Hi all
I am progressing well in the book thanks to all!
I am now doing Voltage Control Current Source,
a) Marked on both images with circles 1 and 2, Are the resistors function are like two current sources(with a PNP transistor)? Am I correct?
b)In figure 20-22 says that when the load resistance increase the transistor become saturated. My difficulty, is the transistor became saturated by the load, independent to the base current supplied by the opamp? Am I correct?
c)In grounded Voltage-to-Current Converter fig 20-23 I have a question regarding the derivation That the author writes that:
" the current through the first transistor is i=Vin/R,
This current produces a collector voltage of Vc=Vcc-Vin"
My thoughts are: if the voltage that is across the lower resistor(on the left Fig20-23) is Vin(inverting input is within micro volts of the noninverting input) supplied by the inverting input of the opamp therefore the current that passes through the lower resistor (on the left of Fig20-23) i=Vin/R, since the same current passes(Ie=Ic) through the upper resistor (on the left Fig20-23) the voltage across the upper resistor is Vin(because it has the same resistance value as the lower resistor and the same current Ie=Ic) ,therefore the voltage of the collector with respect to ground is Vc=Vcc-Vin, Am I correct?
sincere thanks
Sm

 

Hi all
a) Marked on both images with circles 1 and 2, Are the resistors function are like two current sources(with a PNP transistor)? Am I correct?

I don't understand what you mean by saying "two current sources".
But yes, both of these resistors perform the same function in this circuit.
Those resistors are nothing more than a current to voltage converter (error voltage).
They convert Iout current into error voltage for a the op amp. And thanks to this the op amp can "regulate" the BJT in order to maintain output constant and equal to (Vcc - "+")/R. Where "+" is a voltage at non inverting input.

b)In figure 20-22 says that when the load resistance increase the transistor become saturated. My difficulty, is the transistor became saturated by the load, independent to the base current supplied by the opamp? Am I correct?

Yes and no.
If the load resistance increase to such a "big" value so that there is not enough "headroom" voltage (Vce) for the transistor to work properly.
For example for Vcc = 10V ; R = 1k and Vin = 9V the load current is equal to :
Iout = 1V/1kΩ = 1mA the load resistance cannot be larger than RLmax > (Vcc - VR - Vce(sat))/Iout =(10V - 1V - 0.2V)/1mA = 8.8kΩ.
For exampel for RL = 10kΩ ; Vce =(10V - 1V - 1mA*10kΩ) = -1V which is of course impossible.
As you can see, if the load resistance would be greater than 8.8kΩ , the load current no longer is a constant and load independent. So no longer this circuit can can be called a constant current source. Because there is not enough voltage to kept the current unchanged.

http://forum.allaboutcircuits.com/threads/bjt-saturation-question-s.81150/#post-577296
http://forum.allaboutcircuits.com/threads/which-method-npn.97963/#post-731172

c)In grounded Voltage-to-Current Converter fig 20-23 I have a question regarding the derivation That the author writes that:
" the current through the first transistor is i=Vin/R,
This current produces a collector voltage of Vc=Vcc-Vin"
My thoughts are: if the voltage that is across the lower resistor(on the left Fig20-23) is Vin(inverting input is within micro volts of the noninverting input) supplied by the inverting input of the opamp therefore the current that passes through the lower resistor (on the left of Fig20-23) i=Vin/R, since the same current passes(Ie=Ic) through the upper resistor (on the left Fig20-23) the voltage across the upper resistor is Vin(because it has the same resistance value as the lower resistor and the same current Ie=Ic) ,therefore the voltage of the collector with respect to ground is Vc=Vcc-Vin, Am I correct?
sincere thanks
Sm

Yes, exactly.

 

Thanks for your reply
Yes I almost there! Regarding a) the two resistors marked with a circle with two pnp transistors I think they seem like two variable current sources (controlled by Vin)Am I right? just for clarification!

Regarding b) your explanation You wrote I understood it perfectly, Yes now I know can be a limit of load resistance.
You said "Yes and no" is the transistor can became saturated by the load, independent to the base current supplied ( no matter what base current is) by the opamp?
I think Yes if I understood your explanation well, just for clarification! Please can you correct me

 

Thanks for your reply
Yes I almost there! Regarding a) the two resistors marked with a circle with two pnp transistors I think they seem like two variable current sources (controlled by Vin)Am I right? just for clarification!

Well yes, but this resistor are more like a current sense resistors.

Regarding b) your explanation You wrote I understood it perfectly, Yes now I know can be a limit of load resistance.
You said "Yes and no" is the transistor can became saturated by the load, independent to the base current supplied ( no matter what base current is) by the opamp?

For this particular circuit the answer is YES.

 

I learned well thanks to you and all the others of the forum

 

Hi all
thanks for all
I am encountering a problem how Active Positive Clamper works,
The text for me is not much straight forward when says "just beyond the negative input peak(I think when the negative input is decreasing), the diode turns off, the loop opens and the virtual ground is lost" What is it Happening I don't know, kindly can someone put some light for me to override this problem!

 

First understand how a passive version of this circuit works.
http://forum.allaboutcircuits.com/t...diode-and-resistor-circuit.34005/#post-212416

As for the op amp and "losing virtual ground". The answer is simple. If the diode is ON (forward biased) for negative half of a input signal cycle the op amp output can affect the output voltage, because now the negative feedback loop is closed. This means that op amp now can bring the inverting input voltage to non-inverting input voltage, in this case 0V ( V"+" = V"-" = 0V.). So for the negative half cycle of a input signal, the diode is ON and op amp can make Vout = 0V thanks to negative feedback loop through forward biased diode. But for the positive input signal half the diode is OFF (reverse biased). So op amp output no longer has any effect on output voltage. So, the circuit behaves as if there were no op amp in this circuit. Op amps is saturated at the negative supply voltage (Vee).

 

This means that op amp now can bring the Inverting input voltage to non-inverting input voltage, in this case 0V ( V"+" = V"-" = 0V.). So for the negative half of a input signal, the diode is ON and op amp can make Vout = 0V thanks to negative feedback loop through forward biased diode. But for the positive input signal halt diode is OFF (reverse biased) op amp output no longer has any effect on output voltage. So circuit behaves as if there were no op amp in this circuit. Op amps is saturated at negative supply voltage (Vee).

Sorry but Can't understand well the quoted text
thankyou in advance

 

But do you understand that if we have a op amp and there is a "path" (negative feedback loop) from the op amp output to non-inverting input (Vn).
The op amp output will attempts to do whatever is necessary to make the voltage difference between Vp and Vn to zero (Vp = Vn = 0V)?
Vp - Non-inverting input.
Vn - Inverting input.
So for your circuit we have a 0V at the Vp input and the the op amps wants to have the same voltage at it's Vn input. The only way the opamp can do this is by turning ON the diode into conduction. But op amp can "produce" a "positive voltage" at his output only when the voltage at the Vn input is negative (negative half cycle of a input signal). For a "positive" voltage at the Vn input (positive half cycle) the op amp gives "negative voltage", so the diode is OFF. Since the diode is OFF, op amp output has no effect on how this circuit work for positive voltage at the input.

 

Hi
Your explanation was very useful, I searched on the internet and books and I went to a conclusions regarding, first the theory of operation of a simple dc restorer(see attached dc restorer) and than applied on the circuit attached previously(clamper.jpg)Your kindly correction is much appreciated and thanks in advanced!
My thoughts are these
A) ( see attached dc restorer) Provided that the capacitor don't discharge quickly through the load resistor, during the first negative half cycle( from 0V to the negative peak) the diode is forward bias and the capacitor is charged till Vp and the output across the resistor is 0V(please see on the diagram the waveforms), upon beyond the negative peak(for example -0.9Vp of the input) the capacitor starts to discharge through the resistor(because the capacitor is at higher potential than the input signal) and the diode is reverse bias therefor a n output voltage appears across the resistor. During the positive half cycle the capacitor became like a battery and the polarity of the input signal and the capacitor are on the same direction therefore the input signal is added with the capacitor voltage and 2Vp appears across the resistor, beyond the positive peak the input voltage decreases till 0 and capacitor remains charged at Vp(like batteries in series Vin(=0)+VP=VP) the diode is reverse bias, finally the during the half negative cycle the diode is forward bias(like a short) and the capacitor discharges till 0V, and so on

B) during the negative half(before the negative peak) cycle the input capacitor start charging and the current flow from ground through RL and than to capacitor finally to ground , the diode turns and virtual ground appear and the capacitor charges to the peak value with the polarity shown in the previous book figure, sorry but I am getting confused

 

AD1 What about a diode forward voltage drop ?? Don't forget about it. So the capacitor will charge to the voltage equal to:
Vc = Vin_peak - VF. And VF is typical in the range of a 0.5V... 0.8V.
http://forum.allaboutcircuits.com/attachments/2-png.16080/
And in this type of a circuit the time constant is large in comparison to the period of a input signal.
And this means that the capacitor will be quickly charge to the first peak value of a input signal (for a negative half cycle).
And will remain charged to the peak value for the positive half cycle also. In fact the capacitor will slowly discharge via a load resistor for a positive half cycle. So in order to restore this lost charge on the capacitor the diode must conduct a slightly amount of time (negative half cycle) to be able to restore the charge on the capacitor.
Try read this
http://forum.allaboutcircuits.com/threads/voltage-divider-bias.71104/#post-494949(especially panic mode post)
So in summary the capacitor in this circuit act just like a constant voltage source (a battery). And this is why we can "shift" the voltage above the zero. But because the diode voltage drop the output voltage is slightly negative (-0.7V). So to remove this diode drop we add a op amp that "compensate" the diode voltage drop thanks to this property:
"The op amp wants to keep its inverting and non-inverting inputs at the exact same voltage potential at all times (as long as negative feedback loop is closed)".
So we back here
So for your circuit (clamper.jpg) we have a 0V at non-inverting input and the the op amps wants to have the same voltage on its inverting input. The only way the op amp can do this is by turning ON the diode into conduction. But op amp can "produce" a "positive voltage" at his output only when the voltage at the inverting input is negative (negative half cycle of a input signal). We need a positive voltage at the op amp output to turn-on the diode.

 

Dear Jony130
many thanks I understood all you wrote, If I don't bother you, if am demanding so much sorry, please can you tell the theory of operation of the said circuit(clamper.jpg) please can you divide the input signal by every quarter and tell me what is happening in every quarter of the I/p cycle, I don't need diagram just an explanation what is going on

 

Sure I try, but not today.

 

Thank you I am in debt with you, again I understood what you wrote, I don't need elaborated explanation just what is going on in each quarter cycle, take your time

 

Does this description is enough for you.
http://www.nptel.ac.in/courses/117107094//lecturers/lecture_17/lecture17_page2.htm

 

Hi
Thanks read up a I learned the theory of operation now! have one question only regarding to the link given but
"the OPAMP produces a sharp positive going pulse that replaces any change lost by the clamping capacitor between negative input peaks"
I can't understand this only, kindly can you explain this?

 

But haven't we already discussed this ?
The capacitor will slowly discharge via a load resistor for a positive half cycle. So in order to restore this lost charge on the capacitor the diode must conduct a slightly amount of time (negative half cycle) to be able to restore the charge on the capacitor.
Diode will be ON only when OpAmp output voltage is "positive". But OpAmp "produces" positive output only when voltage at inverting input is negative. So when this voltage is negative?
Let as assume that the Vin is at -2.5V (10V peak to peak) and capacitor was previously charge to 5V but was discharge by the load to 4V during positive halt cycle. So, Vo = Vn = 4V - 2.5V = 1.5V, OpAmp is still at negative saturation because the voltage at inverting input (Vn) is positive (1.5V) Vn > Vp.
For Vin = -4V we have Vout = Vn = 4V - 4V = 0V we almost there.
But for Vin > -4V the voltage at output would have been negative (Vin = -4.01V --->Vn = 4V - 4.01V = -0.01V) .
But now OpAmp comes into play and provide a positive voltage. So diode is ON and OpAmp positive voltage charge the capacitor to restore the charge on the capacitor. But as you can see this is happens just before the input signal reach the negative peak value.

 

Thanks a lot
only this why" Vn > Vp"
so can I close this thread

 

Because OpAmp is nothing more than a differential amplifier.
Vout = (Vp - Vn)*gain, So if we connect a non-inverting input (Vp) to ground and apply a inputs signal to inverting input we have this situation:
Case 1
Vin = Vn = +1V---->Vout = (0V - 1V)*10 = -1V*10 = -10V
Case 22
Vin = Vn = -1V ---->Vout = (0V - (-1V))*10 = 1V*10 = 10V
http://forum.allaboutcircuits.com/t...ns-but-im-troubled-by-them.64696/#post-444315

 

sorry I confused Vp with voltage across capacitor sorryyyyyyyyyyyyyy