Sonsivri
 
*
Welcome, Guest. Please login or register.
Did you miss your activation email?
December 06, 2016, 06:47:51 18:47


Login with username, password and session length


Pages: [1]
Print
Author Topic: Photodiode & LM358 problem  (Read 3703 times)
0 Members and 1 Guest are viewing this topic.
koky
Active Member
***
 Warned
Offline Offline

Posts: 157

Thank You
-Given: 83
-Receive: 115


WWW
« on: March 11, 2010, 10:50:09 22:50 »

I have build  an phototransistor amplifier (see schematic)
the phototransistor is the TEMT6000X01 (Farnell 1470161)

i have tried on bread board with an Farchaild LM358N and the circuit work fine, i have make an smd board , i have place LM 358DT (ST  Farnell 1564331) and the ouput is always 3.8mV, i have also changed the diode with a 1Mohm resistor but the result is same, obvius i have also tried to changed the LM358DT, only if i put with pin arrangement the dual in line LM358N the sensor work.

I can't understand the problem !!!!!
please help
Logged
Parmin
Hero Member
*****
Offline Offline

Posts: 544

Thank You
-Given: 357
-Receive: 126


Very Wise (and grouchy) Old Man


« Reply #1 on: March 11, 2010, 11:14:34 23:14 »

Is there any power coming into the collector of Q1?
Logged

If I have said something that offends you, please let me know, so I can say it again later.
solutions
Hero Member
*****
Offline Offline

Posts: 1446

Thank You
-Given: 590
-Receive: 851



« Reply #2 on: March 11, 2010, 11:22:08 23:22 »

The data sheets show a marked difference in input characteristics between the two op amps, meaning the input stages may be quite different in their inherent design and biasing internally.

I'd suggest trying various high valued resistors on the emitter to distance yourself up from ground if you're really bent on using the ST part.

Posted on: March 11, 2010, 11:20:05 23:20 - Automerged

Is there any power coming into the collector of Q1?

You don't need power there - that's a phototransistor
Logged
Ichan
Hero Member
*****
Offline Offline

Posts: 840

Thank You
-Given: 312
-Receive: 387



WWW
« Reply #3 on: March 12, 2010, 05:35:11 05:35 »

I think the photodiode should go to the non-inverting input of the op-amp.

-ichan
Logged

There is Gray, not only Black or White.
hate
Hero Member
*****
Offline Offline

Posts: 556

Thank You
-Given: 156
-Receive: 354


« Reply #4 on: March 12, 2010, 03:41:39 15:41 »

U don't have any type of gain here, it will just work as a Schmidt trigger so very little light falling on the surface of the photo may trigger it. That won't be stable imo.

Regards...
Logged

Regards...
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #5 on: March 12, 2010, 03:44:31 15:44 »

Your problem is simple.  Using a transimpedance amp works fine for photodiodes, but not so well for phototransistors, which require some bias voltage across their terminals.  This is not specified in the datasheet, which is lame.  Your solution is to either use a photodiode, or modify the circuit to put some bias voltage across the phototransistor.

Also, I have no idea why you have a diode in the feedback loop...

edit:  actually, the datasheet does specify the saturation voltage of your device.  It's 0.1V, so it needs at least that much to operate at all.  The sheet specifies its characteristics at 5V, so that would work well.
« Last Edit: March 12, 2010, 03:49:50 15:49 by carbontracks » Logged
bbarney
Moderator
Hero Member
*****
Offline Offline

Posts: 2403

Thank You
-Given: 405
-Receive: 544


Uhm? where did pickit put my mute button


« Reply #6 on: March 12, 2010, 04:34:04 16:34 »

hers a circuit showing the difference along with a link to some other info
http://www.marktechopto.com/Engineering-Services/photo-sensor-application-note.cfm
Logged

Ever wonder why Kamikaze pilot's wore helmet's ?
koky
Active Member
***
 Warned
Offline Offline

Posts: 157

Thank You
-Given: 83
-Receive: 115


WWW
« Reply #7 on: March 12, 2010, 11:05:11 23:05 »

"Your problem is simple.  Using a transimpedance amp works fine for photodiodes, but not so well for phototransistors, which require some bias voltage across their terminals.  This is not specified in the datasheet, which is lame.  Your solution is to either use a photodiode, or modify the circuit to put some bias voltage across the phototransistor.

Also, I have no idea why you have a diode in the feedback loop... "

i have initially used a diode becouse in a schematich i have read the the ouput is similar a the eye sensitivity, but after i have replaced it with 470 Kohm resistor, and the output is more linear.
I have try to polarize the phototransistor with a resistor from pin 2 LM358 and +5V but the output not change.
My preoccupation is the fact that exist LM358 form varius constructor, but inside is not the same component

Next week i will try with photodiode instead phototransistor and i will post the result
Logged
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #8 on: March 12, 2010, 11:50:17 23:50 »

I have try to polarize the phototransistor with a resistor from pin 2 LM358 and +5V but the output not change.
A resistor from +5 to pin 2?  Of course that won't do anything.  Pin 2 will always be at 0V the way you've got it.  The only way to change that is to change the voltage at pin 3.  Try fixing pin 3 at some positive voltage (like 1 volt).  This should allow the phototransistor to operate, but your output will be offset by 1 volt as well.
Quote
My preoccupation is the fact that exist LM358 form varius constructor, but inside is not the same component
that's pretty unlikely.  Its input structure would have to be completely different.  Unless the chip is mislabled or a forgery, it should work fine.
Logged
solutions
Hero Member
*****
Offline Offline

Posts: 1446

Thank You
-Given: 590
-Receive: 851



« Reply #9 on: March 13, 2010, 12:45:22 00:45 »

"My preoccupation is the fact that exist LM358 form varius constructor, but inside is not the same component"

that's pretty unlikely.  Its input structure would have to be completely different.  Unless the chip is mislabled or a forgery, it should work fine.
Not as unlikely as you think.  Semiconductor companies patent their circuits and it is very possible that one company patented the input structure, forcing the second to do a workaround that gets them close enough in the datasheet to call it the same P/N.  Same goes for processing - each fab has its own recipe and Fairchild and ST have their own proprietary processes that the designers have to work with - that also can force their hands to creating a different character input

Look carefully at the input characteristics on the datasheets between the two parts and they are quite different.  The same part number, but a deceptive marketing ploy as the datasheets are not identical, like they should be for the same P/N.

Nobody's asked, so I will - why not just continue to use the Fairchild part?
Logged
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #10 on: March 13, 2010, 01:28:31 01:28 »

Not as unlikely as you think.  Semiconductor companies patent their circuits and it is very possible that one company patented the input structure, forcing the second to do a workaround that gets them close enough in the datasheet to call it the same P/N.  Same goes for processing - each fab has its own recipe and Fairchild and ST have their own proprietary processes that the designers have to work with - that also can force their hands to creating a different character input

Look carefully at the input characteristics on the datasheets between the two parts and they are quite different.  The same part number, but a deceptive marketing ploy as the datasheets are not identical, like they should be for the same P/N.

Nobody's asked, so I will - why not just continue to use the Fairchild part?
Performance related to process parameters, like input bias currents, offset voltages, DC gain, etc are bound to vary, but there's no way it could cause it to not function at all in this particular circuit configuration.  As long as it has a P-channel input stage, its inputs will work at zero volts, and it will work.  And I've looked over half a dozen variants of the lm358 and all of their schematics show pnp inputs.  If it didn't have a pnp input, it would be so drastically different that you couldn't give it the same P/N without essentially lying.
Logged
Parmin
Hero Member
*****
Offline Offline

Posts: 544

Thank You
-Given: 357
-Receive: 126


Very Wise (and grouchy) Old Man


« Reply #11 on: March 13, 2010, 03:36:07 03:36 »

You don't need power there - that's a phototransistor

I am not sure why suddenly there is a lot of "expert" contradicting my replies.

Are you saying that Phototransistor do not need any bias voltage?
Since when? can you cite a source that say so?
Logged

If I have said something that offends you, please let me know, so I can say it again later.
solutions
Hero Member
*****
Offline Offline

Posts: 1446

Thank You
-Given: 590
-Receive: 851



« Reply #12 on: March 13, 2010, 03:58:12 03:58 »

Fairchild AN-3005 Design Fundamentals for Phototransistor Circuits

http://www.fairchildsemi.com/an/AN/AN-3005.pdf

Says that you do need bias on the transistor....

Rather than change the board, try and find a photodiode in the same package.
Logged
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #13 on: March 13, 2010, 04:31:55 04:31 »

This exemplifies the reason photodiodes are so much more common than phototransistors.  Photodiodes can be used very easily in transimpedance amplifiers, and when used is transimpedance amplifiers their capacitance does not slow them down, making them far faster than phototransistors in non-transimpedance circuits.  There are ways to make phototransistors fast as well, but they require additional circuitry.
Logged
hate
Hero Member
*****
Offline Offline

Posts: 556

Thank You
-Given: 156
-Receive: 354


« Reply #14 on: March 13, 2010, 11:13:12 11:13 »

A phototransistor is a normal transistor with a photodiode implemented between its base and collector or it can be modeled like that. That's why it needs a bias for stability.

Same P/N chips from different manufacturers have to comply with the base specifications for that chip otherwise it wouldn't be same chip and same P/N. Improvements can be made though but they won't change the basic properties just improve them for different brands.

@koky: Are u trying to use this circuit in a digital manner? It will be very unstable this way even if it works. I suggest u use an amplifier first by replacing the diode with a resistor then use a comparator for a digital signal.

Regards...
Logged

Regards...
solutions
Hero Member
*****
Offline Offline

Posts: 1446

Thank You
-Given: 590
-Receive: 851



« Reply #15 on: March 13, 2010, 12:12:17 12:12 »

Same P/N chips from different manufacturers have to comply with the base specifications for that chip otherwise it wouldn't be same chip and same P/N. Improvements can be made though but they won't change the basic properties just improve them for different brands.

Tell that one to ST, not me.  You didn't look at the data sheets at all, even though I spoon fed you where to look - takes all of 5 minutes to see what I am talking about.

You're just spouting off idealogy vs the reality that some marketing hack putting a number on a part that tries to take the major market share from a competitor because "not too many people that we're interested in use it that way".

Yes, same number parts should have identical or BETTER specs - but "better" for whom if you are counting on the input circuit to bias up your phototransistor because it worked on the bench, you built a board, and now you're smoked when you picked someone else's part because that was on Farnell's shelf.
Logged
hate
Hero Member
*****
Offline Offline

Posts: 556

Thank You
-Given: 156
-Receive: 354


« Reply #16 on: March 13, 2010, 03:34:29 15:34 »

@solutions:

First of all calm down. Nope I hadn't checked the datasheets, I was talking in general and adding or deleting a couple of transistors in the IC doesn't make it another IC if it meets the specifications of the original.

Yes, same number parts should have identical or BETTER specs - but "better" for whom if you are counting on the input circuit to bias up your phototransistor because it worked on the bench, you built a board, and now you're smoked when you picked someone else's part because that was on Farnell's shelf.

If that's for me u haven't read my post clearly, here is that part again:

Quote
A phototransistor is a normal transistor with a photodiode implemented between its base and collector or it can be modeled like that. That's why it needs a bias for stability.

That says a phototransistor NEEDS a bias(that may be done via a resistor from base to gnd or other way) for a stable operation. It will work without a bias though but not very stable(may trigger with very little light as an example) and I'm talking in general not regarding to his circuit.

Ok as now I've checked both the datasheets of 'Fairchild' and 'ST' and ST version looks better on the specs. Regarding to input circuit there doesn't seem to be much change expect 'Fairchild' detailed its current source biasing the input with multi collector transistors and ST just showed the symbol(6uA). Other then ST has better bias specs they both have can have a 0V input that means both inputs can go down to 0V as that would be the case if the diode in the feedback loop would be substituted with a resistor.

Quote
I'd suggest trying various high valued resistors on the emitter to distance yourself up from ground if you're really bent on using the ST part.

Why would the ST parts input needs to be shifted please be more specific about this. Also please specify the difference in the input circuit more clearly and we'll discuss this.

Regards...
Logged

Regards...
bbarney
Moderator
Hero Member
*****
Offline Offline

Posts: 2403

Thank You
-Given: 405
-Receive: 544


Uhm? where did pickit put my mute button


« Reply #17 on: March 13, 2010, 03:58:13 15:58 »

Just let me warn all of you that any more post's here and anywhere in the forum better look like a discussion and not what I'am seeing here and on some other threads.
This I'am right and your idiot attitude stop's here and now !
Logged

Ever wonder why Kamikaze pilot's wore helmet's ?
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #18 on: March 13, 2010, 09:03:41 21:03 »

Well when someone sees a mistake they shouldn't hesitate to correct it.... of course name calling is bad but I haven't seen any so far.

Anyways, I think its clear that koky's problem is due to the fundamental circuit design, not a faulty component.  Even if the op amp were completely ideal, the circuit, as he presented it, would not work.  It's certainly a simple mistake to mix up phototransistors and photodiodes.  So once he tries fixing the design, we can see if there's anything else amiss that might be due to faulty components.

A phototransistor is a normal transistor with a photodiode implemented between its base and collector or it can be modeled like that. That's why it needs a bias for stability
if that's the case, then you could use a phototransistor as a photodiode by simply using the base-collector junction and leaving the emitter unconnected, correct?  Maybe he could try that to get the circuit working?
« Last Edit: March 13, 2010, 09:09:57 21:09 by carbontracks » Logged
hate
Hero Member
*****
Offline Offline

Posts: 556

Thank You
-Given: 156
-Receive: 354


« Reply #19 on: March 14, 2010, 09:46:33 09:46 »

if that's the case, then you could use a phototransistor as a photodiode by simply using the base-collector junction and leaving the emitter unconnected, correct?  Maybe he could try that to get the circuit working?

May be possible in theory in reverse bias conditions if the leakage current of the base-collector diode of the transistor wouldn't affect anything but in forward bias conditions there would be 2 parallel diodes, photodiode and base-collector diode of transistor so not sure about this.

I still think the diode in the feedback path should be a resistor whether the photo device is a diode or a transistor with a BIAS.

Regards...
Logged

Regards...
DreamCat
Active Member
***
Offline Offline

Posts: 235

Thank You
-Given: 130
-Receive: 77



« Reply #20 on: March 14, 2010, 11:56:08 11:56 »

I agree "hate", whether diode or transistor, all need bias.

but I also think there should change the ground of photo deive to negative supply.
Logged

May be I expressed the wrong meaning, sorry for my bad english. Please correct it for me if you can.
carbontracks
Junior Member
**
Offline Offline

Posts: 67

Thank You
-Given: 20
-Receive: 4


« Reply #21 on: March 14, 2010, 07:37:19 19:37 »

May be possible in theory in reverse bias conditions if the leakage current of the base-collector diode of the transistor wouldn't affect anything but in forward bias conditions there would be 2 parallel diodes, photodiode and base-collector diode of transistor so not sure about this.
Well in reverse bias the voltage across the junction(s) will be 0V, so there shouldn't be any current across the collector, in theory.  I don't see why it wouldn't work.
Quote
I still think the diode in the feedback path should be a resistor whether the photo device is a diode or a transistor with a BIAS.
Yeah, I really don't see the reasoning behind the diode either...
Logged
solutions
Hero Member
*****
Offline Offline

Posts: 1446

Thank You
-Given: 590
-Receive: 851



« Reply #22 on: March 14, 2010, 09:43:36 21:43 »

As I said the diode in the feedback loop linearizes the diode characteristic.  If he has a linear optical signal, vs ones and zeroes (which I suspect he does or he'd be happy to put a classical resistor in), then you need a log response amp.

A solar cell is a photodiode...needs no "bias".
Logged
hate
Hero Member
*****
Offline Offline

Posts: 556

Thank You
-Given: 156
-Receive: 354


« Reply #23 on: March 14, 2010, 11:12:11 23:12 »

Well in reverse bias the voltage across the junction(s) will be 0V, so there shouldn't be any current across the collector, in theory.  I don't see why it wouldn't work.Yeah, I really don't see the reasoning behind the diode either...

I was considering a general case where photodiode may have been pulled up to positive supply like in a transistor amplifier. In that case replacing that photodiode with a phototransistor with collector as cathode, base as anode and emitter open may result in a leakage current flowing through the base-collector diode. Even considering koky's circuit the diode in the feedback loop won't let inverting input to be at ground level, it will be more like a Schmidt trigger than a linear amplifier.

Regards...
Logged

Regards...
Parmin
Hero Member
*****
Offline Offline

Posts: 544

Thank You
-Given: 357
-Receive: 126


Very Wise (and grouchy) Old Man


« Reply #24 on: March 14, 2010, 11:31:38 23:31 »

http://www.fairchildsemi.com/an/AN/AN-3005.pdf
Says that you do need bias on the transistor....

I thought so,  you are not citing anything I did not know before.

I was just wondering why you say

Quote
You don't need power there - that's a phototransistor

Biasing is some sort of Power, so yeah.. my original question stands..

Quote
A solar cell is a photodiode...needs no "bias".

That is correct, but the OP ask for phototransistor circuit.
So, we answer accordingly.
Logged

If I have said something that offends you, please let me know, so I can say it again later.
Pages: [1]
Print
Jump to:  


DISCLAIMER
WE DONT HOST ANY ILLEGAL FILES ON THE SERVER
USE CONTACT US TO REPORT ILLEGAL FILES
ADMINISTRATORS CANNOT BE HELD RESPONSIBLE FOR USERS POSTS AND LINKS

... Copyright 2003-2999 Sonsivri.to ...
Powered by SMF 1.1.18 | SMF © 2006-2009, Simple Machines LLC | HarzeM Dilber MC