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

[EvLer]

I am given an ideal OP-AMP like this (R-resistor, Cap-capacitor):

          ____+Cap-_
         |          |
 ___R____|__|\______|_____ +
|          -|/
+         |
V in      |                  V out
-         |
|         |                 
|_________|______________ -

and V in is given as a step function like so:

/\ (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.

 

[Jeff273]

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

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

so your form,

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

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?

 

[EvLer]

Thanks for reply.
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.

 

[SGT]

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

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

so your form,

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

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 $$V_{in} = 2.u(t-1)$$

 

[Jeff273]

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.

 

[Jeff273]

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.

 

[SGT]

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.

 

[Jeff273]

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:

                      _ 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:

 0 _
-1  \
-2   \____
-3        \
-4         \_

If it discharged at Vin = 0, it would be useless as an integrator.