Strange Op-amp circuit...

MikeML

Joined Oct 2, 2009
5,444
I was just worried that it would draw too much current? Now knowing that is is a receiver, is my wiring still correct?

and with regards to the op-amp... What EXACTLY is the 10M load resistor for? I thought the input bias current wouldn't need a current limiting resistor?
No, your wiring will not work.

As with most circuits you grab off the web, the originator of the video is an inexperienced newbie who doesn't understand how to apply the subject IR receiver.

As I read Fig 4 in my post #18, above, if you put 5V across the receiver, it will source anywhere from 150uA (in the dark) to 20mA when maximally illuminated. I'm guessing that because he used a 10meg load resistor for the detector, he was concerned that he has only a tiny bit of illumination. E=IR, so if the receiver is putting out 150uA, the nominal voltage across the 10meg resistor would be 1.5V.

The opamp is configured as a voltage follower. However, if powered with 5V, the LM324 can only pull its output pin to about 3.5V, so the useful range of inputs to the Arduino's AD converter will be about 1.5V to 3.5V, a crappy design, with very little head-room.
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
No, your wiring will not work.

As with most circuits you grab off the web, the originator of the video is an inexperienced newbie who doesn't understand how to apply the subject IR receiver.

As I read Fig 4 in my post #18, above, if you put 5V across the receiver, it will source anywhere from 150uA (in the dark) to 20mA when maximally illuminated. I'm guessing that because he used a 10meg load resistor for the detector, he was concerned that he has only a tiny bit of illumination. E=IR, so if the receiver is putting out 150uA, the nominal voltage across the 10meg resistor would be 1.5V.

The opamp is configured as a voltage follower. However, if powered with 5V, the LM324 can only pull its output pin to about 3.5V, so the useful range of inputs to the Arduino's AD converter will be about 1.5V to 3.5V, a crappy design, with very little head-room.
Thanks for the eye opener! I will definitely re-configure my circuit tomorrow... Sorry for the late reply. So since the op-amp is configured as a voltage follower and I will be running the LM324 at 5VDC, is it possible to have the range to my Arduino's AD converter as 0v-5V? Or will 1.5V-3.5V be sufficient as I want to map it as an 8-bit resolution out of the PWM pins (0-225). Please advise.

Regards
 

MikeML

Joined Oct 2, 2009
5,444
The Arduino AD is 10bits (1024 steps), using a Vref of 5V the input range is from 0V to 5V. If it only goes from 1.5V to 3.5V, you will only see 1024*2/5 = 409 steps.

Can you power the voltage follower opamp on a supply higher than 5V?
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
The Arduino AD is 10bits (1024 steps), using a Vref of 5V the input range is from 0V to 5V. If it only goes from 1.5V to 3.5V, you will only see 1024*2/5 = 409 steps.

Can you power the voltage follower opamp on a supply higher than 5V?
Yes I can, I will be using a voltage regulator to regulate a 12V powersupply to 5V. But I have just measured the actual voltage out of the unregulated supply and it is 16VDC

Circuit Schem 2.jpg
Would the above circuit suffice?
The gain of the opamp would be 50 using the resistors shown above (25000/500)
 
Last edited:

MikeML

Joined Oct 2, 2009
5,444
Yes I can, I will be using a voltage regulator to regulate a 12V powersupply to 5V. But I have just measured the actual voltage out of the unregulated supply and it is 16VDC
...
Would the above circuit suffice?... (25000/500)
No, that might blow up the detector.

Where did the SFH482 emitter and BPX81 detector come from? I thought you are using a LIR131 and LIR132????
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
No, that might blow up the detector.

Where did the SFH482 emitter and BPX81 detector come from? I thought you are using a LIR131 and LIR132????
No, that might blow up the detector.

Where did the SFH482 emitter and BPX81 detector come from? I thought you are using a LIR131 and LIR132????
Could not find the required parts in CAD Soft... So to not blow up the phototransistor would i need a current limiting resisotr? And is the rest of the circuit okay?
 

MikeML

Joined Oct 2, 2009
5,444
I want you to do this test:

253.gif

I want to know V(out) with the IR path blocked, and unblocked with the air gap and alignment between the LIR131 and LIR132 just as it will be in your final device.

If V(out) is close to 5V for either condition, then change R1 to 10K, and measure again.
 

DickCappels

Joined Aug 21, 2008
10,153
The things I do not understand:
1. Where are all the resistors to set the op-amp's gain?
2. Why is there a 10M resistor going from the non-inverting input to ground?
3. Should there not be a resistor between the inverting input and op-amp output?
http://imgur.com/IhItULj
1. Since the inverting input is connected directly to the opamp's output and the input is applied to the noninverting input (where else?), the circuit has a gain of +1.

2. I think that was answered earlier in the thread -they are the loads for the photo transistors.

3. Yes, adding a resistor in series with the inverting input can help minimize offset and drift. With 20 na of input bias current (and that's for typical input bias current, not the max, of the classy version of the LM324) you would get a 200 mv offset on the non inverting input of the LM324. If you put a resistor of the same value on the inverting input, that resistor will develop a voltage that will partially compensate for the 200 mv drop across the resistor on the noninverting input.

Take note that that 200 mv or more roughly doubles every 10° C. The CMOS opamp would give you much lower input bias current.
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
I want you to do this test:

View attachment 77522

I want to know V(out) with the IR path blocked, and unblocked with the air gap and alignment between the LIR131 and LIR132 just as it will be in your final device.

If V(out) is close to 5V for either condition, then change R1 to 10K, and measure again.
Hi

Thanks for the help! Unfortunately I will only be receiving the additional components needed for my project around the 7th/8th of January 2015 due to the festive season. As soon as I have everything required I will create a new post or resume with the project on this post.

Regards
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
1. Since the inverting input is connected directly to the opamp's output and the input is applied to the noninverting input (where else?), the circuit has a gain of +1.

2. I think that was answered earlier in the thread -they are the loads for the photo transistors.

3. Yes, adding a resistor in series with the inverting input can help minimize offset and drift. With 20 na of input bias current (and that's for typical input bias current, not the max, of the classy version of the LM324) you would get a 200 mv offset on the non inverting input of the LM324. If you put a resistor of the same value on the inverting input, that resistor will develop a voltage that will partially compensate for the 200 mv drop across the resistor on the noninverting input.

Take note that that 200 mv or more roughly doubles every 10° C. The CMOS opamp would give you much lower input bias current.
Hi

Thanks for your input it really cleared a few things up for me!
1. In regard to the gain of plus 1...
Is a gain of +1 normal for an infrared signal booster type circuit? Will my gain of +50 be too much and cause clipping? Please advise.

2. I am still confused as to what the load resistor does. Does is prevent too much current from going into the op-amp?

Regards
 

DickCappels

Joined Aug 21, 2008
10,153
I don't remember seeing a sensor connected exactly this way, so I cannot comment on whether a gain of 1 is typical or not. It appears that the designer intended to set the sensitivity by selecting the right load resistor to go with his photo transistors in his application.

The load resistor is pretty simple. The photo transistor provides current as a function of the amount of light falling on it. This current times the resistance produces a voltage across the resistors. In this case you have 10 meg ohm resistors, so you would get 10 volts per microamp, which sounds like a lot to me. There is a limit to the amount of voltage the phototransistor can get to drop across the load resistor and that maximum voltage is the power supply voltage minus a little bit (probably a couple of tens of millivolts lost in the transistor itself.
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
I don't remember seeing a sensor connected exactly this way, so I cannot comment on whether a gain of 1 is typical or not. It appears that the designer intended to set the sensitivity by selecting the right load resistor to go with his photo transistors in his application.

The load resistor is pretty simple. The photo transistor provides current as a function of the amount of light falling on it. This current times the resistance produces a voltage across the resistors. In this case you have 10 meg ohm resistors, so you would get 10 volts per microamp, which sounds like a lot to me. There is a limit to the amount of voltage the phototransistor can get to drop across the load resistor and that maximum voltage is the power supply voltage minus a little bit (probably a couple of tens of millivolts lost in the transistor itself.
Thank you so much! I now understand the function of the load resistor. I will look at the datasheet of my RX and try to create a circuit accordingly with the correct load resistor
 

MikeML

Joined Oct 2, 2009
5,444
Sort of looks like the test circuit I posted, doesn't it. It matters not if the load resistor is above or below the receiver.

The 10megΩ "load" resistor in the circuit from the video is at least thousand times too high, prompting my "newbie" comment.
You put in a 33Ω resistor, and moved the 10MegΩ to the wrong place. It is supposed to go to ground, and the opamp signal taken from the junction between the receiver and the load resistor, not in-series with the opamp input.

In looking at the plot from the data sheet I posted, it is obvious to me that the load resistor needs to be somewhere between about 10KΩ and 100KΩ. The measurements I want you to do will answer that question. They will also answer the question about how much gain/offset the opamp should supply...
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
Sort of looks like the test circuit I posted, doesn't it. It matters not if the load resistor is above or below the receiver.

The 10megΩ "load" resistor in the circuit from the video is at least thousand times too high, prompting my "newbie" comment.
You put in a 33Ω resistor, and moved the 10MegΩ to the wrong place. It is supposed to go to ground, and the opamp signal taken from the junction between the receiver and the load resistor, not in-series with the opamp input.

In looking at the plot from the data sheet I posted, it is obvious to me that the load resistor needs to be somewhere between about 10KΩ and 100KΩ. The measurements I want you to do will answer that question. They will also answer the question about how much gain/offset the opamp should supply...
Thanks for the help, really do appreciate it good sir! As stated in one of my previous posts, I unfortunately will only be able to carry out the tests you have given me around the 7th/8th of January 2015 as I will be receiving the additional components required for my circuit then.

I will correct my circuit accordingly. How does one deduce that the load resistor is between 10KΩ and 100KΩ simply by reading the plot data? Ohm's law?
 

MikeML

Joined Oct 2, 2009
5,444
...How does one deduce that the load resistor is between 10KΩ and 100KΩ simply by reading the plot data? Ohm's law?
Well, 55 years of experience, for starters...

There is a technique based on Ohms Law and Kirchoff that designers use called "load-line analysis". Follow the link and read while you are waiting for your parts.

btw-I would never commit to a design, pcb, hardware construction, unless I bread-board a circuit first. The little test I am having you do is a bread-boarding step...
 

Thread Starter

T3chNique

Joined Dec 22, 2014
25
Well, 55 years of experience, for starters...

There is a technique based on Ohms Law and Kirchoff that designers use called "load-line analysis". Follow the link and read while you are waiting for your parts.

btw-I would never commit to a design, pcb, hardware construction, unless I bread-board a circuit first. The little test I am having you do is a bread-boarding step...
Thanks I will begin reading it now. In regard to the phototransistor... I will be powering it with 5VDC and it's datasheet states that it will draw 1A? How do I use this data?
 

MikeML

Joined Oct 2, 2009
5,444
Thanks I will begin reading it now. In regard to the phototransistor... I will be powering it with 5VDC and it's datasheet states that it will draw 1A? How do I use this data?
You are still reading the wrong data sheet for the receiver. The 1A number is the peak current pulse allowed for a few usec for the IR-204C emitter that is wrongly linked to from this web page.
I hope that when you finally get your parts, that they send you a real IR photo transistor (Like the LR153), instead of that emitter.
 
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