I need to generate a "pulse" signal with a specific slew rate, period, on, and off time. I can generate any signal I need using a benchtop arbitrary function generator, however, I need to drive a much higher voltage and current than what the function generator is capable of.

Here is an illustration of the pulse that I am generating which needs to be amplified. NOTE: All of the below diagrams show my desired output signal. The input signal is identical except that the amplitude is 0V to <=5V.


The pulse has a period of 1 second (frequency is 1 Hz).

Here are details of a single pulse:

Pulse details

The slope of the rising and falling slew is ~2V/ms. Tpulse is 24ms. Tlow is 16ms. The output voltage needs to be 0V to 24V. The maximum input (i.e., output from the function generator) is 0V to 5V. The load connected to the output (i.e., the 24V side) will draw a current of about 1A (just assume a resistive load for this question).

From my (very meager) understanding, I need some form of linear amplifier. Unfortunately, all my research on the subject has resulted in amplifiers for AC signals (that are AC coupled into biased transistors).

I would really appreciate even just a general solution to this type of problem with some explanation of (or links to) the theory.

  • \$\begingroup\$ Please more adequately define your input frequency, voltage range, and expected rise and fall time variation. Then define the limits that you need the output to be in terms of both positive and negative amplitudes and, the acceptable range of rise and fall times. In other words; what you have written isn't sufficient to design what you want. \$\endgroup\$
    – Andy aka
    Oct 25, 2023 at 19:14
  • \$\begingroup\$ @CristobolPolychronopolis I've added more details to the question. The current and voltage drive capabilities were already listed (24V @ 1A). \$\endgroup\$ Oct 25, 2023 at 19:49
  • \$\begingroup\$ Input slew is also 2V/ms? \$\endgroup\$
    – Reinderien
    Oct 25, 2023 at 19:59
  • 1
    \$\begingroup\$ Maybe try searching for a power op-amp. Many can supply several amps and, are high quality. \$\endgroup\$
    – Andy aka
    Oct 25, 2023 at 20:15
  • 1
    \$\begingroup\$ 2 V/ms is well within the capabilities of just about any op amp. You could probably even do this with the hated μA741. \$\endgroup\$
    – Hearth
    Oct 25, 2023 at 21:43

2 Answers 2


Since the input pulse shape can be tailored by the AFG, you can adjust the amplified high power pulse rise and fall times there. This should be much more accurate than tuning an integrating power amplifier, and way more easy to tune/optimize.

If there is no meaningful energy storage, like an capacitor or inductor, at the load, then the output stage of the power amplifier can be an emitter-follower; there is no need for a full totem-pole output stage.

You don't say what precision of accuracy/precision you need in the output pulse, so for at least the concept stage you could use one section of an LM324, LM358, or any other opamp that has an input common mode range that includes its negative rail (GND in this case). For example, an LM358 running on +30 V and GND, driving a TIP31 NPN power transistor in a non-inverting amplifier circuit with a gain of 5 should produce the desired output. The resistive load is the pull-down for the TIP31 emitter follower.

The concept might have to be tweaked in real life . . .

If the slew rate is to slow, choose a faster opamp.

If the opamp cannot handle the required base current, change the transistor to a darlington.

If the output fall time does not track the input, you might have to add a pull-down resistor to the emitter-follower output stage in parallel with the load.



In your diagram, Vin is somewhat ambiguous and could be interpreted as either 24V or somewhat higher than 24V. You have not described Vlow so I assume it is 0.

With this assumption, your slew is not 2 V/ms:

$$ \frac {24} 2 \div \frac {(24 - 16)\times 10^{-3}} 2 = 3 \text{V/ms} $$

Keep in mind that, in increasing order of complexity,

  1. matching the input and output rise and fall times is very easy (one amplifier stage);
  2. discarding the input slew and recreating a fixed output slew is straightforward and can be done with an integrator; and
  3. matching the output slew to an arbitrary input slew requires a significantly more complex circuit.

I assume (2). You'll need an inverting comparator stage followed by an integration stage. Ideally both stages would be rail-to-rail. There are rail-to-rail opamps, there are opamps supporting > 1A, and there are opamps supporting a supply span of > 24V, but I was not able to find one that does all three (though I didn't look very hard). Your suggestion of an LM675 is very much not rail-to-rail, so if you have a 24 V supply, your output swing will be less than that.

If you use something simple like the TLV930, it's rail-to-rail, and takes up to 40V; but would require an additional power stage. If you can live with a drop of a half a volt or so below Vdd, then a very easy final stage is a single high-current NPN BJT.

A first cut could look like this (power stage not shown):


simulate this circuit – Schematic created using CircuitLab

This would be improved by the addition of some hysteresis feedback on U1.

Caution: U1 is not a generic op-amp, but a comparator. It's designed to handle fast saturation well. This is easier to choose, and something like the LM2903 would probably do fine.

  • \$\begingroup\$ The input signal "amplitude is 0V to <=5V." \$\endgroup\$
    – AnalogKid
    Oct 27, 2023 at 13:23
  • \$\begingroup\$ Cool? This has a comparator to 2.5V. What's your point? \$\endgroup\$
    – Reinderien
    Oct 27, 2023 at 13:25
  • \$\begingroup\$ Your first sentence says Vin is "somewhat ambiguous". \$\endgroup\$
    – AnalogKid
    Oct 27, 2023 at 13:29
  • \$\begingroup\$ I'm talking about the Vin written into OP's diagram, which is written beneath the label "Vin" against the y-axis and not aligned with the maximum level of the curve. \$\endgroup\$
    – Reinderien
    Oct 27, 2023 at 13:40
  • \$\begingroup\$ Agree - the image has issues. But the text is pretty clear. \$\endgroup\$
    – AnalogKid
    Oct 27, 2023 at 15:50

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