Electronic Design

Simple Circuit Controls Pulse Transition Rates

Some applications require that pulse transition rates be controlled to minimize distortion and crosstalk in subsequent systems. The choice of active filtering generally results in elaborate and expensive designs, especially because good transient response is required. The simple circuit presented here needs only a few inexpensive components, yet it yields excellent transient response (Fig. 1). Functioning as a rate limiter, the circuit responds to each input-voltage step with a precise ramping transition at its output.

The topology is an inverting integrator enclosed within an overall negative-feedback loop. The available current for the integration capacitor (C) controls the output voltage's rate of change. This current originates from a pair of coupled differential amplifiers enclosed within the feedback loop. Together, they form a hyperbolic tangent (tanh) function for the integration current. When the input voltage steps negative, the integration current switches to a corresponding limit. The resulting positive-going ramp rate (in V/s) will be I/C, where:

I = (0.7 + VEE)/R3

In this example, VEE = 215 V and VCC = 15 V. If R3 = R4, the rising and falling rates will be equal. As the ramping continues, the junction of R1 and R2 will go toward virtual ground. As this node nears virtual ground, the two differential amplifiers enter their linear region where the integration current is rapidly reduced to zero. This controlled transition from ramping to the steady state along with a large phase margin for the loop gain gives a transient response with no overshoot or ringing.

The output voltage is inverted with respect to the input with a gain of R2/R1. For an inverting gain of one, the output amplitude should be in the range of 2 to 10 V to give a well-defined ramping behavior. Operation within this range also ensures that the reverse base-emitter voltage rating of the transistors isn't exceeded. For a transition at the output of V volts, the rise time will be (0.8VC)/I. The example shown here has a 0.37-V/µs ramp rate. So, a 5-V p-p input yields a rise/fall time of 10.8 µs. Notice how the oscilloscope traces show the ramping behavior of the transient response, which has no overshoot or ringing, and the equal rise/fall time measurements (Fig. 2).

Hide comments

Comments

  • Allowed HTML tags: <em> <strong> <blockquote> <br> <p>

Plain text

  • No HTML tags allowed.
  • Web page addresses and e-mail addresses turn into links automatically.
  • Lines and paragraphs break automatically.
Publish