0
\$\begingroup\$

I use a phototransistor OPB608A in my system,shows as below.After capacitor C5, the voltage is 5.5V, that means it will give 5v supply to Pin 1 and Pin 3 of OPB608A, the resistors values are 7.5K and 150 Ω,respectively.However, when i use multimeter to measure at point A, it always shows 1.71 V no matter there is something or nothing in front of phototransistor. but when i use stabilized voltage supply to do the same test, the result will be normal---that means when i put a paper in front of phototransistor, the voltage will decrease to nearly 0V, if there is nothing, then it will be nearly 5V. Someone can tell me why this happened?thank you so much.enter image description here

\$\endgroup\$
1
\$\begingroup\$

Pin 3 is the anode of an LED whose cathode connects to pin 3 and thence to ground. This means R3 supplies current to the LED and therefore WILL always show about 1.7 volts from pin 3 to pin 4. The forward volt drop of the LED is specified in the data sheet as 1.7 volts so no problems there.

This LED voltage will not change if paper or some other light-blocking material is brought in front of the device - it's a light emitting diode.

Pin 1 is the collector of the phototransistor and it's emitter (pin 2) connects to ground. The collector is connected to 5 volts via the 7.5k resistor R2. This part of the device is sensitive to light and pin 1 will flicker between 5V and 0V depending on the amount of incident light.

My advice is read the data sheet.

\$\endgroup\$
  • \$\begingroup\$ yes,thank you for your comment,i found the problem, i think i need to do some modification on my PCB board. \$\endgroup\$ – Mark0923 Jul 11 '14 at 11:45
0
\$\begingroup\$

The OPB60 has two parts to it - an LED and a phototransistor. Because the LED is a diode, it's forward voltage (Vf) drop is fairly constant with different currents. The datasheet says that vf is 1.7V max at 20mA. You are driving it at around 24mA ((5.5 - 1.7) / 160). This is the voltage at A, so isn't going to change.

It's the voltage at pin 1 that will respond to proximity changes.

If IC2 is not able to supply 25mA (24mA in the diode and under 1mA in the transistor when on), then your 5V supply will drop, which is why it might work with a different supply, but that circuit as it is wouldn't work anyway.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.