# Using an op amp to achieve a voltage gain and a phase shift of 180°

#### Kelko

Joined Oct 1, 2020
33
The question asks to design an amplifier using an op amp which will have a gain of 32 dB and a phase shift of 180° so I used an inverting amplifier and I’m just not sure if I’m doing this right or not? I’m confused on my final answer and if it’s right or not. Any help would be great

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#### ericgibbs

Joined Jan 29, 2010
14,697
hi K,
What are the power supplies to the OPA.?

Have you been given details of the input signal.?
E

Update:
Show your dB to numeric Gain conversion

Last edited:

#### dl324

Joined Mar 30, 2015
13,836
How did you arrive at the current direction for I1?

#### Kelko

Joined Oct 1, 2020
33

this is the exact question and all the details we were given except I've already done part A also I used a similar question from class as reference for the current directions, should I just assume all currents are leaving the node?

#### Kelko

Joined Oct 1, 2020
33
also i just used a converter for changing dB to numeric gain

#### Papabravo

Joined Feb 24, 2006
17,580
also i just used a converter for changing dB to numeric gain
Do you agree with the converter's result, or are you just going to accept it blindly and not bother to understand what is going on?
What did the converter tell you BTW?

#### dl324

Joined Mar 30, 2015
13,836
I used a similar question from class as reference for the current directions, should I just assume all currents are leaving the node?

How much current flows into the inputs of an ideal opamp?

#### Kelko

Joined Oct 1, 2020
33
0 current flows into the inputs of an ideal opamp which is why I2 = - I1 no? at least thats what it says in our notes

#### Kelko

Joined Oct 1, 2020
33
also the equation that it's using for converting dB to numeric is Vout/Vin = 10^(32/20)

#### dl324

Joined Mar 30, 2015
13,836
0 current flows into the inputs of an ideal opamp which is why I2 = - I1 no? at least thats what it says in our notes
Think again about what KCL says.

#### Kelko

Joined Oct 1, 2020
33
the sum of the currents entering a node must be equal to the sum of the currents leaving the node, so I1 + I2 = I3 where I3 is the current entering the amp past the node and so far every time we've used inverting amps our lecturer has said that there is 0 current entering the amp because V+ = V- and V+ is connected to the ground so there is 0 current so I1 + I2 =0 then I1 = -I2 or I2 = -I. What am I missing?

#### dl324

Joined Mar 30, 2015
13,836
the sum of the currents entering a node must be equal to the sum of the currents leaving the node, so I1 + I2 = I3 where I3 is the current entering the amp past the node and so far every time we've used inverting amps our lecturer has said that there is 0 current entering the amp because V+ = V- and V+ is connected to the ground so there is 0 current so I1 + I2 =0 then I1 = -I2 or I2 = -I. What am I missing?
That isn't what you drew. You show both I1 and I2 coming into the node between the resistors and no currents leaving. Is that what KCL says?

#### Kelko

Joined Oct 1, 2020
33
I didnt draw that sorry I meant to draw a another current going from the node to the amplifier thats my bad but I still have the same equations where I1 = - I2 written down. Since there is 0 current going into the amp from the node I1 must be going against I2 so therefore I2=-I1 or I1 = -I2

#### dl324

Joined Mar 30, 2015
13,836
Since there is 0 current going into the amp from the node I1 must be going against I2 so therefore I2=-I1 or I1 = -I2
Think of the words KCL states when you assign the direction for I1. It's okay to assign the wrong direction because the math will still work out, but why assign a wrong direction when you should know the correct direction?

If I2 is entering the node between the resistors and no current enters the opamp, what current is leaving the node?

#### Papabravo

Joined Feb 24, 2006
17,580
also the equation that it's using for converting dB to numeric is Vout/Vin = 10^(32/20)
Yes. Because for an inverting amplifier the voltage gain in dB = 20 * log (Vout/Vin)

#### Kelko

Joined Oct 1, 2020
33
Think of the words KCL states when you assign the direction for I1. It's okay to assign the wrong direction because the math will still work out, but why assign a wrong direction when you should know the correct direction?

If I2 is entering the node between the resistors and no current enters the opamp, what current is leaving the node?
are you saying that I1 should be in the opposite direction? I'm so confused now between what you're saying and what my lecturer has told me to follow. Should I change the direction of I1? which now that you've said about currents leaving the node it makes a lot more sense to do so

#### ericgibbs

Joined Jan 29, 2010
14,697
hi K,
Consider the Vsource is positive, current will flow Via R2 into the INV input trying to drive it positive, this will drive the output of the Amp in a negative direction.
So the R1 feed back resistor will pass a negative current back into the INV input, so balancing the INV close to zero volts.
E

Isum at the INV input = I1-I2 =0

#### dl324

Joined Mar 30, 2015
13,836
Should I change the direction of I1? which now that you've said about currents leaving the node it makes a lot more sense to do so
The formula you've memorized for the gain of an inverting amplifier is derived using the zero differential input theorem.

If the current into the node between the resistors is I2 and the current into the inverting terminal is 0, what does the magnitude and direction of I1 need to be?

Using the direction of I1, what is the polarity of the output voltage?

#### Kelko

Joined Oct 1, 2020
33
hi K,
Consider the Vsource is positive, current will flow Via R2 into the INV input trying to drive it positive, this will drive the output of the Amp in a negative direction.
So the R1 feed back resistor will pass a negative current back into the INV input, so balancing the INV close to zero volts.
E

Isum at the INV input = I1-I2 =0
ahhhh thank you thats helped a lot actually I think I get it all now. Thank you as well dennis for your help

#### ericgibbs

Joined Jan 29, 2010
14,697
hi K,
Just remember that conventional current will flow from the positive potential to a negative potential

Positive Vsrc >>> R2 >>>INV >>> R1 >>> Neg Out