TLV9034-Q1: Slew rate doesn't show difference

Frank,

First off, TLV9304QPWRQ1 is NOT an op amp but comparator so you may not use it in your application. 

When it comes to LM2902B-Q1 (SR of 0.5V/us) vs TLV9354QPWRQ1 (20V/us), you may not see much difference in the output rate of change because you apply a small input step voltage.  LM2902 is a bipolar transistor input op amp where 100mV input step voltage is enough to push it into a slew whereas TLV9354 is a CMOS transistor input op amp where you must apply 1V or higher input step to force it to slew.  Thus, TLV9354 will respond according to a small-signal rise time formula (see below) to the input step in the range of 100mV or less.

In the case of TLV9354 with GBW of 3.5MHz in Gain=-(56k||56k)/1k = -28, and thus effective bandwidth is fc = 3.5MHz/28 = 125kHz and the input step Vin = 2.5V/28 = 89mV.

Therefore, it is a small signal rise time and NOT slew that controls its output rate of change:

the small-signal rise time is:  tr = 0.35/125kHz = 2.8us  for as long as the input signal is less than 1V.

Aside from these, please remove or lower the values of all caps that together with resistors control the time constant, TC=R*C, of the of the signal - see below.

Yes, since the small-signal rise-time is a function of fc, which in turn depends on GBW and gain (fc=GBW/Gain), in order to minimize the settling time you may pick another op amp with higher GBW, lower the gain or do combination of both.

Frank,

This formula arises from analyzing the step response of a first-order system (like an RC circuit) and finding the time it takes for the output to transition from 10% to 90% of its final value. Thus, the rise time of a small-signal circuit, like an RC low-pass filter, is approximated by the formula t_r ≈ 0.35 / BW, where BW is the 3-dB bandwidth, BW=GBW/Gcl.

• Consider a simple RC low-pass filter (resistor R and capacitor C in series). 
• When a step voltage (a sudden change from 0 to a value) is applied to the input, the output voltage across the capacitor rises exponentially. 
• The time constant (τ) of this RC circuit is given by τ = RC. 

• The output voltage (Vout) of an RC circuit to a step input (Vin) is given by: Vout(t) = Vin * (1 - e^(-t/τ)).
• To find the time it takes for the output to reach 10% of the final value (Vf = Vin), we set Vout(t1) = 0.1 * Vf and solve for t1: t1 = -τ * ln(0.9).
• Similarly, to find the time it takes for the output to reach 90% of the final value (Vout(t2) = 0.9 * Vf), we set Vout(t2) = 0.9 * Vf and solve for t2: t2 = -τ * ln(0.1).
• The rise time is then the difference between these two times: t_r = t2 - t1 = τ * (ln(0.9) - ln(0.1)) = τ * ln(9) = 2.2 * τ.
• Since τ = RC, the rise time can be expressed as t_r ≈ 2.2 * RC. 
• For a first-order low-pass filter, the 3-dB bandwidth (BW) is related to the time constant by BW = 1 / (2 * π * τ). 
• Substituting τ = 0.35 / BW into the rise time equation (t_r ≈ 2.2 * τ) gives: t_r ≈ 2.2 * (0.35 / BW) ≈ 0.77 / BW. 
• A more accurate approximation, often used, is t_r ≈ 0.35 / BW. 

In summary: the small-signal rise time formula (t_r ≈ 0.35 / BW) is derived by analyzing the step response of an RC circuit and relating it to the circuit's bandwidth. This approximation is valid for first-order systems and is widely used in electronics and signal processing to characterize the speed of circuits and signals.  The effects of fc (op amp's effective bandwidth) on rise-time, t_r, is clear from its formula - the rise time is inversely proportional to fc, which means that if you increase fc 10x (either by using op amp with 10x higher GBW or using 10x lower gain), you will shorten the rise-time by 10x.