I need to create a test set-up that measures slew rate of a few op-amps. The fastest slew rate among all of the op-amps in my list, belongs to NE5532 with 9 V/us slew rate and output amplitude of ±10 V. the output amplitude is not stated in the test condition of slew rate parameter in the datasheet. But since the power bandwidth has been given, doing the math shown below:

enter image description here

I assume the output voltage is ±10 V. 9 V/us with 10 Vpeak output gives 1.1 us. The question is how do I create an input pulse with so high amplitude and small rise time? the set-up I am thinking of is an inverting buffer with load capacitor and resistance at the output and a window comparator after the load to capture the time between 0.9 Vpeak and -0.1 Vpeak.

  • 1
    \$\begingroup\$ What kind of signal generation are you familiar with, or have available? The NE5532 datasheet doesn't happen to show it, but many do show waveforms for this test, e.g. ti.com/lit/ds/symlink/tl072.pdf page 29 Fig. 6-57. How would you create this waveform and test condition? \$\endgroup\$ Commented Apr 2 at 13:33
  • \$\begingroup\$ I don't believe that image came from the data sheet you linked. \$\endgroup\$
    – Andy aka
    Commented Apr 2 at 13:34
  • \$\begingroup\$ What type of transient are you looking for? If you're trying to adjust the slew rate and your overshoot is too high, you may inadvertently destroy your device. \$\endgroup\$
    – Colin
    Commented Apr 2 at 13:40
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    \$\begingroup\$ Linear Technology AN94 - Slew Rate Verification for Wideband Amplifiers. If you have time, you can build this classic circuit. Rise time is 360 picosecond, which should be enough for most op-amps. Beware that it's a project of its own and may take a long time to build and debug. \$\endgroup\$ Commented Apr 3 at 5:25
  • \$\begingroup\$ @比尔盖子 I think you pretty much answered my question! \$\endgroup\$
    – Fateme
    Commented Apr 3 at 8:46

2 Answers 2


From the 1980s to the 2000s, Linear Technology published a series of application notes on analog circuits, including quick-and-dirty "DIY" circuits for solving practical lab problems, such as low-noise measurements, rise time measurements, fast pulse generators, low-distortion sine wave generation, etc. A full bibliography of these application notes is available here.

Unsurprisingly, slew rate measurement was also covered: a possible solution is described in Linear Technology AN94 - Slew Rate Verification for Wideband Amplifiers for slew rate measurements up to 2500 V/μs.

The heart of this circuit is crude but effective: a picosecond pulse generator based on the 2N2369 avalanche transistor circuit. The original circuit was introduced in Linear Technology AN47: High Speed Amplifier Techniques, Appendix D: Measuring probe-oscilloscope response. By applying a high voltage from a small capacitor between the collector and emitter of a bipolar transistor, a secondary breakdown is intentionally triggered, causing a sharp discharge.

I've already explained the theory of operation of this basic circuit in another answer, including references, schematics, and oscilloscope traces, just click here.

Figure D1. 350 ps rise/fall time avalanche pulse generator

The transistor should be a type designed for fast switching applications (linear RF transistors don't work well even if they have high transition frequency, because they're not designed for non-linear switching). For a 2N3904, the rise time is around 1 to 2 nanoseconds. The 2N2369 has a rise time as short as 350 ps, it's probably related to the fact that it was designed to be a fast switching transistor with gold-doping - today, they're a rare type. 2N2369 is already out of production but you can find some old-new stock. Meanwhile, the SMD variant PMBT2369 & MMBT2369 is still available.

Note that the 2N2369 transistor was never designed to be an avalanche transistor, we're abusing it - but it works, in fact this circuit is quite famous because of its creativity. Today, there's probably a more modern solution based on fast logic circuits. But this trick was how it was done in the 1980s.

However, to make it practical and convenient for lab use, we need much more than a transistor, and the following problems must be overcame.

  1. It's a freerunning relaxation oscillator, and cannot be controlled. We need to modify it so it can be triggered by an external signal.

  2. The output amplitude is high, around 70 V to 150 V depending on the bias voltage and transistor used. One can use a fixed 3-resistor RF attenuator to adjust its amplitude, but ideally we want to adjust it electronically.

  3. The unmodified pulse shape is a narrow Dirac impulse. We need to modify the circuit so it outputs a Heaviside step instead.

According to the application note, after various improvements, the final circuit is the following monstrosity.

Figure 8. Variable Delay Triggers a Subnanosecond Rise Time Pulse Generator. Charge Line at Q5’s Collector Determines
≈10 Nanosecond Output Width. Output Pulse Occurance is Settable from Before-to-After Trigger Outpu

At the top left, LT1553 is a DC/DC converter for supplying the high-voltage bias to the avalanche transistor. By adjusting the feedback signal, the bias voltage can be controlled. At the center of the schematic, LT1394 analog comparators are used for generating an adjustable delay, which is then used to trigger the avalanche transistor. At the bottom right is the modified avalanche transistor pulse generator, a long coax cable (with an open circuit at the far end) is used as a transmission line pulse shaping circuit, changing the pulse shape from a narrow impulse to a step-like pulse.

The rise time of this pulse generator is 360 picoseconds, suitable for testing an opamp with a slew rate as high as 2500 V/μs.

Figure 15. 360 Picosecond Rise Time Monitored in 6GHz Sampled Bandwidth Assures Measurement Integrity (Courtesy of Michael J. Martin, Tektronix, Inc.)

It's not difficult to see that the design and construction of a high-speed pulse generator is a project of its own. It can take a long time to build and debug... If you're on a hurry, the reasonable solution is to simply purchasing a ready-made pulse generator. But if you have a lot of free time, carry on...

If you just need a simple test, I think these elaborate bias, delay and trigger circuits may be unnecessary. You can start with the basic avalanche transistor circuit, then add a coax cable to shape the impulse to a step, then add an RF attenuator to reduce the pulse amplitude. This may be enough for a quick and dirty test. It's also unnecessary to strictly follow the schematics, any circuit with a similar principle of operation can be used. For example, when I first tried to build the avalanche pulse generator, I created the high voltage source using a 555 timer, a 2N5551 switch, and an inductor.

  • \$\begingroup\$ Wow, this one gets an upvote, hoping it becomes a future reference answer :) Luckily, OP's opamp is very very slow, so even standard logic does the job, but this actually does useful output impedances, so it's much much more versatile and precise. Love it! \$\endgroup\$ Commented Apr 3 at 9:36
  • \$\begingroup\$ @比尔盖子 can I not simply use a capacitor instead of the coaxial cable? \$\endgroup\$
    – Fateme
    Commented Apr 3 at 11:01

You'd honestly buy a signal generator. Building a device that reliably generates a slew rate with enough drive strength to test modern opamps is indeed a bit of a challenge (you need to be as good as an opamp developer, minus the fact that they might have intentionally limited slew rate for e.g. stability or noise reasons; however, the NE5532 is slooooow by modern standards).

I have, however, my doubts regarding the formula you added to that screenshot (which doesn't seem to be from a NE5532 datasheet?); this assumes that the maximum slew rate is measure with a sinusoidal input, and I don't think that's true. Sadly, the NE5532 is pretty old, and the datasheet doesn't tell us how slew rate was measured, but we can go by the datasheets of more modern opamps; this is a random pick from the TI website, but the LM7171QML exists in a DIP package, so might be similarly limited in bandwidth by package parasitics.

It has a slew rate of 2400 V/µs (That's freaking fast!), and a "bandwidth", depending of how your define that, of between 170 MHz and 240 MHz or so. It has a Vpeak of 12.7 V, so that according to your formula, if would have a bandwidth of only 2400 V/µs / (6·12.7 V) ~= 30 MHz. So, your formula is … wrong, by a factor of 8, for this specific opamp.

So, the formula assumes slew rate is the maximum derivative w.r.t to to time of a maximum-amplitude sine at GBW as frequency. That's not what slew rate means. The formula assumes you measure slew rate for sinusoidal inputs, whereas it seems more sensible to me to specify it for step inputs, because otherwise, it would be a redundant measure.

You've realized this! You don't try to measure bandwidth, you try to measure response to a step. Now here comes the problem: The NE5532, as ancient and slow as it is, still does 9V/µs according to the datasheet. That means that your input needs to be a solid amount steeper to be able to prove the limiting factor isn't the slowness of the input, but the opamp itself. Now, you're right approaching a step function with a bandwidth approach here, as it gives us indication of the design constraints of such a system.

The classical approach would be to say we need the first 5 harmonics, at least, of a 10V sine wave with 90 µs rise time, as system bandwidth; so that's roughly 122 MHz. That's still pretty doable with digital logic! For example, a simple gate from a ""modern"" "high-voltage" logic family like the NXP HEF4093B can do a low-high transition in less than 60 ns, if run with supply voltage of say 12 V. The trick will be transporting that to you device under test; but seeing the low cost of this: do a one-off measurement board, fabricate it for a handful of euro in China. You'd just literally buy a HEF4093, configure all four gates to be inverters¹, take three to make a slowed-down ring oscillator by chaining them (in a ring) with an RC low pass between the first and the second, connect the output of the second to the trigger input on your oscilloscope, connect the output of the third a) back to the input of the first and b) to the input of the fourth, and use the fourth to drive your op-amp.

You put both the op-amp and your HEF4093b on the same board, make sure that the trace from the output of the fourth inverter to the input of the op-amp is shorter than 5 cm, and that both chips have decoupling capacitors directly adjacent to their supply pins, and that they have low-inductance ground current path, as well as ground below the signal trace. Use vertical SMA or (in this low frequency range, still) BNC connectors you solder onto the board to export trigger and op-amp output to your oscilloscope. We're not impedance-matching here, so make sure you're not interpreting the ringing your high-bandwidth oscilloscope might see for signal properties – these are just reflection along the length of the coax cable between outputs of your device and inputs of your scope. Keep these distances short!


¹ of the two inputs of the four NANDs, you always tie one to VCC

  • \$\begingroup\$ "It has a slew rate of 2400 V/µs (That's freaking fast!), and a "bandwidth", depending "of how your define that, of between 170 MHz and 240 MHz or so. It has a Vpeak of 12.7 V, so that according to your formula, if would have a bandwidth of only 2400 V/µs / (6·12.7 V) ~= 30 MHz. So, your formula is … wrong, by a factor of 8, for this specific op-amp." the bandwidth in the formula I added is the large-signal bandwidth aka full power bandwidth. The bandwidth you are referring to is small-signal bandwidth. \$\endgroup\$
    – Fateme
    Commented Apr 3 at 4:20
  • \$\begingroup\$ "The classical approach would be to say we need the first 5 harmonics, at least, of a 10V sine wave with 90 µs rise time, as system bandwidth; so that's roughly 122 MHz. " can you elaborate on that? what first harmonics?! \$\endgroup\$
    – Fateme
    Commented Apr 3 at 4:26
  • \$\begingroup\$ But it's not the large-signal bandwidth by any means, unless I'm fundamentally missing something; see my explanation, and my counterexample. \$\endgroup\$ Commented Apr 3 at 9:40

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