Op-Amp Output Units With Oscillatory Input Voltage

In summary, the integrating op-amp circuit shown produces an oscillatory output that carries the time units inside the trig function, which doesn't allow those units to cancel. This causes the output voltage to have units of V/s instead of V/Ω.
  • #1
cmmcnamara
122
1
Hi all,

ME major here that's very confused about the integrating op-amp. It's pretty simple what I am confused about, just unit problems.

Suppose you have the following integrating op-amp circuit:

Feedback capacitor: 10μF
Input Resistance: 10kΩ
Input Voltage: 5V[sin(100t)]

My problem arises with the units here. For an integrating op-amp the output voltage can be shown to be vo=-τ∫vi*dt . The time constant obviously has units of Hz or s^-1. However for any oscillatory input, the integration results in another oscillatory output which carries the time units inside the trig function which doesn't allow those units to cancel those of s^-1, leading to output "voltage" having units of V/s. What am I missing here? How am I wrong? I appreciate any help in advance!
 
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  • #2
The [itex]100[/itex] in [itex]sin(100t)[/itex] has units of radians per second. When you multiply [itex]100\ \text{rad/s}[/itex] with [itex]t[/itex] seconds, you get an angle in radians. Then you can take the sine of that. The sine of something that has units of seconds is meaningless.

This is the reason why you should NEVER EVER EVER solve problems with actual numbers. Use variables and plug in the values for those variables at the end. I bet you wouldn't have had this confusion if you or the person who wrote this problem had written the oscillatory input as [itex]V_{\text{in}} = V \sin ( \omega t )[/itex] with [itex]V = 5 \ \text{volts}[/itex] and [itex]\omega = 100 \ \text{rad/s}[/itex]. Radians are not actual units, so it is probably easier to remember the relationship between angular frequency and plain old frequency, which is [itex]\omega = 2 \pi f[/itex]. In other words, [itex]V_{\text{in}} = V \sin ( 2 \pi f t )[/itex], where [itex]f = \frac{100}{2 \pi}\ \text{Hz}[/itex].

Integrate [itex]V \sin ( 2 \pi f t )[/itex] to get [itex]- \frac{V}{2 \pi f} \cos ( \omega t )[/itex], which has units of [itex]\frac{\text{volts}}{\text{Hz}} = \text{volts} \times \text{seconds}[/itex]. Now, when you multiply that by a constant that has units of [itex]\text{Hz}[/itex] or [itex]1/\text{seconds}[/itex], the result has units of [itex]\text{volts}[/itex], which is what you want.
 
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  • #3
Hahah, yea I just caught this about 30 seconds ago and was coming here to close up the thread. I normally go about everything symbolically but I've had my head up my arse with MathCAD this quarter learning to use it. Thanks so much for your response!
 

1. What is an op-amp?

An op-amp, or operational amplifier, is an electronic component that amplifies the difference between two input voltages.

2. How does an op-amp handle oscillatory input voltage?

An op-amp is designed to handle oscillatory input voltage by amplifying the difference between the two input voltages and producing an output voltage that is a scaled version of this difference.

3. What are the output units for an op-amp with oscillatory input voltage?

The output units for an op-amp with oscillatory input voltage are typically measured in volts (V). However, the exact output units will depend on the specific op-amp and its design.

4. Can an op-amp output negative voltage?

Yes, an op-amp can output both positive and negative voltage. The output voltage will depend on the input voltages and the op-amp's configuration.

5. How are op-amp output units calculated?

Op-amp output units are typically calculated by multiplying the difference between the two input voltages by the op-amp's gain. This gain is determined by the op-amp's design and can be found in its datasheet.

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