1. Home
  2. Electronics Design
  3. EL7900

EL7900

Hi,

has anyone used this photodiode?

formatting link

I'm talking about a single sensor, powered at 3.3V with an Rload of 562 ohm.

There are several things I can't understand:

  1. the ds states the output may saturate if the illumination is too high. Instead the output current will suddenly fall to 0V when I increase the light on the sensor too much.

  1. just before that condition, the maximum current read is about 2.7 mA, very far from the 6 mA allowed.

  2. the ds also states the output voltage compliance is 300 mV below the supply voltage but the figure 7 shows this is true only for very small currents (< 400 uA). At 1 mA it raises at 500 mV and there is no indication about higher currents.

I may increase the supply voltage to have the full span I need (2.048 V) but I can't deal with the falling of output current when exposed to a lot of light.

I can't understand why it goes down... it shouldn't *saturate* - that is the output current won't increase anymore? The figure 9 doesn't show any falling edge on the right!

Thanks in advance Marco

Reply to
Marco Trapanese
Loading thread data ...

hm.

It's certainly a very odd-looking circuit, if the block diagram on page 4 of the datasheet is to be believed.

The two P-MOSFETs seem to form a current mirror, with what's drawn as an N-channel MOSFET as the current source. It would make more sense if that MOSFET at the top were another P-channel device. It would then - mostly - be a resistor when enabled with more than 2.7V of gate-to- source voltage - not less than 2.4V according to the data sheet - and guaranteed to not act as a resistor with less than 1.2V of gate-to- source voltage

My guess would be that if the photo-diode draws too much current, the MOSFET at the top comes out of saturation (stops looking like a resistor and starts looking like a constant current source) and the voltage at its drain and the common source connection for the two MOSFETs forming the current mirror would then collapse down to something very close to the negative rail.

It could be that the top MOSFET in your particular device comes out of saturation at a relatively low current when it's only got 3V of gate- to-source voltage.

See what happens if you boost the device supply voltage closer to 5.5V

- perhaps with a battery.

-- Bill Sloman, Nijmegen

Reply to
Bill Sloman

Yup, looks like a bug. Too much light breaks it.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
Reply to
John Larkin

2 ohm.

A,

e

V)

is

ny

Impressively insightful analysis. Care to share your insights as to how the excess of light breaks it?

-- Bill Sloman, Nijmegen

Reply to
Bill Sloman

ohm.

You described it already. Forgotten so soon?

You enjoy being an ass, don't you. I assume you enjoy being unemployed, too. They are related phenomena.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
Reply to
John Larkin

I don't see how this IC is any better than a bare photodiode.

--

John Larkin                  Highland Technology Inc
www.highlandtechnology.com   jlarkin at highlandtechnology dot com   

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom timing and laser controllers
Photonics and fiberoptic TTL data links
VME  analog, thermocouple, LVDT, synchro, tachometer
Multichannel arbitrary waveform generators
Reply to
John Larkin

Il 28/08/2012 18:21, John Larkin ha scritto:

Because it has a pin to put the output in hiZ. I have several hundreds of this IC that share the same output path. With a shift register I enable one-by-one without any other components.

Marco

Reply to
Marco Trapanese

Il 28/08/2012 12:24, Bill Sloman ha scritto:

I requested a clarification to Intersil's engineers.., I'm waiting for their answer.

I'm pretty sure I'm wrong but to activate the output I need to put the enable pin to 0V. So I guess the top MOSFET is a P-channel, despite the drawing.

This is what I measured on the test bench.

Are you talking about my own specific sensor? I have over 800 sensor on my board and all these exhibit this behavior.

I've already done this test. The output current - before falling down close to the negative rail - reaches about 3.5 mA.

Marco

Reply to
Marco Trapanese

Wouldn't this work?

cmos_enable----------+ | | --- photodiode ^ | | +----- out | | R | | | gnd

If you keep the voltage across R well below a junction drop, multiple photodiodes could share R and the downstream stuff. Big matrix.

But in your situation, what does Intersil say?

Are you applying too much light? Daylight can exceed 10K lux; sunlight over 100K.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation
Reply to
John Larkin

Il 28/08/2012 20:06, John Larkin ha scritto:

I guess it would.

Nothing until now. I received several emails but with further requests of details.

Yes, I'm applying too much light because it may happen. It's ok if the output saturates but not if falls down.

Marco

Reply to
Marco Trapanese

If its circuit is as shown on page 4, with all the MOS devices as P-channel, excessive photo current _will_ overwhelm the enable device and the output _will_ collapse. ...Jim Thompson

--
| James E.Thompson, CTO                            |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
I love to cook with wine.     Sometimes I even put it in the food.
Reply to
Jim Thompson

Why use such a horrible photosensor? You can get tiny PDs for a few cents.

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 net
http://electrooptical.net
Reply to
Phil Hobbs

Use regular photodiodes, bias their cathodes at VDD using the summing junction of an op amp or a common-base stage, and drive the anodes from the shift register outputs.

Only the output that's low will produce photocurrent in the output.

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 net
http://electrooptical.net
Reply to
Phil Hobbs

Il 28/08/2012 20:52, Phil Hobbs ha scritto:

But I have to add one mosfet for each sensor in order to enable them selectively. On a single board I have 440 sensors.

Marco

Reply to
Marco Trapanese

Il 28/08/2012 20:57, Marco Trapanese ha scritto:

I've just read your other message.

Marco

Reply to
Marco Trapanese

Probably not clear. Like this: *--RRR---* | | | |\ | *-----*-----*-----*-----*-...*-*---|-\ | | | | | | | | >-*--- | | | | | | *-|+/ --- --- --- --- --- --- | |/ / \ / \ / \ / \ / \ / \ | -t- -t- -t- -t- -t- -t- VDD | | | | | | | | | | | |

*-------------------------------------* | Q0 Q1 Q2 Q3 Q4 QN | | | | | | SHIFT REGISTER (ONE OUTPUT LOW) | | | | | *-------------------------------------*

(Run the op amp from a higher + supply, obviously.)

Zero extra parts.

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 net
http://electrooptical.net
Reply to
Phil Hobbs

Ridiculous brain f*rt. Has to be open drain, of course, in which case you just bias the op amp as usual. I blame it on having spent all day doing business development and taxes!

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 net
http://electrooptical.net
Reply to
Phil Hobbs

The data sheet says 8000 lux max.

I once worked on a thing that had hundreds of lamps and photoresistors in a big egg crate sort of thing. It lit up boxes selectively to put code patterns onto film images somehow. The physicist who did the circuit design took +5, through the CdS sensor, into the base of an NPN transistor. The lights were bright enough to blow out the transistors. The fix was to apply fingernail polish onto all the CdS photoresistors.

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation
Reply to
John Larkin

Hey, if you don't use R but make a matrix, one reverse-biased active PD will dump into all the others on that column in forward-conduction mode. The result will be a voltage that's the log of the light intensity. Matrix selection and log conversion with zero components!

--

John Larkin         Highland Technology, Inc

jlarkin at highlandtechnology dot com
http://www.highlandtechnology.com

Precision electronic instrumentation
Picosecond-resolution Digital Delay and Pulse generators
Custom laser drivers and controllers
Photonics and fiberoptic TTL data links
VME thermocouple, LVDT, synchro   acquisition and simulation
Reply to
John Larkin

Il 28/08/2012 21:22, John Larkin ha scritto:

No. The ds says 8 klux to be "linear".

But there is no an "absolute maximum rating" for illumination. Figure 9 for example shows up to 10 klux with a current over 4 mA.

At page 5 I read: "[...] care must be taken when lux go as high as

10,000lux because the output current rises above 6mA before reaching the device's output compliance".

Table 6 shows the output current (6000 uA) at 10000 lux.

The problem is I cannot get a current above 2-3 mA. And in the whole ds there is no mention about the "over-light" issue.

Marco

Reply to
Marco Trapanese

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.