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What will this yield ???

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saintefi

New Member
I am Electrical Engineering student, with classes in Circuits and Electronics, so I have a theoretical idea as to how passive circuit elements work and a little know how as to how an opamp works.
I have simulated integrators, differentiators, adders and so on, but never actually built any of these circuits.

Well am doing an internship now where I actually need to do some practical work, but without practical experiments from school, I feel lost.

Well I have a circuit which is shown below, what should the output of this circuit be with the input being a pulse generator of 2V amplitude and a frequency of 100 KHz?

7922-circuit.jpg
 

crutschow

Well-Known Member
Most Helpful Member
R3, C12, C13, L2, and R7 form an input low-pass filter. CF and RF form a feedback low-pass filter across the inverting op amp circuit. You need to calculate the response of these filters and then combine them to get the overall response of the circuit to the 100KHz pulse signal.

Easiest thing to do, of course, is just simulate the circuit.
 

saintefi

New Member
Thanks.

So with this circuit, what frequency range would pass ?

I simulated the design and am not sure about the result which is why am confused.
 

crutschow

Well-Known Member
Most Helpful Member
What result did you get?
 

Chippie

Member
Thanks.

So with this circuit, what frequency range would pass ?

I simulated the design and am not sure about the result which is why am confused.

So why not build it and see what actual results you get?

Sometimes the theoretical doesnt always realise the practical
 

schmitt trigger

Well-Known Member
Most Helpful Member
When designing a circuit, ANY circuit, one reaches a point where there is no other path forward but to actually build the circuit and see whether it works or not.
 
Upon inspection of the circuit, several things come to mind:
-- Ignore C9 and C11: they are just there to keep DC out of the output/input
-- R2 looks like output-short protection and can be ignored in frequency-response
-- Cf/Rf will cause the output to decease by a factor of 2 at the frequency where X(Cf)=10K

... and you could waste a lot of time with such observations: just simulate it. This circuit is not that complex: it just has too much AC behavior to solve by inspection. And when you do so, make sure to get the op-amp's power supplies (V1 and V2) connected correctly.

Wade Hassler
 

saintefi

New Member


Thanks guys, like I previously said, this is the first real circuit am working on and am so confused, so assuming we cut off the components from r3 up to r7, and introduce the pulse generator at C11, what would determine the gain or attenuation of the op-amp ? I have seen differentiator, integrator, inverting and non-inverting amplifiers, but I simply just dont understand this circuit, I dont even know what to look for.

Can someone just put me through completely?
 

crutschow

Well-Known Member
Most Helpful Member
You don't want to eliminate R7. R7 and RF determine the mid frequency gain of the op amp (RF/R7), which is connected as an inverter (in this case the gain is -1).

The capacitor across RF provides high frequency rolloff (low pass filter). The -3dB point is 1/(2 x Pi x R x C) = 15.9MHz. Since this is well above the frequency response of most op amps it will have no practical effect on the circuit response.

C11 blocks any DC from the input and acts as a high pass filter. It's -3dB point is 1.59Hz.
 

saintefi

New Member
Also guys, I have someone coming over to show me how to use a spectrum analyzer , what questions should I ask him ? What measurements would be important to know so I can stress what I need to know.
Thanks again
 
Last edited:

ericgibbs

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Also guys, I have someone coming over to show me how to use a spectrum analyzer , what questions should I ask him ? What measurements would be important to know so I can stress what I need to know.
Thanks again

hi,
Can you confirm the component values on the circuit are correct.

A frequency sweep of the circuit show, using LTSpice, shows just a simple amp.???
 

saintefi

New Member
Simulating the whole circuit above, the circuit seemed like an attenuator, but after putting input right before C11 and deleting the elements before input, it served as a low pass filter, but i still didnt figure out the cut off frequency.
 

ericgibbs

Well-Known Member
Most Helpful Member
Simulating the whole circuit above, the circuit seemed like an attenuator, but after putting input right before C11 and deleting the elements before input, it served as a low pass filter, but i still didnt figure out the cut off frequency.

hi,
With the original circuit comps it was just a simple amp.

I'll try it with the C11 change.

EDIT:
Added simulations from LTS
 

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crutschow

Well-Known Member
Most Helpful Member
My simulation of the whole circuit looks like ericgibbs' with -3dB points of 1.59Hz and 1.6kHz and a midband gain of -.84db (0.909). The inductor apparently has no significant effect on the response.
 
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saintefi

New Member
From ericgibbs' circuit diagram, I see a 100k resistor labeled R5, this wasnt in my diagram. So wrong output.

Crutshow, could you post your simulation results?
 

crutschow

Well-Known Member
Most Helpful Member
From ericgibbs' circuit diagram, I see a 100k resistor labeled R5, this wasnt in my diagram. So wrong output.

Crutshow, could you post your simulation results?
It has the correct output.

Coincidentally I put the same resistor at the output for my simulation. A resistor was required since SPICE will not simulate with an open capacitor terminal. The 100kΩ is too high a value to affect the measured circuit frequency response (in conjunction with the 10µf output capacitor it forms a high-pass filter with a -3dB point of .159Hz).
 
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