Op-Amp Output Units With Oscillatory Input Voltage

AI Thread Summary
The discussion centers on confusion regarding the output units of an integrating op-amp circuit with an oscillatory input voltage. The main issue arises from the integration of the input voltage, which is expressed in terms of time, leading to an apparent output unit of V/s. It is clarified that using angular frequency (ω) instead of direct numerical values helps avoid confusion, as radians are not actual units. The correct approach involves expressing the input as V sin(ωt) and integrating it, resulting in units of volts when multiplied by the appropriate constants. The participant acknowledges their oversight and thanks others for the clarification.
cmmcnamara
Messages
121
Reaction score
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!
 
Engineering news on Phys.org
The 100 in sin(100t) has units of radians per second. When you multiply 100\ \text{rad/s} with t 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 V_{\text{in}} = V \sin ( \omega t ) with V = 5 \ \text{volts} and \omega = 100 \ \text{rad/s}. Radians are not actual units, so it is probably easier to remember the relationship between angular frequency and plain old frequency, which is \omega = 2 \pi f. In other words, V_{\text{in}} = V \sin ( 2 \pi f t ), where f = \frac{100}{2 \pi}\ \text{Hz}.

Integrate V \sin ( 2 \pi f t ) to get - \frac{V}{2 \pi f} \cos ( \omega t ), which has units of \frac{\text{volts}}{\text{Hz}} = \text{volts} \times \text{seconds}. Now, when you multiply that by a constant that has units of \text{Hz} or 1/\text{seconds}, the result has units of \text{volts}, which is what you want.
 
Last edited:
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!
 
While I was rolling out a shielded cable, a though came to my mind - what happens to the current flow in the cable if there came a short between the wire and the shield in both ends of the cable? For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the shield has the same cross section area. There are insulating material between them, and in both ends there is a short between them. My first thought, the total resistance of the cable would be reduced...
Hi all I have some confusion about piezoelectrical sensors combination. If i have three acoustic piezoelectrical sensors (with same receive sensitivity in dB ref V/1uPa) placed at specific distance, these sensors receive acoustic signal from a sound source placed at far field distance (Plane Wave) and from broadside. I receive output of these sensors through individual preamplifiers, add them through hardware like summer circuit adder or in software after digitization and in this way got an...
I am not an electrical engineering student, but a lowly apprentice electrician. I learn both on the job and also take classes for my apprenticeship. I recently wired my first transformer and I understand that the neutral and ground are bonded together in the transformer or in the service. What I don't understand is, if the neutral is a current carrying conductor, which is then bonded to the ground conductor, why does current only flow back to its source and not on the ground path...
Back
Top