Basic electronic question

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tony ennis

New Member
Ok, here is the standard photo-transistor circuit:

Now, looking at the one on the left, what is the purpose of the resistor?

One source I read said that as that resistor got larger (about 5kΩ) the circuit was in 'switching' mode which sounds better for my application.

Nigel Goodwin

Super Moderator
It's a 'pulldown' resistor, just like the other is a 'pullup' resistor.

Generally the common emitter one is by FAR the most commonly used, and I would suggest far better.

jbeng

Member
Actually, the one on the left has the pull-up resistor. When light shines on the transistor, the output of the circuit is low. On the one on the right, the output goes high when light shines on it.

Jeff

tony ennis

New Member
I can sort of understand the one on the right. Without the resistor, the circuit would be shorted to ground and Vout spoiled.

But the one on the left... if light shines on the phototransistor, the circuit grounds out. If it doesn't, however, you get... some voltage less than Vcc on Vout. There's something very fundamental I don't get.

And what makes the right one better? That Vout gets all Vcc, undiluted by a resistor?

Hero999

Banned
Nothing makes the right one any better, both circuits are as good as each other.

With a normal emitter follower there's a higher voltage loss and a unity gain but with a photo transistor it makes no difference.

JCLrd

New Member
The circuit on the left is used when you want a negative going signal to be your output when light is inputed to the base.

Circuit on the right is used when you want a positive going signal to be your output when light hits the base.

example.
Say you wanted to drive a PNP transistor stage at the output of the photo detector stage, then you could use the circuit on the left. (CE)

If you needed to drive an NPN stage than you could use the circuit on the right (CC).

These resistors are needed to make the output signals clean and concise. (they give a definite predetermined output current and voltage, before and after the input signal occurs)

These are just quick examples. But resistors are used to set currents and voltages to predetermined values to make circuits operate properly,.

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tony ennis

New Member
Ok, I'll try a different tac. What would happen in the left diagram if the resistor was positioned by the ground, as it is in the right circuit?

Banned
That would mean there's no resistance between VCC and VOUT so VOUT would always equal VCC minus the loading effect of the photo transistor through the resistor, which would be very low. It would essential do nothing.

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JCLrd

New Member
Ok, I'll try a different tac. What would happen in the left diagram if the resistor was positioned by the ground, as it is in the right circuit?
I'm not sure I understand your question...

If you moved the resistor to the emitter lead as with the circuit on the right, and kept the left circuit output in the same place than you would have a constant volt output of Vcc no matter what the input is.

If you kept the resistor at the output and added a second resistor to the emitter lead than you would have an output voltage determined by the amount of current flow through the circuit which depends on the value of resistors as well as the amount of voltage drop across the transistor ect...

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tony ennis

New Member
No, you understand my question correctly, I was being a moron.

colin mac

New Member
When there is no light shining on the one at the left, the transistor is off and it acts like an open switch, no current flowing, so all of Vcc is across the transistor and the output is high. When light shines on the transistor, it is on and acts like a closed switch, the output is low.

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