enter image description here

I am designing the circuit shown above using mosfet NDS355AN. MCU is 5 V.
However, I have become stuck as I am unsure of how to calculate the gate charge time of the mosfet. Hence what is the formula to calculate the charge time for the gate of the mosfet?

Is it just essentially tau = RC = 1500 * Ciss?

Note LED contains resistor inside (is a very large all in one)

  • 1
    \$\begingroup\$ Calculating gate charge time misses the point - you probably want to know how long it takes to reach an adequate gate-source voltage for the MOSFET to be reasonably conducting. \$\endgroup\$ – Andy aka May 9 '19 at 9:13
  • \$\begingroup\$ What currents do you expect the FET to handle? and how much delta_V do you expect on the drain? The charging of drain-gate capacitance (Miller Capacitance) may define your speed. \$\endgroup\$ – analogsystemsrf May 9 '19 at 9:17
  • \$\begingroup\$ @Andy aka how would I do that then? \$\endgroup\$ – Student May 9 '19 at 9:19
  • \$\begingroup\$ @analogsystemrf, gate current would be 3.33 mA, LED current is 30 mA \$\endgroup\$ – Student May 9 '19 at 9:19
  • 1
    \$\begingroup\$ Couldn't read it clearly. You're right that 1M\$ \Omega\$ will have almost no effect. \$\endgroup\$ – Huisman May 9 '19 at 9:31

The step by step time constants

enter image description here

I coloured zones for dynamic losses.

When using a driver with matched impedance to the gate resistance, Rg ;

  • During turn-on Ids almost peaks THEN Vds falls
  • During turn-off Vds almost peaks THEN Ids falls.
    • during these dynamic switching times
    • Pd= Vds*Ids is greatest.

As once written, seek and ye shall find. or re-Search, next time DIY.

From your datasheet...

spec: VGS = 4.5V, RGEN = 6 Ω You are driving gate with 1k5 instead of 6 Ω , so we use total Gate Charge equation and datasheet specs instead using worst case.

So effectively with a high source R you are driving with a current source Ig=Vgs/R then from $$ I_g = ΔQ_g/Δt_s , (ΔQ=Q) , Δt_s = Q_g/I_g = Q_gR_g/V_{gs} = 5nC_{max}\cdot 1k5/5V=1us$$

Yet if we simply used T=RC=1k5*195pf=293ns (typ) (0.3us) is only 30% typ of the more accurate worst case gate charge times.

The Miller plateau for a current source gate drive means that from C=Q/V is Vgs is flat and Q jumps to the right, that means C rises until Vds drops to minimum. This is a negative feedback effect.

  • \$\begingroup\$ Thanks, so can tau = RC = 1500 * Ciss be used as a general rule or no? \$\endgroup\$ – Student May 9 '19 at 12:41
  • \$\begingroup\$ Why don't you compute the formula and tell me the Vgs/Vth ratio that gives you a 10% error. \$\endgroup\$ – Tony Stewart EE75 May 9 '19 at 12:47
  • \$\begingroup\$ Sorry im just confused with what the total turn on time would be, is it t1 + t2 + t3? \$\endgroup\$ – Student May 9 '19 at 12:49
  • \$\begingroup\$ which part is confusing? \$\endgroup\$ – Tony Stewart EE75 May 9 '19 at 12:51
  • \$\begingroup\$ just specifically what t1 t2 and t3 actually mean potentially. As in is t1 when the mosfet fully switches on as the threshold voltage is reached then? Then what does t2 and t3 refer to? \$\endgroup\$ – Student May 9 '19 at 12:52

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