Schematic analysis for voltage drops

Jony130

Joined Feb 17, 2009
5,005
Yes, if R4 = 9.1kΩ then the T2's collector voltage = 1.9 + 0.344*9.1 = 5.0304V
=> Vcollector > Vemitter => T2 should be OFF
if R4 = 22kΩ then the T2's collector voltage = 1.9 + 0.344*22 = 9.468V
=> Vcollector > Vemitter => T2 should be OFF
No , you are completely wrong. How can Vce voltage determined whatever BJY is turn ON or Off? From what I know if the base current has closed path for his current to ground ( for PNP transistor) and Veb > 0.5, the BJT will be ON.
 

screen1988

Joined Mar 7, 2013
310
No , you are completely wrong. How can Vce voltage determined whatever BJY is turn ON or Off? From what I know if the base current has closed path for his current to ground ( for PNP transistor) and Veb > 0.5, the BJT will be ON.
I based on the arrow indicating current direction. Then I thought the voltage (VCE) also has to match with the current.
I know that emitter and collector are not identical and not exchangeable.
But does the trans PNP can operate if collector voltage is larger than the emitter one?
 

Jony130

Joined Feb 17, 2009
5,005
I based on the arrow indicating current direction. Then I thought the voltage (VCE) also has to match with the current.
I know that emitter and collector are not identical and not exchangeable.
But does the trans PNP can operate if collector voltage is larger than the emitter one?
OK I see your problem. So let's try this simple circuit.




Pleas do DC analysis. Assume Vcc = 10.6V; Vbe = 0.6V and β = 100.

Just find Vec and VRc voltage. And also don't forget that Kirchhoff's law always holds.
 

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screen1988

Joined Mar 7, 2013
310
The first circuit:
Ib = (Vcc -Veb)/Rb = (10.6 - 0.6)/1MΩ = 10uA
Ic = β*Ib = 100*10uA = 1mA
VRc = Ic*Rc = 1mA * 1KΩ = 1V
Vec = Vcc - VRc = 10.6V - 1V = 9.6V

The second one:
Ib = (Vcc -Veb)/Rb = (10.6 - 0.6)/1MΩ = 10uA
Ic = β*Ib = 100*10uA = 1mA
VRc = Ic*Rc = 1mA * 5KΩ = 5V
Vec = Vcc - VRc = 10.6V - 5V = 5.6V

The third one:
Ib = (Vcc -Veb)/Rb = (10.6 - 0.6)/1MΩ = 10uA
Ic = β*Ib = 100*10uA = 1mA
VRc = Ic*Rc = 1mA * 20KΩ = 20V
Vec = Vcc - VRc = 10.6V - 20V = -9.4V


I think in the third circuit the transistor is OFF.
 

Jony130

Joined Feb 17, 2009
5,005
Ib = (Vcc -Veb)/Rb = (10.6 - 0.6)/1MΩ = 10uA
Ic = β*Ib = 100*10uA = 1mA
VRc = Ic*Rc = 1mA * 20KΩ = 20V
Vec = Vcc - VRc = 10.6V - 20V = -9.4V


I think in the third circuit the transistor is OFF.
But how can transistor be in cut-off mode if there is a IB current flow in this circuit? Do you ever herd about the saturation?
Also how can Vec be equal to -9.4V? II Kirchhoffs law (KVL) don't hold any more ?

The base current is equal 10μA and transistor will do whatever is necessary to sustain Ic = β*IB = 1mA. But our load resistor don't allow Ic current to be equal to 1mA. In this circuit the collector current cannot be greater than
Ic_max = Vcc /Rc = 10.6V/20K = 530μA
So the only think that transistor can do in this situation is to enter the saturation region. And in saturation transistor will be full ON.
And transistor will act as a "switch". Shorting emitter with collector. So in our case. The collector will be short to the Vcc.
http://forum.allaboutcircuits.com/showpost.php?p=577296&postcount=7
 
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WBahn

Joined Mar 31, 2012
24,850
Yes, if R4 = 9.1kΩ then the T2's collector voltage = 1.9 + 0.344*9.1 = 5.0304V
=> Vcollector > Vemitter => T2 should be OFF
if R4 = 22kΩ then the T2's collector voltage = 1.9 + 0.344*22 = 9.468V
=> Vcollector > Vemitter => T2 should be OFF
By the way, how can you know that ignore Ib will not have big effect to the result?
I am not confident about it, maybe a small current has important role to other part.
Remember that "ON" has two flavors -- "active" and "saturated". If you using the term "ON" to mean "active", then there are two alternatives to "ON", namely "cutoff" and "saturated".

In both of these cases the transistor is not "OFF", but rather it is saturated.
 

Jony130

Joined Feb 17, 2009
5,005
If we change R4 from 2.2kΩ to 22kΩ we know that T2 will entry saturation region. And let as try to find Ie1 current. I use a Thevenine and replace R1 and R2 with his Thevenine equivalent resistance Rth = 50kΩ and add Eth = 2.5V. Additional I assume Vbe = 0.6V ; β = 150; Vce(sat) = - 0.1V.



So I write this equations using kirchhoff's laws.

Eth = Ib1*Rth + Vbe + (Ie1 + IR4)*R5 (1)

And

Ib1 = Ie1/(β + 1) (2)

IR4 = ((Vcc - Vce) - (Eth - Ib1*Rth - Vbe))/R4 (3)

And after solving this equations I get this result:

Ie1 = 246.79μA ; Ic1 = 245.2μA; Ib1 = 1.634μA; Ib2 = 187.2μA;
Ic2 =140.44
μA and Ve = 1.82V ; Vb1 = 2.42V; Vc1 = 4.4V;

We can do the same thing and we can finding Ie1 current in station where both BJT are in active region (R4 = 2.2K).

Eth = Ie/(β + 1) * Rth + Vbe + (Ie + Ic2)*R5

Ic2 = β * Ib2

Ib2 = Ie*β/(β + 1) - Vbe/R6

and solve for Ie1
 

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Jony130

Joined Feb 17, 2009
5,005
Let us try to find a voltage gain for the circuit under discussion.

0.1.PNG
As you can see R4 resistor provide a negative feedback. And if you compare it with a classic feedback "topologies", you will see that we have a voltage-series feedback (Series-Shunt).

http://forum.allaboutcircuits.com/showthread.php?p=430967#post430967

And we are very lucky here because the classic op-amp non-inverting amplifier also has voltage-series feedback. And what is the gain for a non-inverting amplifier?

Av = 1 + R2/R1 so in our case Av = 1 + R4/R3 = 1 + 2.2K/4.7K ≈ 1.46V/V

Of course full closed loop gain equations look like this

Acl = Aol/(1 + Aol*B)


where

Aol - open-loop voltage gain. The gain without the feedback.

B - feedback factor (feedback gain) in our example

B = Ve1/Vc2 = R3/(R4 + R3) = 0.681 V/V

Ve1 - T1 emitter voltage

Vc2 - T2 collector voltage

http://forum.allaboutcircuits.com/showthread.php?p=152087#post152087


Now let us try to find the open-loop gain (Aol).

All we need is to redraw the circuit to this form

0.2.PNG

And from our super simplified DC analysis, we get this DC bias condition for a orginal circuit:

Ic1 ≈ Vbe2/R6 ≈ 60μA


and

Ic2 ≈ (Vb1-Vbe1)/R3 - Ic1 = (2.5V - 0.6V)/4.7kΩ - 60μA ≈344μA.

Therefore

re1 = 26mV/Ic1 = 433Ω ;
re2 = 26mV/Ic2 = 75.6Ω ;

Additional I assume the beta values as β1 = β2 = 150;

So let's find a open loop gain.

The first stage (T1) work as a CE amplifier.
So we can find the gain expression by inspection

Av1 = Rc/Re = (R6||RinT2)/(re1 + R4||R4) ≈ 5K/1.9K ≈ 2.63V/V

Where

RinT2 = (β2 + 1)*re2

The second stage also work as a CE amplifier.

So we have

Av2 = Rc/Re = ((R4 + R3)||RL)/re2 ≈ 4KΩ/75Ω ≈ 53V/V


and finally, the open loop gain is equal to

Ao = Av1 * Av2 ≈ 140V/V

So after we close the feedback loop the gain will drop to:

Acl = Aol/(1 + Aol*B) = 140/ (1 + 140*0.681) = 1.453V/V

Of course we can change Acl without changing DC bias pointy.
We simply can add capacitor + resistor in parallel to R3 resistor.

And here you have some more negative feedback examples.
http://forum.allaboutcircuits.com/showthread.php?p=313234#post313234
http://forum.allaboutcircuits.com/showthread.php?p=420624#post420624
 
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screen1988

Joined Mar 7, 2013
310
There is a clear way to differentiate between the two type of feedback. If you disconnect your load, and the feedback signal goes to zero (output current goes zero), that's current feedback (load current > zero, feedback > zero ). If, on the other hand, you short the load, and the feedback goes to zero (output voltage goes zero), that's voltage feedback (load voltage > zero, feedback > zero).
The input summing is shunt when feedback connects to the same node that the input source is connects. Otherwise we have series feedback.
This method is great. I can check feedback types by disconnecting the load or shorting the load.
I think the words "feedback signal" refers to the output of feedback block, right?
Then the feedback block is similar to linear systems. If input is zero its output also has to be zero.
Figures 16-2a and 16-2b represent voltage feedback since the signal fed back is proportional to the output voltage. Now, look at the inputs to the amplifiers. In Fig. 16-2a the feedback signal is placed in series with the input signal, so V3=V1+bV2
Is this a mistake? I think it should be V3 = V1 - bV2
 

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screen1988

Joined Mar 7, 2013
310
Hi Jony!
Can you explain about the part R3//R4 in the redrawn schematic? I can't recognize this is an opamp and don't know where is inverting/non-inverting input.
In the redrawn schematic, you exam open loop gain. I think you remove the feedback part and short out the capacitor C2. Then I get the part R3//R4. And how about the part R3 in series R4?
 

screen1988

Joined Mar 7, 2013
310
Hi Jony,
How can you find your old posts (from 2009, for example)?
I just tried to find all of your posts by going to "profile" --> "Find all posts by Jony130", but the forum was only able to generate 476 posts, which didn't include all of your posts.


I can use Google, but it is not convenient. You joined this site in Feb 2009. Now it's March 2013. On this page, I input "2009-03-01" in the left upper corner. Scroll down, and click "Convert". The "Julian Day" value is what's interesting to us. Copy the "2454891". Then we do the same with "2013-03-30", and we get "2456381". Now we can compose our Google query:

site:forum.allaboutcircuits.com/ daterange:2454891-2456381 "Jony130"
The results won't be usefully ordered but can manipulate the dates.

I wonder if there is an easier way to do this?
Maybe, I should post this in "Feedback and Suggestions " forum.
 

Jony130

Joined Feb 17, 2009
5,005
I think the words "feedback signal" refers to the output of feedback block, right?
Then the feedback block is similar to linear systems. If input is zero its output also has to be zero.
Yes, you are right

Is this a mistake? I think it should be V3 = V1 - bV2
Yes it is a mistake. It should be V3 = V1 - bV2


Hi Jony!
Can you explain about the part R3//R4 in the redrawn schematic? I can't recognize this is an opamp and don't know where is inverting/non-inverting input.
You will not see R3||R4 part on the op amp circuit. Why? Because we assume that there is no current flow in or flow out from the op amp inputs. So this resistor will have no effect on the op amp open loop gain.
But in our circuit we have a slightly different situation. We cannot ignore the loading effect. Because R3||R4 loads our amplifier. And this is why we need to include them in open loop calculation.
As for the input simply T1 base is a on-inverting input and emitter is a low impedance inverting input. How can we know that?
Let as assume the we have a constant Vb voltage at base.
Then we change emitter voltage. If we increase Ve voltage the Vbe and Ic current will decreases. But when we decrease Ve voltage, Vbe voltage and Ic current increase. So emitter behaves just like a inverting input.
On the other hand if we connect emitter to the fix voltage and the input signal to the base. The base will behaves just like a non-inverting input.
If Vb voltage increase Ic current also increase. But when Vb decrease, Ic current decrease also.
I hope that now you can see why base is non-inverting input and emitter is a inverting input.

Then I get the part R3//R4. And how about the part R3 in series R4?
For the voltage type of a feedback. To find resistance seen by T1 emitter.
We simply disconnect the the feedback network from the output and we short it to ground. So we have R3||R4.
But from the output perspective we disconnect T1 emitter. And now our R3 and R4 resistors appears in series connection .
 

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Jony130

Joined Feb 17, 2009
5,005
Hi Jony,
How can you find your old posts (from 2009, for example)?
I just tried to find all of your posts by going to "profile" --> "Find all posts by Jony130", but the forum was only able to generate 476 posts, which didn't include all of your posts.
Try use a search and keywords and Search by User Name.
 
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