# How does a step function affect the output voltage of an ideal OP-AMP?

• EvLer
In summary: The cap holds the charge and ramps until -Vin goes to zero.In summary, the conversation discusses an ideal OP-AMP circuit with a resistor and capacitor, and a given input voltage that switches from -2V to 0V. The goal is to find the output voltage at t = 2 seconds. The formula for V(out) is -Vc, with Vc being the voltage across the capacitor. The conversation also discusses how to handle a step function in this case of integration. The summary explains that the capacitor will hold its charge and the output voltage will stop ramping when the current goes to zero, and this is demonstrated with a sample calculation.
EvLer
I am given an ideal OP-AMP like this (R-resistor, Cap-capacitor):
Code:
          ____+Cap-_
|          |
___R____|__|\______|_____ +
|          -|/
+         |
V in      |                  V out
-         |
|         |
|_________|______________ -
and V in is given as a step function like so:
Code:
/\ (volts)
|
|____1____2_______> (t, sec)
|____| (-2)
|
|
that is V(in) is -2 on [0,1] and then 0 further. I need to find V(out) at 2sec.
So far I think I got the expression I need to work with.
V(out) = -Vc
Vc = 1/(RC) * integral(V in dt)
But my problem is how I deal with a step function in this case of integration. Any help is appreciated.
Thanks.

Last edited:
$$Vc = \frac{1}{C} \int_{0}^{t} i_c dt$$

In your circuit $$i_c = V_{in}/R$$

$$Vc = \frac{1}{RC} \int_{0}^{t} V_{in} dt$$

looks right. So, assuming Vc started at zero at To with Vin = -2V, you have,

$$Vc = \left[\frac{1}{RC} \times -2(t)\right]_{0}^{1}$$

When you switch from Vin = -2V to Vin = 0V the output will stop ramping because the current goes to zero.

Does that make sense?

So, to find V(out) at t = 2, I need to add integral from 0 to 1s and then from 1 to 2s? That is my question.

Jeff273 said:
$$Vc = \frac{1}{C} \int_{0}^{t} i_c dt$$

In your circuit $$i_c = V_{in}/R$$

$$Vc = \frac{1}{RC} \int_{0}^{t} V_{in} dt$$

looks right. So, assuming Vc started at zero at To with Vin = -2V, you have,

$$Vc = \left[\frac{1}{RC} \times -2(t)\right]_{0}^{1}$$

When you switch from Vin = -2V to Vin = 0V the output will stop ramping because the current goes to zero.

Does that make sense?

Not really! When the input voltage switchs to zero the capacitor is charged. So it will start discharging towards zero
$$Vc(t) = Vc(1) + \frac{1}{RC} \int_{1}^{t} V_{in} dt$$
Where $$V_{in} = 2.u(t-1)$$

SGT,

Ignore the math for a second and check the circuit. When Vin is at zero the current through R is zero. Since there is no current flow at the virtual node (which is always at zero volts - the reference - in this circuit) the voltage across the cap as measured at Vout will hold at whatever voltage is across it.

EvLer,

Yes. Evaluating, you will have:

$$Vc = \left[\frac{1}{RC} \times -2(t)\right]_{0}^{1} \plus+ \left[\frac{1}{RC} \times 0(t)\right]_{1}^{2}$$

You can see that the second term is zero forever with Vin = 0.

Jeff273 said:
SGT,

Ignore the math for a second and check the circuit. When Vin is at zero the current through R is zero. Since there is no current flow at the virtual node (which is always at zero volts - the reference - in this circuit) the voltage across the cap as measured at Vout will hold at whatever voltage is across it.
Nope! The capacitor is negatively charged and since there is a path to ground through R, the capacitor will discharge.

SGT:

Nope! The capacitor is negatively charged and since there is a path to ground through R, the capacitor will discharge.
Nope. The voltage at the RC node is always zero (this is a virtual gnd). With Vin at zero, the circuit looks like this:

Code:
                      _ Vout
|
| -
I           C
<---         | +
gnd ____ R __________|
|
Vgnd
I = zero, and no current can flow into or out of a virtual gnd, therefore the cap has no discharge path and holds at whatever the last ramp voltage was.

Think about it. This is an ideal integrator. Assume R and C are 1 so it sums -Vin x (t).

t...Vin...Vout
--------------
0...0...0
1...1...-1
2...1...-2
3...0...-2
4...0...-2
5...1...-3
6...1...-4
7...0...-4

Looks like this:

Code:
 0 _
-1  \
-2   \____
-3        \
-4         \_
If it discharged at Vin = 0, it would be useless as an integrator.

Last edited:

## 1. What is an amplifier?

An amplifier is an electronic device that increases the amplitude or power of a signal. It takes a weak signal as its input and produces a stronger signal as its output.

## 2. How does an amplifier work?

An amplifier works by taking an input signal and using an external power source to increase its amplitude. The amplified signal is then output through a speaker, allowing the signal to be heard more clearly and at a higher volume.

## 3. What is the difference between analog and digital amplifiers?

An analog amplifier works by varying the strength of a continuous electrical signal, while a digital amplifier works by converting the input signal into a series of discrete values and then amplifying these values. Digital amplifiers are more efficient and have better sound quality, but they can also introduce distortion and noise to the signal.

## 4. What is a step function in amplifiers?

A step function is a type of input signal that is characterized by an abrupt change in voltage. In amplifiers, a step function can be used to test the response of the amplifier to sudden changes in input signals.

## 5. How do I choose the right amplifier for my needs?

Choosing the right amplifier depends on several factors such as the type of signal you want to amplify, the power requirements, and the desired sound quality. It is important to consider these factors and do research on different amplifier models before making a decision.

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