Homework about a simple inverting amplifier

In summary, the conversation involved finding the closed-loop voltage amplifier of a circuit with a closed loop amplifier. The formula for the closed loop voltage amplifier is Av = v0/v in = -(R2/R1). The challenge was understanding how the part connected to the ground affected the formula. The solution involved giving each resistor the same value, assuming an output voltage of 10 volts, and starting at the input of the opamp to determine the total impedance connected to the output. The gain was found by working out the input voltage and the current going towards the negative input. The final gain was an integer number.
  • #1
Learner123
16
0

Homework Statement



inverting amplifier.jpg

The picture is an closed loop amplifier, the question is simple: find the closed-loop voltage amplifier of the circuit

Homework Equations



Av = v0/v in = - (R2/R1)

The Attempt at a Solution



If R2 is just simply a bunch of resistors connecting crazily, then it is no problem to me, but in here, R2 is partly connected to the ground (2 connected to the ground), so if anyone could explain how the part connecting to the ground affecting the closed loop formula, i would very thankful.
 
Physics news on Phys.org
  • #2
Give each resistor the same value and assume an output voltage.
I assumed 10K for the resistors and 10 volts output.

Because the output is positive and this is an inverting amplifier, the input must be negative.

The other thing we know is that the inverting input is at ground potential, even though there is no current flowing into this input.

Start at the input of the opamp and take the series 10 K and the 10 K to ground in parallel. (the inverting input is at ground potential, remember).
so, that is 5 K in series with the next 10 K ie 15 K to the left of the RH T junction.

Continue like this to get the total impedance connected to the output.

Then work out the total current using the 10 V figure.

Then it splits with 10 K to ground and 15 K to the left.

When you have all the currents, the last one is the current going towards the -Ve input. This continues into the input resistor and the negative signal source.

Knowing the size of the series resistor, you can work out the input voltage and hence the gain because this voltage is causing the 10 volts output.

I have a set of answers when you try this. The gain is an integer number.
 

1. What is a simple inverting amplifier?

A simple inverting amplifier is an electronic circuit that amplifies an input signal while inverting its polarity. This means that the output signal will have the opposite polarity of the input signal.

2. How does a simple inverting amplifier work?

A simple inverting amplifier works by using an operational amplifier (op-amp) to amplify the input signal. The op-amp has two inputs, a positive and a negative, and one output. The input signal is connected to the negative input, while the positive input is grounded. The op-amp amplifies the difference between the two inputs and outputs the result.

3. What are the components of a simple inverting amplifier?

The components of a simple inverting amplifier include an operational amplifier, two resistors, and a power supply. The op-amp is the main component responsible for amplifying the input signal. The two resistors are used to set the gain (amplification) of the amplifier. The power supply provides the necessary voltage for the circuit to operate.

4. How do you calculate the gain of a simple inverting amplifier?

The gain of a simple inverting amplifier can be calculated using the following formula: Gain = -(R2/R1), where R1 is the resistor connected to the input and R2 is the feedback resistor connected to the output. For example, if R1 is 10kΩ and R2 is 20kΩ, the gain would be -2, meaning the output signal will be twice the amplitude of the input signal but with the opposite polarity.

5. What are the applications of a simple inverting amplifier?

Simple inverting amplifiers are commonly used in audio amplifiers, filters, and signal processing circuits. They can also be used to convert a non-inverting signal into an inverting one, which can be useful in certain applications. Additionally, they are used in instrumentation and control systems to amplify small signals for measurement and control purposes.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
5
Views
800
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
34
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
34
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
967
Back
Top