# Non Inverted Operational Amp

#### MaxRobotics

Joined Mar 13, 2014
4
Hey AAC,

I am stuck with a non inverted operational amp as included below. V1 and V2 are constant voltages and the photo resistor above is from a OSRAM FSH optical sensor. The question is: What will the output voltage be for a given collector current I_c? The collector current is the current trough the photo resistor.

When I leave out this sensor the scheme left is just a non inverted operational amplifier where V_f = ( V_plus - V_min )* (1 + R2/R1) and V_min = V_1 * (R2 / R1 + R2) and V_plus = V_2.

But how can I calculate V_f when the sensor is present?

Regards Max

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#### shteii01

Joined Feb 19, 2010
4,647
There are some rectangles labeled R1 and R2. Any of them mean anything?

#### MaxRobotics

Joined Mar 13, 2014
4
Those are resistances.

#### shteii01

Joined Feb 19, 2010
4,647
Those are resistances.
I see.

You are not being very forthcoming with information. You mention photo resistor, yet the rest of us are required to figure out where it is. I hope someone else is able to read your mind and help you.

#### crutschow

Joined Mar 14, 2008
24,108
Is it a photo resistor or a high impedance photo current source?

Since current from a high impedance photo sensor goes to the summing junction it will then all go through R2 due to the negative feedback from the op amp. So it would be amplified by the value of R2.

#### BytetoEat

Joined Mar 5, 2014
25
This is actually a trick question made to reinforce some "rules of thumb"
one is that the input pins take virtually 0 current, and the other is that the op amp will strive to make both its input pins at the same voltage(within its limits of saturation of course)
I'm going to assume that the photoresistor you are talking about is the BJT transistor and also that we are not going to operate the op amp outside its limits of supply voltage(saturation)..

We can see that the positive pin is at V2, which would make the negative pin at V2 as well since the op amp will force this with Vf. That makes the current across R1: V1-V2/R1 (which can be negative if V1 is less than V2)

Next lets think about extremes.. one where the photoresistor is basically an open circuit, and the other where its very low resistance. When its very high resistance, we can pretty much ignore it and will have our previous calculated current going through R2. Please try to calculate the gain in this case and post back.. When its a very loooow resistance, it will basically be pulling the negative pin to ground while the op amp is trying to force that pin to V2, so it will probably saturate to +Vsupply

In the inbetween cases, the current flowing through R1 will now have two paths.. The way to solve this is using nodal analysis at the negative pin.(aka sum the currents going in and out of the node to zero).

$$(V1-V2)/R1$$+$$(Vf-V2)/R2$$+$$V2/Rphoto$$=0

(Also, the last term is also Ic the collector current)

#### LvW

Joined Jun 13, 2013
890

The output voltage is (Rp=resistance of the transistor)

Vout=V2*(R1*Rp+R2*Rp+R1*R2)/(R1*Rp) - V1*R2/R1.

As can be seen, for Rp=0 there is no negative feedback and Vout approaches infinity (supply rail).

#### MaxRobotics

Joined Mar 13, 2014
4
Thanks, BytetoEat for your elaborate explanation,

I made indeed the mistake to call the photosensitive element a resistor while it is a transistor.
I can follow your calculation completely however the first assumption, that the op amp will strive to a situation where V+ = V- is one that I cannot match with a previous exercise.

I am not sure if my calculation in this previous exercise is correct, But if it is, than I think it conflicts with the rule that the op amp will strive to V+ = V-.

Again a circuit is enclosed. This time the photosensitve transisor is attached differently. again I want to calculate V_f for a given I_c.

My calculations were as follows:
V+ = Ic*R3
V- = V1*(R2 / (R1 + R2))
Vf = (V+ - V-) * (1 + R2/R1)

Where the gain (1 + R2/R1) is the non inverted amplifier gain, and I calculated the voltages on the minus and the plus port. However this approach is not aplicable when I can say V- = V+,

So what am I doing wrong here, or does the V- = V+ law only applies to a op amp in the configuration in my original post?

Regards Max

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#### LvW

Joined Jun 13, 2013
890
My calculations were as follows:
V+ = Ic*R3
V- = V1*(R2 / (R1 + R2))
Vf = (V+ - V-) * (1 + R2/R1)
V+ is correct.
V- is not.
V- consists of two parts:
V-=V1*R2/(R1+R2)+Vf*R1/(R1+R2)

Combination with V-=V+ results in
Vf=Ic*R3*(R1+R2)/R1 - V1*R2/R1=Ic*R3*(1+R2/R1) - V1*R2/R1.

(Edit: Sorry, perhaps too much information for homework. I realized it only now)

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#### BytetoEat

Joined Mar 5, 2014
25
NP MaxRobotics, happy to help.

As for this different circuit, your solutions are not correct.

First you are right on V+ = Ic*R3

To find Vf, use superposition for the fastest way.

First we imagine that V+ is ground and that the photoresistor is super high resistance. Then we can see how much V1 contributes to Vf. It ends up being a simple inverting amplifier and the gain is -R2/R1*V1

Now, Lets imagine V1 is off and see how much V+ contributes to Vf. V1 is ground, so now you have a noninverting amplifier of V+*(1+R2/R1)

Now we just add them and set V+ to Ic*R3 and you get
Vf=-(R2/R1)*V1 + (1+R2/R1)*Ic*R3

#### AnalogKid

Joined Aug 1, 2013
8,252
Coming at this from a different direction, your original circuit is a classic conditioning circuit for a photo diode. In this case, the current through the diode is the *starting point* for the design, not something that can be calculated from a schematic. The current through the diode (and your phototransistor) is proportional to the light impinging on the sensor. The opamp takes this current and works with it, usually by amplifying it and presenting it as a low impedance voltage source.

The "collector current" isn't set by the opamp or by the resistor values. It is set by the illumination of the sensor.

ak