Photodiode or phototransistor for receiving ~15KHz IR signal?

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uoficowboy

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Hi - I need to send a PWM signal through IR. My plan is to PWM an IR LED at 15KHz and then receive it on the other end with an IR phototransistor. The output from the phototransistor would then be buffered, band pass filtered, and rectified (for peak detection). My goal is to get a DC voltage representative of the strength of the IR signal. To be clear - the whole goal here is to figure out how much the IR is attenuated. So there is no data being put on the IR signal - just a simple PWM.

My questions are as follow:

Is an IR LED or an IR phototransistor better for this?

I understand phototransistors have more limited bandwidth - but they seem to be up near ~100KHz+, so maybe I'm OK? Further, the way their ft is speced is confusing - for example - the Vishay TEST2600 quotes a ft of 110KHz with Vs = 5V, Rl = 100 ohms, and IC = 5ma. I'm assuming this means they have a 100 ohm resistor and the TEST2600 in series, with 5V across the series combination of them. But does it also mean that it has enough light shining on it to already have 5ma flowing through it? So does that mean it's better to drive my LED with a PWM signal that does not have 0ma during the low state? (in other words - PWM would switch between 10 and 20ma instead of between 0 and 10ma)

I understand that IR photodiodes typically need more conditioning electronics, but I don't view that as a particularly large problem as there will already be plenty of electronics in the circuit regardless, and it is not super sensitive to BOM cost.

Thank you for any light you can shine on this!!
 

Hi,
As far as I know, photodiodes are faster than phototransistors. So if you are interested to obtain good response for fast signals, use a photodiode followed by an amplifier. Phototransistors are more sensitive, they are able to generate an electric signal from a weaker optical signal (because of the amplification of the transistor), but they operate better with slower signals. I consider 15kHz is not a very fast signal anyway, so I'd first give a try to a phototransistor.
You should drive your led between zero current to nominal current, this way the signal will be more easily recognised by photoelement. Of course, the highest sensitivity of the optical signal appears in darkness, ambient light will have a negative impact. Usually they use IR filters in front of the receiver, which block significantly the visible light, but let IR pass through. The phototransistor you indicated has IR filter, so that makes things easier. Also, fT is about 7 times higher than your signal, so you can use that phototransistor.
You should do some tests, you only need a few components for transmitter and receiver (could build the transimitter with a LM555 and the receiver with a photoreceiver and an opamp or two transistors for example and check the signal amplitude with an oscilloscope, if you have access to one).

Regards,
Nicolae
 
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Hi Nicolae! I have the same understanding that photodiodes are faster than phototransistors. What I'm not sure of is if a phototransistor has any advantages over a photodiode that has an amplifier on it.

Indeed, I already have done some minimal tests and got very low signal amplitude out of the phototransistor I mentioned. With a 100 ohm resistor, I was seeing something along the lines of 50mV when I had a matched IR LED just a few cm away from it. It was very worrisome! I will do more testing soon to see if I can improve the situation.
 

We call it a specification of test conditions. What's confusing with it? Bandwidth respectively cut-off frequency is a small-signal parameter, there must be a DC bias to measure it. The specification doesn't say, how fc will be different for a different bias point or suggest a specific mode to operate the device.

The advantage of a photo transistor is that it saves an amplifier in some applications.

I understand that you want a simple square wave modulation. Strictly spoken, it hasn't to do with PWM (pulse width modulation), it's just the most simple way to generate modulated light.
 

Hi FvM - I agree that you need to specify test conditions! My confusion is that it seems to me that 5ma is quite a lot - but perhaps I am mistaken. Do you have any feel for if I should be trying to pre-bias the photo-transistor or not? My instinct says yes, but I am not positive about that. But I'm also thinking that I should just forget about the phototransistor and switch to a photodiode since I've already seen bandwidth issues.

You're right that it's not actually PWM. Apologies for the mistake!

-Michael
 

I think, Figure 13 in the below datasheet can shine a light on the speed versus Ic properties of phototransistors. Although it's from an optocoupler datasheet, you can most likely expect similar behaviour from all silicon photo transistors.

https://www.vishay.com/docs/83606/cny17.pdf
 
Interesting - it looks like you definitely want to bias the parts! I do not believe photodiodes show the same property, correct?

one more thing: If I wanted to calculate the irradiance that the phototransistor is seeing, and the resulting collector current, how does this look? I'm using the Vishay TSSS2600. It has a typical radiant intensity of 2.6mw/sr at 100ma. So let's assume I'm operating at 100ma. If my photodiode is directly facing my phototransistor and is 10cm away, I believe the irradiance is 2.6mw/(10cm)^2 = 26uw/cm^2. Does that look right? That comes from the spherical surface area of a single steradian being, I believe, r^2. That means my phototransistor, the TEST2600, with a typical response of 2.5ma @ 1mw/cm^2, would have a current of 2.5ma * 26uw/1mw = 26ua. Does that all seem somewhat accurate? I must admit I am not very experienced with this sort of calculation.
 


I would expect a better amplitude than 50mV when using an IR led so close to it. An IR led has a typical current of 50...100mA, more than visible leds (the old generation). Are you applying the recommended level of current to the IR led? Is the wavelength of the led you are using close to the peak wavelength of your phototransistor?
In order to identify if the poor amplitude is related to limited bandwidth of the phototransistor, you could reduce the modulating frequency and see if the amplitude improves (I don't expect so). Try to use a 50% duty cycle of the squarewave signal. If you use a larger value resistor in collector of your phototransistor (1kohm...4.7kohm), you can probably get higher amplitude.

Regards,
Nicolae
 

Hi uoficowboy,

I've done some simple tests with an infrared led (CQY89) and a very old phototransistor (FPT1010) and also an infrared diode (type unknown). My phototransistor is said to have a raise time of 6us and a fall time of 7us. It was receiving fine a 2kHz squarewave signal, but it was pretty bad at 15kHz (it was too slow). At 10cm and 2kHz, the amplitude was about 3Vpp. The photodiode was significantly faster, although at 10cm the signal amplitude was of the order of 100mVpp. After I added a transistor, the photodiode signal has improved, although the response was also slower (not as slow as the phototransistor though).
You can see some photos at **broken link removed**

Regards,
Nicolae
 

At 10cm and 2kHz, the amplitude was about 3Vpp.
With 100 ohm load resistor as used by the OP, you are in the same order of magnitude. 100 ohm is actualy reasonable with a phototransistor to achieve the intended bandwidth. But a cascode circuit can achieve low phototransistor load impedance and higher output voltage.

it looks like you definitely want to bias the parts! I do not believe photodiodes show the same property, correct?
The internal photodiode is seeing rbe as load impedance, it's inverse proportional to Ic, the cut-off frequency formed by Cbe || Ccb and rbe is respectively increasing with Ic.

A photodiode would be preferably operated with constant load impedance or "zero input impedance" transconductance amplifier.
 

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