Amplifying and filtering the Phototransistor output for pulse signals

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patan.gova

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Hello,
I am using NJL5501R phototransistor for the pulse signal purpose.I used the circuit as shown in the second link but used NJL5501R phototransistor instead of MCT6 and got the response as this **broken link removed**
But the output signal is of low amplitude and noisy so I want to amplify the signal for which I used the circuit **broken link removed** but was unable to amplify the signal instead the signal was very bad compared to the original phototransistor output .
So can someone help with the amplification or the filtering of the phototransistor signal.
And any related links for the pulse signal extraction using the phototransisor.
Thanks in Advance.
 

Datasheet Archive dot com could not find the datasheet for an NJL5501R photo-transistor.

The circuit you used has a very high IR diode current so the photo-transistor is probably saturated (turned on as hard as possible). Then the opamp is also saturated and does not amplify.
 

Hi ,
Sorry for not providing the datasheet and here I have attached it.
So how to make the signal better.
Thnks
 

Attachments

  • 8508.NJL5501R_E_20130418.pdf
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The current in the red LED is (3V - 2.1V)/100 ohms= 9mA to (3V - 1.7V)/100 ohms= 13mA. Then the output of the photo-transistor is 2.25mA to 14mA and the output voltage of the photo-transistor is extremely over-saturated. The spec's in the datasheet used an LED current of 4mA, maybe from an accurate current source circuit.

You can calculate how much the IR LED over-saturates the photo-transistor.

Your photo-transistor is over-saturated because the LED currents are too high and because the 5.1k collector resistor value is much too high. The circuit in the datasheet has the load resistor at the emitter, not at the collector.
 

Sorry, I am not that good in datasheet analysis can I know how you related the red LED current of 9ma-13ma to the phototransistor current of 2.25mA-14mA and how to relate this current to phototransistor output.I mean form which plot.
The IR current is (3V-1.3)/100ohms=17mA to (3-1)/100=20mA.
And also can u suggest how to overcome this saturation problem.
 

Please note that, in your circuit (second link), the opamp output is always close to zero.
As you know, when the output is in its linear range (between upper and lower limits), the inverting input is equal to the non-inverting one.
So let us assume that the opamp is in its linear state. IN+ is 0.5V. Therefore IN- should be also 0.5 V.
If no light, a dc positive current passes from V3 to IN- thru R2 (5K1) and to the output thru R2 (120K). But Vout (of the opamp) cannot be lower than 0V. The voltage at IN- (Vn) is therefore:
Vn = V3 * R1 / ( R1 + R2 )
Vn = 3 * 120 / ( 120 + 5.1 )
Vn is close to 3V.
It is clear that the opamp is saturated already and its output is close to ground.

About the opto part, I will read the datasheet to find out the practical values of R2 and R3 for V3 = 3V. Let us not forget that the response amplitude depends also on the distance between the device and the reflective surface (also on the material of the latter).

Kerim
 
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Thanks a lot kerim for the information.
I will be waiting for your suggestions to overcome this.
 

Let us start with the RED LED.

From "Output Current vs. Forward Current, RED LED (Ta=25°C)"
The collector current is 2mA @IF=6mA and Vce=0.5V.
Rc = ( Vcc - Vce ) / Ic
Rc = ( 3 - 0.5 ) / 2 = 1.25 K
We choose R2 = 1K2

From "Forward Voltage vs. Forward Current"
The forward voltage of the RED LED is about 1.82V @6mA.
R3 = ( 3 - 1.82 ) / 5 = 0.197 K
We choose 180R for R3.

Added:
I guess you like doing the calculations for the IR LED by yourself now
Please note that the above Ic is for a distance d=0.7 mm.
Ic (%) versus distance is plotted on "Output Current vs. Distance, RED LED (Ta=25°C)".
 
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Hi Kerim!!
Thanks for this neat and clean explanation. If you don't mind can I know why you have selected IF=6ma and Ic=2ma for Vce=0.5 why not the other combinations of IF and Ic corresponding to the Vce=0.5. similar for the selection of IF for the LED resistor calculation.
 

Good question. I mean you can select any other point if you like
In my case, I started with Vce=0.5V which is low enough without requiring a relatively high input forward current. It happens that at Vce=0.5V the lowest IF is about 6mA where Ic is 2 mA which is, in general, good enough in noisy environment.
So you can experiment other points anytime you like.
I mean you can see the calculated resistors (earlier) as a starting step.
As you know, the main goal is to finally get what the circuit is supposed to do in a practical situation (including mechanical and electrical factors, as vibration and noise for example).
 

Thnks Karim and the calculation for IR goes like this
Taking Vce=0.5 into account
Collector Resistor:
So, from Output Current vs. Forward Current, IR LED (Ta=25°C)"
If IF=4ma then Ic is around 0.2ma ; Rc=(3-0.5)/0.2=12.5k
If IF=6ma then Ic is around 0.36ma ; Rc=(3-0.5)/0.36=6.9k
If IF=8ma then Ic is around 0.5ma ; Rc=(3-0.5)/0.2=5k
LED resistor:
So, "Forward Voltage vs. Forward Current"
The forward voltage of the IR LED is about 1.1V @4mA;R3=(3-1.1)/4=0.47k
The forward voltage of the IR LED is about 1.15V @6mA; R3=(3-1.15)/6=0.3k
The forward voltage of the IR LED is about 1.19V @8mA; R3=(3-1.19)/8=0.22k

Isn't it what you explained above.
So, now I am confused with the selection of practical resistor value because for a single of Vce(0.5) there are two possible combinations of resistor values corresponding to the each particular IF value.How the IF value can be determined to select a pair of calculated resistor values to implement practically onboard.
 

I am not familiar with applications that need such an opto device.
I think one may need using one LED only. The choice between the two may depends on the environment in which the device works.
I guess, in a practical situation, one tests each case to find out which one seems having a more reliable response. If both are good, either one could be used
 

Ok then I am planning to use the IR LED and I will practically check the results of different combinations of calculated resistor pair values with Vin+=0.5.
Can you please explain a bit with the selection process of feedback of omamp to amplify the signal with the mentioned phototransistor so that the opamp amplifies (unlike the getting saturated has happened practically with the circuit shown in the link above).
As my final goal is to amplify the photo transistor signal and filter if needed.
 

To design an amplifier properly, one needs first knowing the minimum level at its input (estimated or better be known by a basic test).
Also the required level of the output voltage (its upper and lower limits) that is suitable for the next stage (as an MCU pin for example) should be given.
I hope you agree with me that a problem without given data couldn't be solved to get a sure final answer
 

An IR LED plus a red LED with a photo-sensor are used in a "Reflective Pulse Oximeter" to measure blood pulses and blood oxygen levels in a patient.

The spec's for the device vary a lot with different ones so the "typical" graphs probably need some tweaking with each one since nobody sells "typical" devices. The written spec's show the wide range for some of the spec's.
 

@kerim :I am planning to use MSP430 microcontroller for reading the pulse output signals and the OPA380 as my opamp.The OPA380 datsheet is attached below.

@Audioguru:Yes, I am building a reflective pulse oximeter and as a first part I am building the pulse signal circuit with IR Led.But I was unable to amplify the signal because of the opamp saturating problem .So, need help with the amplification part design as I am not that good in analog circuit design.
 

Attachments

  • opa380.pdf
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Thank you for attaching the datasheet.
Please ignore my previous remarks about the opamp since I didn’t know it is a transimpedance amplifier.
 

For instance, I wasn't familiar with oximeters so I read about them and the one that uses the RED and IR LED.
I thought first that the project is for an industrial application.

You are right, in case of reflective pulse oximeter, the signal at the output of the opto IC is likely weak and should be amplified.
I prefer not to go on because I have no idea how this signal (its strengh, shape... etc) should be analysed and interpreted (mostly by the MCU) to get useful data.
 

@kerim: thanks for the information about the phototransistor.
And I hope someone will help with this.
 

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