i assume diode D1 will conduct for negative half cycle so voltage at + terminal of opamp will be -1.5V peak.Some questions you can ask yourself -
1) What happens to V1 sinewave at the OpAmp input, eg. how does D1
affect the signal presented to OpAmp input ?
2) An OpAmp configured with 100% feedback does what to a signal ?
3) What has to happen at OpAmp output for Vo to increase ?
Just some clues....
Regards, Dana.
Apparently this is not homework so i can tell you.View attachment 183592
Hi Guys,
Could anybody explain me, how to analyze the above circuit.
What is out put of this circuit ?
I expect theoretical calculation and explanation ?
Regards,
Thanks thats probably interpretation of circuit ?Apparently this is not homework so i can tell you.
When the input goes negative the cap charges up to nearly the input voltage level with negative to the left and positive to the right assuming the input voltage is reasonable. Now when the input goes positive, the input to the op amp sees about 2 times the input voltage because the previous cap voltage adds to the input voltage. The op amp is set up as a voltage follower, so the output goes up to about 2 times the input voltage. This view ignores the diode voltage drops. If you want to include those, then subtract as needed.
There is not much to the equations to do this circuit. Just subtracting voltage drops. That is unless you want to use actual diode models similar to what we find in spice then it gets more complicated.
Hi,View attachment 183594
what is Vo of this circuit.
When D3 will go in breakdown ?
How do i calculate Vo ?
Yes your logic is correct but probably that is not correct answer.Hi,
Do you know the gain of an inverting op amp circuit?
If so you know that it is G=-R7/R6 ignoring the diode for now.
Since the input is always positive, the output is always negative. That means the diode never conducts therefore the output is -10 volts.
If we had a negative input it would be a little more complicated because we would have to figure in the diode. With an ideal diode the gain would drop once the diode conducts to about one half of 10 which is about 5 (ignoring the voltage drop of the diode that is). But since the input is always positive the output is always negative.
Well let me know when you know whyYes your logic is correct but probably that is not correct answer.
it could be 6.6V but i do not know why ?
Sorry there was correction in answer it should be -6.6VWell let me know when you know why
Seriously though if you only have 1v positive input then the output must be -10v because it inverts and gain of 10.
The only way it could be 6.6v is if the input signal was negative by enough and the power supply was limited to maybe around 6.6 volts. with ideal op amp the power supply would be 6.6v and then it would clip at 6.6v. With only -1v input though i dont think you can get that but i will do another calculation in a minute to find out for sure.
Ok here is the expression for the output voltage:
Vout=1.6-10*Vin
and in order to get 6.6v output you have to input -0.5v (that is minus one half of a volt).
The output can be derived as follows...
The voltage at the inverting terminal is:
vn=(Vout-Vin)*1/11+Vin+(Vout-Vd-Vin)*1/11+Vin
vp=0 (at the non inverting terminal)
so
Vout=(vp-vn)*A (with A=open loop gain)
solve that for Vout we get:
Vout=-((20*Vin-Vd)*A)/(2*A+11)
and allowing A to go to infinity we get:
Vout=Vd/2-10*Vin
and with Vd=3.2v we get:
Vout=1.6-10*Vin
and then you can calculate with any input or output voltage as long as the input is negative and enough to forward bias the zener diode.
If the power supply causes clipping, then we have just:
Vout=1.6-10*Vin {Vee<=Vout<=Vcc}
where Vcc is the positive supply voltage and Vee is the negative supply voltage.
Basically, this converts a sine wave to a steady output. If you have a 3Vpp input, you wind up with that voltage output (assuming the A/C signal is also powering the OpAmp) that is the same. Minus the diode drop of the output diode (about .6V), and you wind up with a slightly increasing 1.76-1.80+V due to the cap on the output and the diode clipping the waveform to keep the cap charged. The current through the output cap is exactly twice the current through the input cap, and you will find that current steadily declines while it's running, due to the caps. Since you left it unspecified, I use 47uF caps for input and output.View attachment 183592
Hi Guys,
Could anybody explain me, how to analyze the above circuit.
What is out put of this circuit ?
I expect theoretical calculation and explanation ?
Regards,
Ok well with -6.6v on the output you would need +0.66v on the input. The diode does not conduct when the output is negativeSorry there was correction in answer it should be -6.6V
That's not a diode it's suposed to be a 3.2v zenerWhat?
When the output magnitude is more that -3.2V the Zener will start to conduct and the gain will drop.
LTspice simulation below.
View attachment 183624
Yes.That's not a diode it's suposed to be a 3.2v zener
Oh yes for some reaspn i thought it was pointing to the rightYes.
How does that change what I said?
yes, simulation shows output is 6.6V.What?
When the output magnitude is more that -3.2V the Zener will start to conduct and the gain will drop.
LTspice simulation below.
View attachment 183624
Nodal analysis for one.yes, simulation shows output is 6.6V.
but how can be proved theoretically .
The easiest is probably to write equations for all the current into the summing node (whose sum has to be zero).yes, simulation shows output is 6.6V.
but how can be proved theoretically .