• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

How to reduce the voltage needed for a photointerrupter

Status
Not open for further replies.

Mr RB

Well-Known Member
Sounds about right, you can characterise the circuit by connecting 24 volts, then measure the voltage at the bottom of that resistor, where it connects to the LED and phototransistor.

Then choose a new resistor value so that voltage remains the same with supply of +5v and the new resistor.
 

kchriste

New Member
Forum Supporter
What I'm sure Mr_RB really meant to say was, "Measure the voltage across the 3.3K resistor with 24V input and calculate the LED current. At the same time measure the voltage across the LED (VfLED). Then choose a resistor that will give you the same LED current at 5V - VfLED"
 
Last edited:

vipsam10

New Member
How can we convert -48V to +12V

Hi.
I recently got a project about power conversion. I have an equipment which requires +24V DC. I am currently using two 12V batteries to get the required voltage. I need to replace the two batteries which get damage too often with an alternate source. I have as option to convert the output of a rectifier (230VAC to 48VDC) to +24V. The issue here is that the +ve terminal of the rectifier output has been earthed and only the positive terminal is accessible. As such i need to convert -48V to +24V. Note that i cannot use an additional power supply, i cannot use another rectifier and i cannot use an inverter as well. Is such a project feasible ladies and gentlemen?
Any proposal will be most appreciated.

Thank you
 
Last edited:

Mr RB

Well-Known Member
What I'm sure Mr_RB really meant to say was, "Measure the voltage across the 3.3K resistor with 24V input and calculate the LED current. At the same time measure the voltage across the LED (VfLED). Then choose a resistor that will give you the same LED current at 5V - VfLED"
That's not what I said or what i meant to say at all! :mad:

The LED current is fairly irrelevant. The thing he needs to maintain is the VOLTAGE at the phototransistor to maintain correct biasing. An opto-interruptor like that will operate fine with a very large range of LED current values.

Blueroom already gave the answer, then I added a simple logical method to determine the resistor value and maintain reliable operation. At no point was I incompetent either with the advice I was providing or with my ability to provide the advice.
 

AlainB

Member
Hi,

Well, I did change the resistor for a 300 Ohms one. I did not had a 330 Ohms on hand. It is working nicely at 5 volts.



I plugged the +5 volts to Vcc pin and the ground (0 volt) to the GND pin. On the datasheet they say that when detecting a substance (??) the output become high level. What I experiment is that if it is not triggered (no substance) I have 5 volts between Vcc and out. If trigered, I have 0 volt. The contrary of what I am expecting. Am I doing something wrong? They say that it is an open collector output. Would that mean that I should use another voltage or circuit to get the expected behaviour?
 

Attachments

Last edited:

Chippie

Member
Hi.
I recently got a project about power conversion. I have an equipment which requires +24V DC. I am currently using two 12V batteries to get the required voltage. I need to replace the two batteries which get damage too often with an alternate source. I have as option to convert the output of a rectifier (230VAC to 48VDC) to +24V. The issue here is that the +ve terminal of the rectifier output has been earthed and only the positive terminal is accessible. As such i need to convert -48V to +24V. Note that i cannot use an additional power supply, i cannot use another rectifier and i cannot use an inverter as well. Is such a project feasible ladies and gentlemen?
Any proposal will be most appreciated.

Thank you
You need to start a new thread rather than hijacking this one......
 

AlainB

Member
You can make the voltage go high instead of low by adding a PNP transistor on the output like this:
Thanks!

But It will not be needed after all. The CNC software I use (TurboCNC) has a provision to inverse a high trigger to a low. But anyway, having a normal high is better for a limit switch. It will get low and stop the machine if tripped but it will also stop the machine if the connexion is lost due to the wire or the switch itself being damaged or disconnected.

As I said the switch is working fine at 5 volts with a 300 Ohms resistor. It is working fine also with a 1K Ohms resistor. From the datasheet, the max current comsuption of the hole thing is 9,5 mA. I can't figure out wy they use 3K3 Ohms at 24 volts but I think that maybe 300 Omhs at 5 volts is too near the limit. Would it be better to use a value around 500 or 600 Ohms?

Alain
 

kchriste

New Member
Forum Supporter
That's not what I said or what i meant to say at all! :mad:
The LED current is fairly irrelevant. The thing he needs to maintain is the VOLTAGE at the phototransistor to maintain correct biasing.
No. The LED current is VERY relevant! VfLED is less so. As you can see from the diagrams posted above, the LED is actually performing two functions:
1) It emits light.
2) It acts as a shunt regulator for the photo-diode AMPlifier.
The best way to duplicate the operating point of the circuit is to go by the LED/AMP current into pin3 and not the voltage across the LED.
 
Last edited:

mneary

New Member
That's not what I said or what i meant to say at all! :mad:
The LED current is fairly irrelevant. The thing he needs to maintain is the VOLTAGE at the phototransistor to maintain correct biasing.
Unfortunately the phototransistor is inaccessible. The output is buffered and is a logic signal (see data sheet page 5/6).

We have access to two other pieces of information; the LED/opamp voltage and the sensor input current. As Blueroom/kchriste have suggested, the current is easy to manage.
 

Mr RB

Well-Known Member
I was well aware of the relationship between current and voltage on the 3k3 resistor.

The easiest and most logical test was to measure the operating voltage on the 3-pin sensor with the original input voltage and original resistor (with reference to ground). Then replicate that 3-pin sensor operating voltage with the correctly chosen new resistor. That was important factor to replicate. Kchistie focused on the LED current, saying that was what I "meant" which it was not, as the LED current was barely relevant. I have worked with these opto-interruptors for years, the critical factor is the biasing voltage of the detector.

If kchristie wanted to to a clumsier test and measure current, or convert voltages to current for some reason, then that is his perogative, what I objected to was his assumption that he knew what I MEANT better than I myself did. And explaining my post as if to correct my "incompetence" when he didn't even seem to grasp that the critical factor was the biasing voltage fo the detector, not the transmitter current.

As for the 3-pin device being integrated or just a phototransistor I had no way of knowing that without reading the datasheet which was not necessary or relevant to the good advice I gave re measuring the device and choosing a specific resistor value.

If you think I'm worng then just say so, I'm wrong as often as the next guy. But don't explain what my words "mean" as if I was too stupid to choose the right words myself. :(
 

mneary

New Member
The sensor data sheet shows an internal resistor in series with the LED and op amp. It would be unclear where that resistor is, until we study the photograph carefully and compare it to the data sheet electrical and mechanical drawing.

After some study, some of us decided that the unmarked resistor in the data sheet is the 3.3k resistor that's visible on the underside of the sensor assembly. I, (and apparently kchriste) felt that matching the current consumption into the V+ pin was the best way to choose the new resistor value.

Since this resistor is internal to the sensor, choosing its value does require careful study of the sensor data sheet. I think that suggesting a change to the part without looking at the data sheet was hasty.

If you had some other resistor in mind, (external to the sensor?), I was confused and please accept my apology. Actually I still am confused.
 
Last edited:

AlainB

Member
This is another part that I have. Its number is S7275-1 and I did not found any datasheet for it.

The circuit being more complex I just tried it without modification at 5 volts and to my surprise It is working. I then tried an unmodified one like the first one and it is working too at 5 volts. Same crisp and clean trigger.

For the usage of this part, as I understand it now, the CNC software keep the photointerrupter dedicated parallel port pin high. Since its output (of the part) is floatting when not trigered, the port pin stay high. When triggered, It connect the ground to the port pin forcing a low witch indicate to the software to stop everything. Does it make sense?

Alain

 

Attachments

Last edited:

AlainB

Member
"303","122","122" and "PA"

I suppose that because of the PA being there this part has a much higher output capacity then the other one??

What is that PA? Is it a PNP transistor?
 
Last edited:

mneary

New Member
PA is probably either a Toshiba 2SC4666A or 2SC3295A (which probably doesn't matter since I don't see any difference between them anyways). It would be a high gain (600-1200) NPN transistor, 150 mA, 50V. [edit] Actually it appears that it's probably the 2SC4666A because it's the narrower one.[/edit]

Although the collector is rated at 150 mA, it's not wise not to push it because it could make the crisp operation much softer.
 
Last edited:
Status
Not open for further replies.

EE World Online Articles

Loading
Top