Help needed for circuit analysis Homework

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engineer2006

Joined May 22, 2013
5
Hi,

First of all I am in highschool and part of my higher physics course is based on electronics. I have been asked to find out what this circuit is used for and how it works. My friend has said that it is a 'constant current load'. What does this mean? How does it work?

Thanks,
Eng06
 

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Dodgydave

Joined Jun 22, 2012
9,923
The voltage at pin 3 is the reference voltage for the op amp, when the op amp turns off it will turn the transistor on, this will put voltage across the the two series collector resistors one of which is connected to pin 1,

when this reaches slightly more than the voltage as pin 3, the op amp turns on, so transistor turns off and no voltage at pin 1, this cycle is repeated several times a second and thus maintains a fixed current through the collector and also through the resistors.
 

screen1988

Joined Mar 7, 2013
310
Dodgydave,
I am not quite understand this circut.
Can you explain a bit more?
What is the role of the capacitor?
And also why we need to put a resistor between emitter and base of the transistor?
When op amp turn off, its output voltage is 0V but how can you see the transistor will be on? I know that PNP is on when Veb > 0.7 V(si) but can you make it clear?
Thanks, and hope you don't mind when I asked too many question.:)
 

Jony130

Joined Feb 17, 2009
5,228
I am not quite understand this circut.
Can you explain a bit more?
Which part you don't understand? Do you know how op-amp works?
http://forum.allaboutcircuits.com/showpost.php?p=444315&postcount=9
As you should see we have a negative feedback in this circuit.
And when the op amp see 0V difference between his inputs terminals V(+) - V(-) = 0. And output voltage will stop changing.

What is the role of the capacitor?
Low-pass Filter

And also why we need to put a resistor between emitter and base of the transistor?
Do you ever herd about pull-up resistor? This resistor ensure that BJT is cut-off when op-amp output is in "high state".

When op amp turn off, its output voltage is 0V but how can you see the transistor will be on? I know that PNP is on when Veb > 0.7 V(si) but can you make it clear?
When op-amp is in "low state". There is a path for the PNP transistor base current from Vcc do gnd.
Vcc --->Emitter---->Base -->through resistor between BJT base and op-amp output ---> to GND.

Also notice that if we assume that resistor between emitter and base of the transistor = R1 (10K). And resistor between BJT base and op-amp output = R2 (10K). The transistor will be ON when op-amp output will be lower then:
V_on < Vcc - (Vbe * (1 + R2/R1)) = (5V - 0.6V * (1 + 10K/10K)) = 5 - 0.6V*2 ) = 5V - 1.2V = 3.8V
So if op amp output voltage is smaller than 3.8V the PNP transistor for sure will be ON.
 
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Jony130

Joined Feb 17, 2009
5,228
Could this type of circuit be used with a lamp attached to the output?
Not the best idea if you want constant current. The load should be connect instead Rc1 resistor. But this circuit can also work as a voltage regulator.
If we connect the load between BJT collector and ground. Then Vout = (1 + Rc1/Rc2) * Voltage at pin 1
 

screen1988

Joined Mar 7, 2013
310
Thanks Jony,
As you should see we have a negative feedback in this circuit.
I usually mess up when a circuit has many parts and I can't see the relation between them.
In the circuit which part you consider a feedback? The resistor between emitter and base R1 or both R1 and the transistor?
Low-pass Filter
OK, that part is clear now!
Do you ever heard about pull-up resistor? This resistor ensure that BJT is cut-off when op-amp output is in "high state".
Jony, I am still not clear this part. When op-amp output is in "high state" (I think it is 5V) then emitter and output of op amp has the same voltage. But how can the resistor ensure BJT cut off? If the base current is 0A, I can get Vbe = 0 and the transistor cut off but because Ib is not zero and I got stuck.
When op-amp is in "low state". There is a path for the PNP transistor base current from Vcc do gnd.
Vcc --->Emitter---->Base -->through resistor between BJT base and op-amp output ---> to GND.
Did you have to look into the op amp? Is it mean that when the op amp is in low state two pins of op amp 4 and 6 will be connected together?

Also notice that if we assume that resistor between emitter and base of the transistor = R1 (10K). And resistor between BJT base and op-amp output = R2 (10K). The transistor will be ON when op-amp output will be lower then:
V_on < Vcc - (Vbe * (1 + R2/R1)) = (5V - 0.6V * (1 + 10K/10K)) = 5 - 0.6V*2 ) = 5V - 1.2V = 3.8V
So if op amp output voltage is smaller than 3.8V the PNP transistor for sure will be ON.
I think you ignored the base current? How about if we take Ib into consideration?
 

Jony130

Joined Feb 17, 2009
5,228
Thanks Jony,
I usually mess up when a circuit has many parts and I can't see the relation between them.
In the circuit which part you consider a feedback? The resistor between emitter and base R1 or both R1 and the transistor?
The small part of a output voltage is feed back to the input.


As you can see on the diagram we have a feedback loop in this circuit.
If the voltage at (+) input rise the op amp output also will rise.
This increase in op amp output voltage will decrease the base current and as a result the collector current also will decrease. And this means that less current will be flow through the R4 resistor, this decrease voltage at (+) input.
So as we can see we have a negative feedback in the circuit.
http://www.allaboutcircuits.com/vol_3/chpt_8/4.html

Jony, I am still not clear this part. When op-amp output is in "high state" (I think it is 5V) then emitter and output of op amp has the same voltage. But how can the resistor ensure BJT cut off? If the base current is 0A, I can get Vbe = 0 and the transistor cut off but because Ib is not zero and I got stuck.
Why Ib current is not zero ?
Also kept in mind that most op amps cannot produce an output voltage equal to their supply voltage.
So in "high state" (positive saturation) op amp don't give us 5V at output.
The output voltage will be lower 5V. Sometimes Vout_max is 2V lower then supply voltage.
http://e2e.ti.com/blogs_/b/thesigna...input-and-output-clearing-some-confusion.aspx
And now I hope you see why we need pull-up resistor.

Did you have to look into the op amp? Is it mean that when the op amp is in low state two pins of op amp 4 and 6 will be connected together?
Yes, but you also have to remember about "negative saturation output voltage"

I think you ignored the base current? How about if we take Ib into consideration?
But if BJT is in cut-off (no base current). And we treat Vbe as a threshold voltage. We don't need to care about base current anymore.
 

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screen1988

Joined Mar 7, 2013
310
Jony, I will put what I understand base on what you said here and hope you can check it.
The small part of a output voltage is feed back to the input.
As you can see on the diagram we have a feedback loop in this circuit.
If the voltage at (+) input rise the op amp output also will rise.
This increase in op amp output voltage will decrease the base current and as a result the collector current also will decrease. And this means that less current will be flow through the R4 resistor, this decrease voltage at (+) input.
So as we can see we have a negative feedback in the circuit.
I am surprise that this is negative feedback. At first, it appears to be a positive feedback.
As you can see on the diagram we have a feedback loop in this circuit.
If the voltage at (+) input rise the op amp output also will rise.
Vout = A* (V+ - V-), and because V+ increases and V- = Vin = constant => Vout increases.
This increase in op amp output voltage will decrease the base current and as a result the collector current also will decrease.
Let's call the voltage at base of transistor is Vb and current is Ib.
Ib = (Vb - Vout)/R2 - (Vcc - Vb)/R1
Because the transistor is ON Veb = 0.7 V ,for example. => Vb = Vcc - Veb = Vcc -0.7 V =constant.
Now if Vout increases => Ib decreases.
And this means that less current will be flow through the R4 resistor, this decrease voltage at (+) input.
So as we can see we have a negative feedback in the circuit.
This part is OK.
And now I hope you see why we need pull-up resistor.
I read the article and think that I understand it but still don't know why the pull up resistor here ensure the trans is OFF when op amp output is in high state.
Could you explain it more?
But if BJT is in cut-off (no base current). And we treat Vbe as a threshold voltage. We don't need to care about base current anymore.
This assume is only right for this characteristic:

(at the threshold voltage Vt, I =0)
but not right for this diode characteristic:

In the case, at threshold Vt = Vbe I can't say that Ib =0.
 

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Jony130

Joined Feb 17, 2009
5,228
I am surprise that this is negative feedback. At first, it appears to be a positive feedback.
Yes, at first glance this circuit looks like a positive feedback one.
Because we connect output signal to V(+) input. But this circuit has a negative feedback thanks to BJT. The BJT work here as a CE amplifier.
And CE amplifier add 180° phase shift. And this is why we have a negative feedback. But in reality this typo of a feedback is very prone to parasitics osculation. And this circuit is very common to use in low-dropout voltage regulator.
Vout = A* (V+ - V-), and because V+ increases and V- = Vin = constant => Vout increases.

Let's call the voltage at base of transistor is Vb and current is Ib.
Ib = (Vb - Vout)/R2 - (Vcc - Vb)/R1
Because the transistor is ON Veb = 0.7 V ,for example. => Vb = Vcc - Veb = Vcc -0.7 V =constant.
Now if Vout increases => Ib decreases.

This part is OK.
Yes the pat is OK.

I read the article and think that I understand it but still don't know why the pull up resistor here ensure the trans is OFF when op amp output is in high state.
Could you explain it more?
See this example.
I Assume Vcc = 5V and the op amp output positive saturation voltage = Vo_sat = 4V
And I hope that this diagram explains all




This assume is only right for this characteristic:

(at the threshold voltage Vt, I =0)
but not right for this diode characteristic:

In the case, at threshold Vt = Vbe I can't say that Ib =0.
I ignore the Ib current because I'm too lazy. The math is much simpler when we ignore Ib current. Also in real world we don't know exact Vbe and Ib value. So why even bother to include this in calculations.
 

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screen1988

Joined Mar 7, 2013
310
I ignore the Ib current because I'm too lazy. The math is much simpler when we ignore Ib current. Also in real world we don't know exact Vbe and Ib value. So why even bother to include this in calculations.
I also want to ignore it but each time I try to include Ib I see the equations quite different and it makes me think that the result will aslo different.
I am going to understand it. :)
One more question.

In the lower picture, I think at first you assume the transistor OFF. Then you calculate Veb = VR1 = 0.33V < Vt = 0.7 V
Now you check your assumption and see it right.
Is this the way that you did?

PS. I amaze that you can quickly understand how the circuit work as a whole. I have to go analyse many times and now I am starting to understand it.
 

Jony130

Joined Feb 17, 2009
5,228
I also want to ignore it but each time I try to include Ib I see the equations quite different and it makes me think that the result will aslo different.
I am going to understand it. :)
The solution is not so obvious. If you want include Ib current in the calculations. You are force to use Shockley diode equation. So we end up with non-linear equation impassible to solve by using algebra only.
Also we don't know Is current for a given BJT as well as thermal voltage. Also this turn ON voltage will also depend on Rc resistance.
So as you can see there is no simple answer.

One more question.

In the lower picture, I think at first you assume the transistor OFF. Then you calculate Veb = VR1 = 0.33V < Vt = 0.7 V
Now you check your assumption and see it right.
Is this the way that you did?
Yes, first thing is to check if voltage divider output voltage is high enough to open the BJT.
 
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