The Transistor — Question 10

Section 41-10 · Physics pages 6990–6991

Question 10 Charging an insulated gate consumes energy even without dc gate current Model a transistor gate as constant capacitance Cg = 20.0 pF charged from 0 to VDD = 3.30 V through a resistive driver, then discharged to ground. There is no energy recovery or leakage. driver Cg 0 ↔VDD Schematic; labels, not drawing lengths, specify the data. (a) Derive energy drawn from the supply during charging and energy stored at the final gate voltage. (b) Find total energy dissipated per complete charge-discharge cycle and average dynamic power at 10.0 MHz complete cycles per second.
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Question 10 - Solution Charging an insulated gate consumes energy even without dc gate current driver Cg 0 ↔VDD (a) Method and calculation Esupply = ∫ VDDI dt = VDDQ = CgV 2 DD . Ug = ∫CgVDD 0 (q/Cg)dq = 1 2CgV 2 DD . (b) Interpretation and further calculation Ecycle = 1 2CgV 2 DD + 1 2CgV 2 DD = CgV 2 DD = 217.8 pJ . Pdyn = fCgV 2 DD = 2.178 mW . Physical check Half the supply energy is dissipated during charging and the stored half during discharge. Zero steady gate current does not mean zero switching power; actual gate capacitance can vary with bias.

Question 10

Charging an insulated gate consumes energy even without dc gate current

Model a transistor gate as constant capacitance Cg=20.0pFC_g=20.0\,\mathrm{pF} charged from 0 to VDD=3.30VV_{DD}=3.30\,\mathrm V through a resistive driver, then discharged to ground. There is no energy recovery or leakage.

Diagram shown in the original SVG worksheet.

  1. Derive energy drawn from the supply during charging and energy stored at the final gate voltage.

  2. Find total energy dissipated per complete charge-discharge cycle and average dynamic power at 10.0MHz10.0\,\mathrm{MHz} complete cycles per second.

Question 10 - Solution

Charging an insulated gate consumes energy even without dc gate current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Esupply=∫VDDIdt=VDDQ=CgVDD2.E_{\mathrm{supply}}=\int V_{DD}I\,dt=V_{DD}Q=\boxed{C_gV_{DD}^2}.Ug=∫0CgVDD(q/Cg)dq=12CgVDD2.U_g=\int_0^{C_gV_{DD}}(q/C_g)dq=\boxed{\tfrac12C_gV_{DD}^2}.

(b) Interpretation and further calculation

Ecycle=12CgVDD2+12CgVDD2=CgVDD2=217.8pJ.E_{\mathrm{cycle}}=\tfrac12C_gV_{DD}^2+\tfrac12C_gV_{DD}^2=\boxed{C_gV_{DD}^2=217.8\,\mathrm{pJ}}.Pdyn=fCgVDD2=2.178mW.\boxed{P_{\mathrm{dyn}}=fC_gV_{DD}^2=2.178\,\mathrm{mW}}.

Physical check

Half the supply energy is dissipated during charging and the stored half during discharge. Zero steady gate current does not mean zero switching power; actual gate capacitance can vary with bias.