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dnmosdcprob

# dnmosdcprob - RE 255 Sample Problem — D-mode NMOS dc...

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Unformatted text preview: RE 255 Sample Problem — D-mode, NMOS, dc Calculations VD]: ill-“J For the circuit at the right, assume VDD = 10 volts, VT 2 —2 volts and KN = l mAfV'". Find the dc operating point for the circuit, i.e., find V03, and ID for RS = l K, 2 K and 3k ohms. Since the transistor has the gate tied to the source, V65 is greater than VT : -2 and the transistor is not cut off. If the IDS RS voltage drop is small, the transistor may be ' saturated. If saturated, then ID : Kancs — VJZ : 1[0 — (.2)]2 = 4 mA. Now check to see if the transistor is saturated. The IDS RS product is 4 mA times 1 K :4 volts. VDS is 10 — 4 : 6 volts. The transistor is saturated so we have a solution. Try R8 = 2 K. Assume the transistor is still saturated so ID : 4 mA. ID,S R8 = 4 mA times 2 K equals 8 volts. VDS : 10 — 8 = 2V. Check V05 — VI = 0 —(—2) : 2V. The transistor is biased at the boundary so the solution is correct. Try Rs : 3K. Assume the transistor is still saturated so ID = 4 mA. IDS RS ': 4 mA times 3 K equals 12 volts. VDS : 10 — 12 = — 2V. The drain—source voltage is now reversed. The transistor is not in saturation so the solution is not correct! With RS :_3 K, assume the transistor is not saturated. Then: ID:KN[2(VGS _ VT)VDS — Vosgl : 4VDS ' Vosi Also, from the circuit we know: V135 : Von — ID Rs : 10 F 311) Then: in = 4(10 — 310) — (10 —- 3L3): Collecting terms, we have: 911,3 — 4THD + 60 : 0 Which gives ID : 3.0 or 2.222 mA. Choose IU : 3.0 mA (the smaller current would put the transistor back in saturationl). Vm : 1.0 V. Everything checks ok. — l — [..l,.()g‘.}tirn 3-star} ...
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