# Homework Help: OP-Amp connected with Transistor whose Base and Collector are short

1. Jun 4, 2013

### darkxponent

1. The problem statement, all variables and given/known data

Problem number 6a in the picture.

2. Relevant equations
Transistor:
Ic = βIb

Ic = αIe

Ic = Is*e^(Vbe/nVt)

OP-Amp:

Vo = Acl*Vi

Acl = 1 + (Rf/Ri)

3. The attempt at a solution

I tried to solve but just couldn't understand the behavior the transistor.

2. Jun 4, 2013

### rude man

First, find Vo1.

Then find Vo.

Note that Q1 and Q2 behave like p-n junction diodes.

3. Jun 4, 2013

### darkxponent

I tried that.
Now Q1 must be forward biased so that the constant current Iref flows through Q1 and the voltage drop across Q1 is 0.7 Volts that is

Vbe1 =.7 Volts

Vo1 = Vi*(R4/R4+R3) - .7

Now what will be the behavior of transistor Q2? The Voltage across it depends on Vi

Vbe2 = .7 - Vi*(R4/R4+R3)

So how can i take diode 2 as forward or reverse biased without knowing Vi?

If the diode 2 is forward biased then where does the current go at NODE Vo1. Does it go into the OP-Amp1?

Taking the Diode 2 reverse biased the circuit is open and Vo2 becomes 0 Volts

PS: I don't have a camera right now or i could have shown you the full attempt.

4. Jun 4, 2013

### rude man

You can't!
What current? Vo1 sinks currents from both diodes. Obviously, all net diode current has to go into op amp 1's output.

Q1 current is always = I_ref.
That is correct.

********************

Hint: You got Vo1 right.

This circuit is a Rube Goldberg comparator. The output Vo tells you whether the input is above or below a certain threshold voltage. What is the correspondence between input voltage and output voltage? Vo is one voltage when Vin is < threshold and another when Vin > threshold.

Last edited by a moderator: Jun 4, 2013
5. Jun 5, 2013

### darkxponent

Thanks for help. Now tell me is this the answer

1) When Vi < .7*(R4 + R3)/R4, Vo = 0 Volts

2) When Vi > .7*(R4 + R3)/R4, Vo = R3*Is*e^((.7 - Vi*(R4/R4+R3))/nVt)

6. Jun 5, 2013

### rude man

'Fraid not.

First, what is Vo1 as a function of Vi?

Second question, what happens if Vo1 is < -0.7V or > -0.7V?

BTW your expression for Vo must be compared to the saturation output voltage. Your op amp can't go more positive than a certain voltage, say +12V for a 741 op amp running on +/- 15V supplies. This applies to terms like your Vo = R3*Is*e^((.7 - Vi*(R4/R4+R3))/nVt) which might be correct if your op amp had no saturation limit.

7. Jun 5, 2013

### darkxponent

Vo1 = Vi*(R4/R4+R3) - .7

1) Vo1> -.7 that is Vi > 0, Vo = 0 Volts

2) Vo1 < -.7, Vi < 0

In this case the diode would be forward biased and second op-amp would behave as a negative comparator and then op amp will saturate at positive saturation voltage that is 12 volts.