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Hello there,
Let me see if i can get you started and then you can take it from there or come back and ask more questions if you like.
Assumptions:
Vs is the source voltage positive, -Vs is the source voltage negative (for the op amp or comparator).
Vd is the clamp voltage of either diode D2 or D3.
CL will be used to indicate C(large) and the other cap will just be called C.
The reference voltage -Vr is greater than -Vd (ie more positive), and +Vd is less than +Vs, and -Vd is greater than -Vs.
Notes:
Note that the output Vo gets clamped to either +Vd or -Vd, except during the recovery period where Vo is changing.
For a very narrow pulse CL looks like a short circuit because it has a large value so the voltage across it changes very little during the pulse period and after the pulse is over.
Quiescent:
V2 is pulled to -Vr, and C is charged so V1=0 (ground), and since V1>V2 then Vo is high which means D3 conducts and clamps Vo to +Vd, and C has a voltage across it equal to Vd with the left side at 0 and the right side at +Vd.
Pulse Applied:
With the narrow positive pulse at Vt, V2 becomes very positive very quickly (t=0+) because CL conducts which causes the output of the op amp to go to -Vs but Vo clamps at -Vd. This puts V1 at negative twice output clamp voltage or -2*Vd because it was charged before this already. That, in combination with V2 being at -Vr, keeps the output of the op amp at -Vs for now. A very short time later after the pulse at Vt goes back low, V2 goes back to -Vr but now V1 is at a very negative level -2*Vd so the output stays positive for now. Next R starts to discharge C back up toward 0v and this takes time T to get to -Vr volts and you can calculate T knowing the voltages and R*C as shown on the question sheet. Just after V1 gets up to -Vr the output goes high again, causing the output Vo to move toward the clamp voltage of Vd. This also causes C to conduct through D1 which means the circuit starts to reset. The time it takes to reset (recover) will be Tr=ln((Vs+Vr)/(Vs-Vd))*R1*C because it takes this time to charge the right side of C back up to +Vd. The circuit is then ready for another input trigger pulse.
The overall effect of the short input pulse is the circuit stretches it out and makes it longer (although of opposite polarity) so it would be considered a one shot circuit.
I can not understand from here:
"This puts V1 at negative twice output clamp voltage or -2*Vd because it was charged before this already. That, in combination with V2 being at -Vr, keeps the output of the op amp at -Vs for now."
firstly, why V1=-2*Vd ?
secondly, you said " in combination with V2 being at -Vr", why the V2 is at -Vr? V2 becomes very positive very quickly at t=0+, doesn't it?
Lastly, should the output of op amp be clamped at -Vd?
Sorry I am a new learner..
And also....
you said "A very short time later after the pulse at Vt goes back low, V2 goes back to -Vr but now V1 is at a very negative level -2*Vd so the output stays positive for now."
If V2= -Vr and V1= -2*Vd
since V2>V1,
shouldn't the output become negative?
Hello there,
Let me see if i can get you started and then you can take it from there or come back and ask more questions if you like.
Assumptions:
Vs is the source voltage positive, -Vs is the source voltage negative (for the op amp or comparator).
Vd is the clamp voltage of either diode D2 or D3.
CL will be used to indicate C(large) and the other cap will just be called C.
The reference voltage -Vr is greater than -Vd (ie more positive), and +Vd is less than +Vs, and -Vd is greater than -Vs.
Notes:
Note that the output Vo gets clamped to either +Vd or -Vd, except during the recovery period where Vo is changing.
For a very narrow pulse CL looks like a short circuit because it has a large value so the voltage across it changes very little during the pulse period and after the pulse is over.
Quiescent:
V2 is pulled to -Vr, and C is charged so V1=0 (ground), and since V1>V2 then Vo is high which means D3 conducts and clamps Vo to +Vd, and C has a voltage across it equal to Vd with the left side at 0 and the right side at +Vd.
Pulse Applied:
With the narrow positive pulse at Vt, V2 becomes very positive very quickly (t=0+) because CL conducts which causes the output of the op amp to go to -Vs but Vo clamps at -Vd. This puts V1 at negative twice output clamp voltage or -2*Vd because it was charged before this already. That, in combination with V2 being at -Vr, keeps the output of the op amp at -Vs for now. A very short time later after the pulse at Vt goes back low, V2 goes back to -Vr but now V1 is at a very negative level -2*Vd so the output stays negative for now. Next R starts to discharge C back up toward 0v and this takes time T to get to -Vr volts and you can calculate T knowing the voltages and R*C as shown on the question sheet. Just after V1 gets up to -Vr the output goes high again, causing the output Vo to move toward the clamp voltage of Vd. This also causes C to conduct through D1 which means the circuit starts to reset. The time it takes to reset (recover) will be Tr=ln((Vs+Vr)/(Vs-Vd))*R1*C because it takes this time to charge the right side of C back up to +Vd. The circuit is then ready for another input trigger pulse.
The overall effect of the short input pulse is the circuit stretches it out and makes it longer (although of opposite polarity) so it would be considered a one shot circuit.
WoW..Thank you so much MrAl.
It is much clearer now!
last two question:
"Next R starts to discharge C back up toward 0v and this takes time T to get to -Vr volts"
1. Why R would discharge C back up toward 0V?
"The time it takes to reset (recover) will be Tr=ln((Vs+Vr)/(Vs-Vd))*R1*C because it takes this time to charge the right side of C back up to +Vd"
2. how do you calculate this time? Is it the recovery time constant asked by the part(d)?
Hello again,
The time constant is simply R1 *C, because C is charging through R1 from the output of the op amp.
Here is the equation for a cap charging through a resistor where the cap has some initial voltage V2 and V1 is the source voltage:
V3=(V1-V2)*(1-e^(-t/RC))+V2
and the solution for t is:
t=RC*ln((V2-V1)/(V3-V1))
Try that.
Hi again,
Yes V3 would be -Vr.
We'll have to come back to this tomorrow, sorry, running out of time here.
Hello again,
It's another day
Below we'll call R1 just R because it makes the text easier to read...
Ok, we start with:
V3=(1-e^(-t/RC))*(V1-V2)+V2
solve for t:
t=RC*ln((V2-V1)/(V3-V1))
replace V2 with -Vr:
t=RC*ln((-V1-Vr)/(V3-V1))
inside the ln() multiply top and bottom by -1:
t=RC*ln((V1+Vr)/(V1-V3))
replace V1 with Vs:
t=RC*ln((Vs+Vr)/(Vs-V3))
replace V3 with Vd:
t=RC*ln((Vs+Vr)/(Vs-Vd))
replace R with R1:
t=R1*C*ln((Vs+Vr)/(Vs-Vd))
and we have the required equation.
There's a problem however. The initial cap voltage here is not -Vr it is actually Vd-Vr because just before this period starts the right side is at -Vd and the left side is at -Vr, which means the total voltage across the cap is Vd-Vr. Thus if Vr=3v and Vd=8v, the initial cap voltage is -5v not -3v. Sorry about that.
This means you will have to rework the equations above using this new information.
We will ultimately test all the equations in a circuit simulator in the end too.
Hello again,
The cap C is originally charged with 0v on the left and +Vd on the right. That means it has Vd volts across it, with the left side negative and the right side positive. Now when the output of the op amp goes to -Vs, the right side of C gets clamped to -Vd and that means the left side jumps to -2*Vd (which they show as 2*Vo). This means we now have a cap with the right side -Vd and the left side -2*Vd, so we are now discharging the left side of the cap back up toward 0v, so we can use the equation:
Vr=2*Vd*e^(-t/RC)
and solving that for t we get:
t=R*C*ln(2*Vd/Vr)
and since R is really R1 here we end up with:
t=R1*C*ln(2*Vd/Vr)
Let me double check this first though.
Yes ok this looks correct, although the left side of the cap would look like it was charging from -2*Vd to -Vr.