What is slew rate?

The slew rate is the maximum rate at which the output voltage can change. This is especially relevant when amplifying sound waves using an op amp.

slew rate is the maximum change in voltage that can be achieved per second. The unit typically used is V/μs.

For example, a slew rate of 0.5 V/μs would be calculated using this formula:

2 x π x f x V = Slew rate (V/s)

So in this case you’d first convert 0.5 V/μs to 500 000 V/s.

Then to find the maximum voltage for a 1KHz signal you’d use this formula:

500 000/ (2 x π x 1000) = 15.71 volts.

This also means that a 1KHz signal can have an amplitude of 15.71 volts.

To put this into context, the human range of hearing is 20Hz to 20KHz. Therefore when amplifying audible sound you could have an amplitude of up to 3.98volts with this slew rate. Meaning that this op amp would be suitable for a 20KHz signal with an amplitude of 3.98 volts or less.

However when amplifying a 40KHz signal (a typical ultrasonic frequency), you’d only be able to achieve an amplitude of 1.989 volts. meaning that this op amp would only be suitable for 40KHz frequencies with an amplitude below 1.989 volts.




Diode bridge


This is a diode bridge. It converts alternating current (AC), into direct current (DC). The circuit is set up so that current is always flowing towards the input of RL, and away from the output. More information about the properties of diode can be found here:’https://tphelectronics.com/2015/09/29/diodes/


AC signal output is most commonly just connect to ground (0v).

RL represents the circuit to which you’re applying the DC voltage.


Signal diagram explanation


This shows one full oscillation of the AC signal at the input. Whilst the signal is above the horizontal line, the voltage at the signal input is positive. Whilst the signal is below the horizontal line, the signal input is positive


Flow of current during positive half of the oscillation.


During the section labelled ‘Input is positive’, here’s how the current flows around the circuit:


The reason why the current can’t flow through D4, after passing through RL, is because D4 has already been reverse biased by current flowing from the AC input.


Flow of current during negative half of the oscillation.


During the section labelled ‘Input is negative’, here’s how the current flows around the circuit:




As you can see, regardless of whether the AC input is positive or negative, the current flowing through RL never changes. If we could see what the current looked like at RL, It’d look something like this:



Notes for future.

I’ll add in a diagram and explanation as to how to dampen the rippling effect seen in the graph above.