spot where my lack of attention in Electrical Fundamentals class has caught up with me.

The main change I have made is to use some more modern PMTs. The 931b units that come stock in the Howtek really fall off in sensitivity in red. Going to a multi-alkali style fixes that and provides a lot more signal overall to work with.

The problem is the new PMTs are too good, I lose the bottom half of my histogram. Scans (Velvia included) bunch up so far to the right that even with DPL Pro I can't get full use of the 12 bits available to me. The endpoints will not adjust far enough.

I think the best way to fix this is to reduce the gain in the front end where the current to voltage conversion takes place. Some other fixes that seem less desirable (or harder) include reducing the available light or modifying the circuit that adjusts the lower endpoint of the scan.

I have included a schematic of the front end that I came up with by inspection and a multi-meter, I think it is accurate. I also have a photo of the circuit board included if that helps.

The op amp is an OPA 602 and I have a machine where I swapped those out for OPA 827s. I don't know if the 827s helped because I swapped the PMTs at the same time.

I would like to shoot for a 5x reduction in gain as a starting point. I think the best way to do that would be to change resistors but with my limited experience I am unsure where to start. Can anyone help me figure this out?

Andy
120074
120075

[wombat2go]

The present steady state transfer function is about 0.65 Volt per microAmp

To reduce that by about 5, you could increase R53 from 500 Ohm to between 2700 ~ 3000 Ohm.
The 47 Ohm and 17800 Ohm resistors should not be changed.

[polyglot]

I mostly agree with wombat2go. Your current DC transfer function (assuming R38=R37) is:
H = R372(2/R37 + 1/R53) / R50

which is 14267 voltage gain with an input impedance of 47 (R50), or 0.67V/uA, treating the PMT as a perfect current source. You probably don't want to change the input impedance, assuming that you're operating the tube with the correct loading. It's possible your newer tube has higher or lower load requirements than the old one, which might imply that a change to R50 is called for. Do you have a datasheet for the tube?

Decreasing the gain means either decreasing R37 and R38 (keep them equal) or increasing R53. Increasing resistance means increasing noise, but decreasing resistance can mean overloading the op-amp, not that you're anywhere near that point. Increasing R53 will increase your noise but not very much because its contribution is both swamped and attenuated by the R37/R38 voltage divider. Changing R53 to 3k is therefore the simplest solution.

You could reduce R37/R38 and arrive at a lower-noise amplifier, but I suspect it wouldn't really help. You would also need to adjust C2 in order to preserve the stability characteristics.

[ah693973]

Polyglot and Wombat,

Thanks so much for your help. This was a little past my meager understanding.

One of the main points of this exercise is to learn more about electricity. Thanks for expanding on the answer Polyglot. I had thought that R37, R38, R53 might be a voltage divider, but there was enough other stuff going on that I was not sure.

Andy

[Bernard_L]

I would recommend to decrease the gain by increasing R53. If you change R38, you need to change C2 such that their product remains constant: it defines the cutoff frequency (fc = 1/2piRC) of the current-voltage converter. If fc is too low, pixel transistions wil be smeared, if it is too high, noise will be above optimum.

[polyglot]

Yeah, it's a bit more messy than a voltage divider. To figure out what's happening around an op-amp, you make two assumptions:
- no current into the input pins
- the input pins are at the same voltage (0 in this case because one is grounded)

That means you have the same current through R50 as through R38 and then through the parallel combination of R37 and R53. You write down expressions for each the currents in terms of the various voltages and that gives you sufficient simultaneous equations to allow you to solve for all the unknowns (output voltage, voltage at the R37/R38/R53 junction) in terms of one free variable (the input voltage). Do a bunch of substitutions and rearranging, and you arrive at the formula I posted above. Then you figure out what values to change to get the answer you want.

[wombat2go]

While I know almost nothing about photo-multipliers, I can contribute to the OA bit.

[ghostcount]

wombat2go just went "Alpha Nerd" on everyone.:cool:

How about a frequency domain plot to see where that cap shorts out.

[Jody_S]

Yeah, it's a bit more messy than a voltage divider. To figure out what's happening around an op-amp, you make two assumptions:
- no current into the input pins
- the input pins are at the same voltage (0 in this case because one is grounded)

That means you have the same current through R50 as through R38 and then through the parallel combination of R37 and R53. You write down expressions for each the currents in terms of the various voltages and that gives you sufficient simultaneous equations to allow you to solve for all the unknowns (output voltage, voltage at the R37/R38/R53 junction) in terms of one free variable (the input voltage). Do a bunch of substitutions and rearranging, and you arrive at the formula I posted above. Then you figure out what values to change to get the answer you want.

Crap. I just had a brief flashback to something called 'linear algebra'.

[polyglot]

Crap. I just had a brief flashback to something called 'linear algebra'.

lol do we need to start putting trigger warnings on maths posts for those with PTSD?

[ah693973]

So, I switched out the R53 resistors to 3.3k Ohms.

I had also decided to try just removing the c2 capacitors. "Filters? We don't need no stinking filters." The test scans I've done look great so I think that might have been a fine idea. There is very little noise with the upgraded PMTs.

I did run into trouble when I dropped one of my 928 PMTs during assembly. I have several 3896 PMTs and so I had to use those. They are even more sensitive so I need to turn down the "gain" even more. Without the C2 caps I should be able to decrease R37 and R38 without changing a filter frequency. I should be able to get close to the value I need by "piggybacking" a resistor. I don't want to tear up the board with too many value swaps.

Here is a screenshot of the new setup using wide gamut linear chrome on DPL:
120471

And here is wide gamut log chrome:
120472

The PMTs are so good that the black electricians tape I used to provide a "zero" point is actually somewhat transparent in the red. When the log amp is used the velvia underneath is visible. That is the tape going from just under the target to the bottom of the image. I was impressed.

You can see that the red and green need to come down substantially still to move the histogram to the left. I may end up using different "gains" for each channel. There is a ton of information in the blue channel of velvia that most scanners will not pick up.

Thanks again for all the help with this.

Andy

[polyglot]

You want to keep the C2 capacitors. They're there to help ensure stability of the op-amp, if you take them out you risk having ringing and oscillations where there are very sharp transitions in the image.

A bit of a change to the filter frequency is far preferable to removing the stability control entirely!

[wombat2go]

Changing R38 & R38 will not do anything to the system transfer function that can be done by only adjusting R 53.
The C2 capacitors should not be removed because the value is already down around the track capacitance value.

You could add a trim pot in place of R53
http://electronics.mcmelectronics.com/electronics/Trim-Pot

I am using fast Fourier transform to boost the sharpness (reduce spread function generally) of some 35mm C41 negs scanned back in 1994. It is working well but needs a powerful computer.

Convert by Fourier Transform to complex frequency domain, subtract a variable Gaussian black hole centred on Amplitude_0 then convert back to Cartesian

http://www.pentaxforums.com/forums/32-digital-processing-software-printing/268073-lens-diffraction-compensation-computer-fft.html

[onnect17]

I think for small adjustments in the gain it could be better to use the high voltage power supply for each PMT. Also I am a little concern about the bandwidth in the OPA827 with that level of gain.