EL7900

Rload of 562 ohm.

too

about 2.7 mA,

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(2.048 V)

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and

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the

the

Rload of 562 ohm.

is too

about 2.7 mA,

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gate-to-

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Some of it's grunt work, like manuals and parts lists and checking stuff. But overall it's a lot more fun that retiring and watching TV and playing golf or whatever.

But if you're saying that I have to work forever, out of obligation to somebody or other, that's just silly. Any time I want I can turn the company over to the kids and just tell them where to send the checks. But not any time soon.

One of my engineering lackeys just barged into my office with a six pack of Mirror Pond Pale Ale and forced me to drink one. He even made me open the bottle myself. Can't get no respect around here.

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

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

Jamie is one of Larkin's sycophants (or is it "sickophants" ?:-) and should be ignored/kill-filed.

Any response back from Intersil?

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. ...Jim Thompson

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

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

It's easy to demonstrate why the "folding" occurs... if anyone cares. ...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.

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

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

Not yet.

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

Marco

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

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

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

I did, in...

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

(1) It's ALL P-channel, mirror AND enable devices (2) The mirror is 1:M

No it's not obvious. You need to know at least the physical size of the PD, and what material it's made of... see for instance the number of DIFFERENT types available on XFAB processes... ah, yes, I see that Larkin has no clue... just lots of mouth but no brains.

BS... that's all you do... the only contradictions are in the vagueness of the datasheet... is it constructed like the "cartoon"? probably is. Now, tell me, Mr. Smartass, details of the enable device and the mirror... you can't.

BS, I ran it a couple of days ago.

Do you eat them, or just "stuff" them directly ?:-) ...Jim Thompson

saying

I SWAG'd M at 10, but the actual value looks to be closer to 11. Not a bad initial guess.

A self-proclaimed Master Circuit Designer like you should be able to look at the data sheet for a minute and estimate M, without using a calculator. It *is* obvious, and it is simple.

Do it. Now.

I can do a bunch of math analysis on this, and point out some interesting contradictions. And point out some additional measurements that Marco could do to maybe untangle things a bit. I will. Buy I'll allow you to do it first, if you can.

Do something real! Last chance!

Take a look at this sketch...

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

ohm.

=A0 =A0 ...Jim Thompson

=A0 =A0| =A0 =A0mens =A0 =A0 |

=A0 | =A0 =A0 et =A0 =A0 =A0|

=A0|

=A0 =A0 =A0 |

ide 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.