If there exists a cascade of op amps, does the previous op amp's slew rate affect the later op amp's slew rates?

All op amps used are the same Texas Instrument LM348n. The datasheet says the op amp has a nominal slew rate of 0.5 V/µs, but gives no tolerance.

I have a square signal with an amplitude of 500 mV peak to peak being sent into a two op amps system. The first op amps is a unity gain op amp. It exists to prevent the wave generator's output impedance from affecting the second op amp's gain. The second op amp is a simple inverting op amp with a gain of -10.

When I measured the slew rate of the inverting op amp alone, I found a slew rate of 0.62 V/µs. However when the cascade circuit is created, the slew rate measured at the second op amp's output is now 0.589 V/µs.

Is this decrease in slew rate due to the fact that the signal was sent through two op amps, each with its own slewing effect? Do the slew rates combine in some way?

Lastly when I set the second op amp to a gain of -1, the slew rate decreases even more to 0.414 V/µs. Is unity gain instability the reason for the second decrease in slew rate?

I thought the slew rate was an intrinsic property of an op amp. Is there a variable that affects slew rate I am not considering? • How are you measuring the slew rate of the circuit? I am asking this since the measure of this parameter is not easy, and is heavily influenced by the stray capacitances in the measure circuit and by the scope and signal generator rise time characteristics. Even if we are not dealing with high slew rate OpAmps, inaccuracies in the measuring process can nevertheless lead to erroneous conclusions – Daniele Tampieri May 14 '18 at 4:35
• @DanieleTampieri Very good point. I used an oscilloscope's built in slope measurement function to measure the output's slew rate. However, could this alone be the reason for such a large change? Even if we ignore my personal experiment, I am still very curious about the main question of multiple slew rates interacting. – Jonathan Riley May 14 '18 at 4:45
• It is critical to report all component and test impedances. Please specify R1,R2 and probe capacitance. – Tony Stewart Sunnyskyguy EE75 May 14 '18 at 5:15
• The slew rate of a string of op-amps can be no faster than the slowest op-amp. And yes, like series resistors the effects accumulate, if you look at the op-amps as RCL networks in series. – Sparky256 May 14 '18 at 5:25
• The fact that you use your scope measurement function certainly rules out its contribution to slew rate variations, since it affects equally all measures you do with it (it is a "systematic error"). Echoing what Tony Stewart asked, I would like to ask one more question: did you lower the gain of the second stage by increasing $R_2$ to the value of $R_1$ or, on the contrary, by lowering this one to the value of the former? – Daniele Tampieri May 14 '18 at 7:04

Inspired by the many correct considerations of Tony Stewart, I'll give my contribution and answer in the same order to your questions.

Is this decrease in slew rate due to the fact that the signal was sent through two op amps, each with its own slewing effect?

• Yes: cascading many stages of amplification lowers the overall speed of their response. Therefore if, for a given input signal step, you get the same output by using a two-stage amplifier instead of a single stage one, you slow down the output signal by rising its rise time $t_{r_o}$, and thus the slew rate of the output, defined as $$\mathrm{SR} = \max\left(\left|\frac{dv_\mathrm{out}(t)}{dt}\right|\right)$$

Have a look at the Wikipedia entry on rise time, precisely the section on cascaded blocks, for more infos.

Do the slew rates combine in some way?

• Yes, they combine precisely via the respective output rise times $t_{R_1}$ and $t_{R_2}$ of their stages (in the presently analyzed two stage amplifier). Roughly speaking, by using the expression for the rise time of the output of cascaded stages found in the reference given above, we get $$\mathrm{SR}_1\approx \frac{V_{o_{pk}}}{\sqrt{t_{r_i}^2+t_{r_{o1}}^2}}$$ for the single inverting gain stage, and $$\mathrm{SR}_2\approx \frac{V_{o_{pk}}}{\sqrt{t_{r_i}^2+t_{r_{o2}}^2}}$$ for the single inverting output stage, where

• $t_{r_i}$ is the rise time of the input signal, and
• $V_{o_{pk}}$ is the peak-to-peak maximum output voltage

When the two stages are cascaded, we obtain roughly $$\mathrm{SR}\approx \frac{V_{o_{pk}}}{\sqrt{t_{r_i}^2+t_{r_{o1}}^2+t_{r_{o2}}^2}}$$ and thus $\mathrm{SR}<\mathrm{SR}_1$, $\mathrm{SR}<\mathrm{SR}_2$

Is unity gain instability the reason for the second decrease in slew rate?

• No, as long as this instability does not affect the rise time of a given stage of the chain, and I suspect it is exactly this that happens when you change the gain of the second stage by rising $R_2$ to $10\mathrm{k}\Omega$ instead of lowering $R_1$ to $1\mathrm{k}\Omega$ (given the charateristics of the LM324, perhaps $R_1=R_2=2\mathrm{k}\Omega$ would be the best choice, see the discussion at the next point or Tony Stewart's answer). I have not tried to analyze your circuit according to the standard theory, but I suspect that rising $R_2$ rises the rise time of the second stage, thus giving you the sensible drop of slew rate you noticed.

I thought the slew rate was an intrinsic property of an op amp. Is there a variable that affects slew rate I am not considering?

• Yes: slew rate is an intrinsic property of every operational amplifier in the sense that it is the best performance you can get from it. Loading effects, as those explained by Tony Stewart, parasitics and design choices not aimed at obtaining the highest speed of response can and will give you a lower effective slew rate.

Do the slew rates combine in some way?

YES

Is unity gain instability the reason for the second decrease in slew rate?

NO

Is there a variable that affects slew rate I am not considering?

YES
Follow the recommended Figure 1 test conditions in the datasheet then understand why.

Slew rate is current limited into either Miller capacitance or load capacitance or limited by excessive low load resistance or any combination that reduces drive current available to create the virtual ground at the input.

Normally this ancient 1MHz uA741 style Op Amp, the LM348 is expected to have a nominal rise time of ;

$$t_R=0.35/f-3dB=0.35us$$
This is defined by the 10% to 90% slope of a small step pulse for a "small signal" response.

Max Slew rate spec of 0.5V type will reduce when cascaded by equal devices. This is a 'large signal" response and is limited by the current limited output and load. The datasheet defines it for 2k//100pF to achieve 0.5V/us large swing.

• If you choose 50 Ohms for R1 and R2 then the output will be driving R2 as the load for +/2.5V then it needs 2.5V/50R=50mA. which is current limiting the OA.

• If you used 500 Ohms and 5V then it would demand 10mA and the balance of available current of 15mA into 100pF can drive is as follows;

$$dV/dt = Ic/C = 15mA/100pF = 0.15V/us$$

This is key to understand how current limit reduces slew rate from R as well as C load effects.

The proper way to design is to define the input, output characteristics and impedances then choose the best parts to meet that criteria, rather than choose whatever is around and guess on the wrong R values to use.

A 50 Ohm driver with 5V square wave needs at least 10x the BW as the applied frequency to have the 9th harmonic and be able to source the current to drive the load Vp/R+ CdV/dt (load capacitance).

So I suspect the LM348 is not a good choice.

In future, specify output slew rate required, Voltage and load impedance. Then consider BW, drive current, load resistance and cable capacitance in your design. ( e.g. 100pF/m )