0
\$\begingroup\$

Is there way to calculate slew rate of following circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

Datasheet for Transistor

Datasheet for Diode

I chose the PMOS and Diode arbitrarily from what I could choose from Circuit Lab.

If VTRIG goes from 5V to 0V, how can I calculate the slew rate (V/us)?

\$\endgroup\$
4
  • \$\begingroup\$ Uhhh datasheet? \$\endgroup\$
    – Bradman175
    Commented Dec 26, 2016 at 1:43
  • \$\begingroup\$ @Bradman175 I've added link to datasheet \$\endgroup\$
    – Steve
    Commented Dec 26, 2016 at 2:19
  • \$\begingroup\$ Now go into dynamic under specifications and look at the delay and rise fall times? \$\endgroup\$
    – Bradman175
    Commented Dec 26, 2016 at 2:52
  • \$\begingroup\$ poor choice of FET for -5V Vgs \$\endgroup\$
    – D.A.S.
    Commented Dec 26, 2016 at 3:27

1 Answer 1

2
\$\begingroup\$
  • A rough approximate is to use the RdsOn for -5V.
  • It is guaranteed to be 0.3Ω but that is for -10V @25'C with 7.2A Pulse, width = 300 μs; duty cycle = 2 %.
  • your situation is different and vague;
    • Initial conditions :
    • Vcap (unknown )
    • Cap part number (unknown ) and
    • trace inductance (unknown )
    • cap ESR or Dissipation Factor (unknown )
    • Diode ESR 1N4148 0.1W ~ 10ohm ( assumed to be larger than 1nF
    • Cap ESR which is expected to have ESR*C value=T < 0.1us for ceramic
    • The RdsOn starts at a high value due to Vdes =5V and Vgs=-5V and not being a "logic level" gate controlled FET requires a certain amount of calculations from the datasheet below enter image description here

With the slew rate being dV/dt=Ic/C for the cap and Ic = (V+-Vcap)/(RdsOn+ESR(diode)) it becomes highly nonlinear.

But with initial conditions of ESR diode=10, and ESR or RdsOn of FET=10Ω then dropping to 0.34 Ω as Vds drops below 3V.

  • Ic=C dV/dt=1e-9 * (5V)/(10Ω+10Ω) and C= 1e-9F for Vds=3~5V

slew rate dV/dt= Ic/C= 5V/20Ω*1e9= = 0.25V/ns

Low confidence in results due to inadequate info.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.