EL7900

[Marco Trapanese]

Il 30/08/2012 21:50, [email protected] ha scritto:

what the resistance of the ammeter?

It's a Fluke 115, it should have a negligible input resistance for this
measurement.
Anyway, the sensor is a current source - that current will flow no
matter the resistance of the ammeter.

Marco

 

[Marco Trapanese]

Il 30/08/2012 22:02, Jim Thompson ha scritto:

Can you use a small value load resistor, such as 100 Ohms, use a +5V
supply, ENbar grounded, and _gradually_ increase the light level,
tabulating points as you go?

When I did this experiment I got these results:

Vs = 3.3V
Rload = 562 ohm
Imax = 2.7 mA

Vs = 3.3V
Rload = 180 ohm
Imax = 2.7 mA

Vs = 5.0V
Rload = 180 ohm
Imax = 4.2 mA


But I wrote down only the maximum value of the output current reached.
If you think it may be useful I can repeat the test tabulating the points.

My guess is the enable device is too weak to support the current and
collapses.

A really bad configuration! It should have been done by shunting the
mirror input device. In series, all kinds of strange things can
happen.

I see...

If you have access to a curve tracer, sweep output to Vcc, everything
else open and report back.

Unfortunately I haven't a curve tracer, nor my friends.

If you're lucky, it'll be one diode drop.
If you're not lucky, the device will NOT be suitable for paralleling
outputs.

Before select this component and start developing the board I explicitly
asked to Intersil if it is suitable for this purpose. I have their emails
Last year I made a prototype with ISL29000: very similar, but obsolete
now. There were no problems.


Now, a workaround should be:

- raise the Vsupply to 5V
- decrease the Rload

And for the next release, throw out this s&/*$t and use a bare pd instead

Marco

 

[George Herold]

That shouldn't matter. He's measuring output current, and it folds
over before the spec limits.

Sure, it's broken. I was just trying to explain how even a desk lamp
breaks it.

George H.

 

[George Herold]

Probably the passivation is half a wave thick at that wavelength.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot nethttp://electrooptical.net- Hide quoted text -

- Show quoted text -

Ahh, cute. Now they just need an NIR filter.
Years ago I bought a PD from photonic detectors that had a built in
filter to block visible light. I thought it'd be great for doing NIR
stuff at trade shows, dang filter was polarization sensitive.

George H.

 

[Marco Trapanese]

Il 31/08/2012 00:23, John Larkin ha scritto:

No response from Intersil?

Not yet.
They said "we don't know why it happens". After providing them further
details: "we're going to forward the issue to the Product Line Engineers".

Marco

 

[Jamie]

Il 30/08/2012 18:20, Jim Thompson ha scritto:




What other information should I provide?
Even if the actual board has a chain of 74HC595 to enable one EL7900 at
time, the tests I reported was conducted with only one sensor soldered
on an empty PCB.

Vsupply: 3.3V (and 5V as described)
Enable: tied to GND
Rload: 562 ohm (I tried other values)

I think such a setup should be very close to the one the Intersil's guys
used to characterize the pd.

I use a desktop lamp to illuminate the sensor and I have both a
voltmeter and a ammeter to measure the output current and the voltage
across the Rload.

Marco

I don't know if it makes much differences however, a desk lamp may be
producing 100/120 hz of light and your measuring methods could be bogus..

THis is assuming the PD will respond to that, which it should.

I would put a scope on that to see if this is pulsing. If you don't
have a scope, then put a AC voltage meter at the load to see if you
can see any measured signal.


Jamie

 

[whit3rd]

Il 29/08/2012 01:08, John Fields ha scritto:

As said to Phil Hobbs I just need to scan an array of several hundred of

photodiodes in order to get their output.

Well, of course a vidicon tube scans thousands of photoresistors to
get their output, and a CCD does the same with photodiodes.

Have you considered running 'several hundred' optical fibers as
lightpipes, from the to-be-sensed locations to a linear CCD array?

Usually, one doesn't want a photodiode to charge up (because it
has significant capacitance), but CCDs make that into a virtue,
and you get a time-integration feature for free.

 

[Marco Trapanese]

Il 31/08/2012 19:14, Jamie ha scritto:

I don't know if it makes much differences however, a desk lamp may be
producing 100/120 hz of light and your measuring methods could be bogus..

THis is assuming the PD will respond to that, which it should.

I would put a scope on that to see if this is pulsing. If you don't
have a scope, then put a AC voltage meter at the load to see if you
can see any measured signal.

It depends of the kind of lamp. Anyway that noise is clearly visible (if
any) and has nothing to do with the issue described.

Marco

 

[Jamie]

Il 31/08/2012 19:14, Jamie ha scritto:




It depends of the kind of lamp. Anyway that noise is clearly visible (if
any) and has nothing to do with the issue described.

Marco

Oh really? How would you know that?

If I were to use a 50% duty cycle light source on a PD, my DMM would
register only 50% of the peek current and there for would lead to saying
that the PD device is broke.

QUESTION :

DID YOU PUT A SCOPE ON THE OUTPUT TO VERIFY ?

If not, then you're just wasting peoples time.

Jamie

 

[Marco Trapanese]

Il 01/09/2012 18:44, Jamie ha scritto:

If I were to use a 50% duty cycle light source on a PD, my DMM would
register only 50% of the peek current and there for would lead to saying
that the PD device is broke.

QUESTION :

DID YOU PUT A SCOPE ON THE OUTPUT TO VERIFY ?

Obviously. Otherwise how can I say "it is *clearly visible* ?

If not, then you're just wasting peoples time.

Thank you, it's very kind of you.

Marco

 

[Jamie]

Il 01/09/2012 18:44, Jamie ha scritto:




Obviously. Otherwise how can I say "it is *clearly visible* ?





Thank you, it's very kind of you.

Marco

Dodging the question does not answer it..

Your definition of clearly visible is still in the fog..

I can use a DMM a say it's clearly visible but it isn't

I'll just go with my feelings on this, and you don't need to
know what they are.


Jamie

 

[Jamie]

Some modulation of the light input wouldn't explain the output
foldback he's seeing.

I understand that however, his measuring tools most likely is not
reporting the actual current he is getting before it folds back.

The point is, the device may actually be doing as advertised but may
not be DC and there for the device being used to measure this will not
report a DC current but one with a pulse which will just generate some
average reading on DMM, which will be less than the expected value
that was already expected to happen but only at a higher level.

It's all about the tools you use for doing this that makes a big
difference in determining the factors.

We have current meters in line with devices that have a pulse on the
line. In order to get a peak current reading we need to have a filter
circuit on the meter to show that otherwise, it's just a calculated
reading summed between the on and off and if you watch the meter it's
all over the place.

I know my FLuke 289 I use when going out in the field will do this also
in current mode. And even if I use the Peak function it still is a
little flaky...


Jamie

 

[Marco Trapanese]

Il 01/09/2012 19:01, Jim Thompson ha scritto:

Any response back from Intersil?

Not yet.

My guess is it's either a really bad batch, on the "slow" side of the
process, or maybe it's just a dud design period... as drawn it _has_
to collapse at some high input level.

I'm afraid it's a good guess...

Marco

 

[Marco Trapanese]

Il 01/09/2012 20:05, Jamie ha scritto:

The point is, the device may actually be doing as advertised but may
not be DC and there for the device being used to measure this will not
report a DC current but one with a pulse which will just generate some
average reading on DMM, which will be less than the expected value
that was already expected to happen but only at a higher level.

I exclude that.
I measured the output signal with an oscilloscope first, and I saw a
little 50/100 Hz noise only if I use a neon lamp. With incandescent
lamps there is no ripple at all.

Furthermore, I experience the same odd behavior with sunlight, using
some filters to increase the illumination gradually.

Marco

 

[George Herold]

Take a look at this sketch...

http://www.analog-innovations.com/SED/EL7900_IDSS.pdf

I replaced the upper P-channel "enable" device with a current source
of value IDSS.  (One of my special Spice tools... can't pull to
infinity, stops at zero volts across it

Clearly, when the sum of Iphoto and M×Iphoto exceeds IDSS, the output
will collapse.

(There's a multiplication symbol between "M" and "Iphoto"... in my
fonts, maybe not in yours.)

The shape of that collapse will depend on the particular CMOS process
it was made from.  (A very sharp collapse would indicate that the
"mirror" is actually made up as a cascode (maybe even combined with
the enable device)... which is likely, if it was done properly.)

Without conducting testing with a calibrated light source, I can't
tell whether it is you simply overloading the device above its design
range, or if it is defective.

                                        ...Jim Thompson
--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
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I love to cook with wine.     Sometimes I even put it in the food.- Hide quoted text -

- Show quoted text -

Thanks Jim, I didn't know you could make a current mirror out of cmos.

There's a spec on the device page where it shows output current vs
Lux.
you'd have to convert lux to some 'real' unit of flux and calculate
photons per second for the PD area. PITA, I can never remember the
definition of lux.
For any decent PD it is a 'pretty good' guess to assume every photon
makes an electron.

George H.

 

[[email protected]]

You haven't figured out how to find M yet, which is a necessary first
step. It's obvious. Pitiful.

any M would work for a simple sim to illustrate the problem I think


Version 4
SHEET 1 1744 680
WIRE 576 -80 448 -80
WIRE 768 -80 576 -80
WIRE 16 -64 -112 -64
WIRE 208 -64 16 -64
WIRE 576 -48 576 -80
WIRE 16 -32 16 -64
WIRE 448 48 448 -80
WIRE 768 48 768 -80
WIRE -112 64 -112 -64
WIRE 208 64 208 -64
WIRE 576 64 576 32
WIRE 720 64 576 64
WIRE 16 80 16 48
WIRE 160 80 16 80
WIRE 768 176 768 144
WIRE 896 176 768 176
WIRE 1040 176 896 176
WIRE 208 192 208 160
WIRE 336 192 208 192
WIRE 768 192 768 176
WIRE 896 192 896 176
WIRE 1040 192 1040 176
WIRE 208 208 208 192
WIRE 336 208 336 192
WIRE 832 208 816 208
WIRE 848 208 832 208
WIRE 992 208 848 208
WIRE 272 224 256 224
WIRE 288 224 272 224
WIRE 1040 288 896 288
WIRE 768 304 768 288
WIRE 832 304 832 208
WIRE 832 304 768 304
WIRE 208 320 208 304
WIRE 272 320 272 224
WIRE 272 320 208 320
WIRE 768 320 768 304
WIRE 896 320 896 288
WIRE 208 336 208 320
WIRE 336 336 336 304
WIRE 768 432 768 400
WIRE 896 432 896 400
WIRE 208 448 208 416
WIRE 336 448 336 416
FLAG 208 448 0
FLAG 336 448 0
FLAG -112 144 0
FLAG 768 432 0
FLAG 896 432 0
FLAG 448 128 0
SYMBOL pmos 160 160 M180
SYMATTR InstName M1
SYMATTR Value BSS84
SYMBOL pmos 288 304 M180
SYMATTR InstName M2
SYMATTR Value BSS84
SYMBOL pmos 256 304 R180
SYMATTR InstName M3
SYMATTR Value BSS84
SYMBOL voltage -112 48 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL current 208 336 R0
WINDOW 3 -261 36 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName I1
SYMATTR Value PULSE(0 10m 0 1 0)
SYMBOL res 320 320 R0
SYMATTR InstName R1
SYMATTR Value 1k
SYMBOL voltage 16 -48 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V2
SYMATTR Value 2.355
SYMBOL pmos 720 144 M180
SYMATTR InstName M4
SYMATTR Value BSS84
SYMBOL pmos 848 288 M180
SYMATTR InstName M5
SYMATTR Value BSS84
SYMBOL pmos 816 288 R180
SYMATTR InstName M6
SYMATTR Value BSS84
SYMBOL voltage 448 32 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V3
SYMATTR Value 5
SYMBOL current 768 320 R0
WINDOW 3 -261 36 Left 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName I2
SYMATTR Value PULSE(0 10m 0 1 0)
SYMBOL res 880 304 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL voltage 576 -64 R0
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V4
SYMATTR Value 2.355
SYMBOL pmos 992 288 M180
SYMATTR InstName M7
SYMATTR Value BSS84
TEXT -50 472 Left 2 !.tran 1

 

[Marco Trapanese]

Il 02/09/2012 04:01, Jim Thompson ha scritto:

Of course the OP's measurements may be "loose" ;-)

I don't have a calibrated light source, indeed.
What I've noticed is that if I approach the lamp slowly the current
increases slowly too.

After reaching the maximum values (reported in other posts, depending on
Vsupply and Rload) closing further the lamp (few cm) leads to a very
fast collapse of the output current.

I know it's a qualitative measurement, but the collapse seems to have a
quite sharp edge.

Marco

 

[Marco Trapanese]

Il 02/09/2012 17:14, John Larkin ha scritto:

With a fairly hi-Z load to ground, 10K maybe, does the output voltage
collapse at high illumination?

Tomorrow I'll try and I'll report the results.

You should probably dump this part and use matrixed photodiodes.
That's a board redesign, but probably a good long-term investment.

That's what I'm going to do.

Marco

 

[Marco Trapanese]

Il 02/09/2012 19:27, John Larkin ha scritto:

My log-matrix thing has a fatal flaw: the unselected PDs are still
working in PV mode. Pity. Phil's architecture, with open-drain
switches, works because it really isolates the unselected diodes. Or
use one 8-input analog mux per 8 PDs, 4051s maybe. How fast do you
need to scan the array?

The minimum scan frequency is about 4 Hz for the whole array/matrix.

I don't see an obvious way to do a diode matrix out of the PDs, but I
haven't had my quota of coffee and sugar yet.

Marco

 

[josephkk]

...Jim Thompson

So demonstrate it. You never say anything substantive any more; you
just brag about what you can do without ever actually doing it.

Take into accout the probable mirror ratio.

^^^^^^
Larkin needs to learn to spell as well ;-)

I did, in...

Date: Thu, 30 Aug 2012 07:40:59 -0700
Message-ID: <[email protected]>
[snip]

(1) It's ALL P-channel, mirror AND enable devices
(2) The mirror is 1:M <<<<<<<<<<<<<<<<<<< LARKIN TAKE NOTE
(3) Crank it up to Vcc=+5V and see what happens
(4) Refer to:

Date: Tue, 28 Aug 2012 11:36:15 -0700
Message-ID: <[email protected]>

Learn to read, you maroon (that's maroon=Larkin, for others that are
also reading compromised

Can someone hazard a guess on M? Hobbs? Presuming an on-chip PD,
what current from the PD itself?

...Jim Thompson

With the small size of the PD i hazard a wild guess of about 100 for M.

?-)