L3_1 5.9.2 Part II CE High Frequency Response Exercises

# L3_1 5.9.2 Part II CE High Frequency Response Exercises -...

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5.9 CE High Frequency Response Lecture 3 Part 1 Jan. 15, 2010 Zubair Rehman

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2 Example 5.18 Solution
3 β O = 100 V A = 100 V C μ = 1 pF f T = 800 MHz r x = 50 A m = V o / V sig = -0.952 * 0.357 * 120 = - 40.75 R’ sig = r π || (r x + R B || R sig ) = 2.5 k || (.05 k + 100 k || 5 k) = 1.645 k C T = C π + C μ (1 + g m R L ’) = 7 pF + 1 pF(1 + 120) = 128 pF R sig C T = 1.645 k * 128 pF = 210.6 ns ω H = 1 / R sig C T = .004749 Grad/s = 4.749 M rad/s f H = 1 / (2 π R sig C T ) = 0.756 MHz = 756 kHz f H = 0.756 MHz << 53 MHz The assumption that V O = -g m R’ L is valid. = = = = = = = = = k mS g r mV q kT V mS mV mA V I V I g m T T DC T C m 5 . 2 40 100 25 / 40 25 1 β π 120 3 * 40 952 . 0 5 100 100 = = = + = + k mS R g k k k R R R L m sig B B 375 . 0 5 || 100 05 . 0 5 . 2 5 . 2 || = + + = + + k k k k k R R r r r sig B x k k k k R R r R R xg R R r r r x R R R V V A L C O L L m sig B x sig B B sig O m 3 5 || 8 || 100 || || || = = = + + + - = = pF pF MHz mS C f g C k mA V I V I V r T m DC A C A O 7 1 800 * 2 40 2 100 1 100 = - = - = = = = = μ Is ? 2 1 C R f L H < MHz GHz pF k C R L 53 053 . 0 1 * 3 * 2 1 2 1 = = = ′ π

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4 Exercise 5.51 For the amplifier in Example 5.18, find the value of R L that reduces the midband gain to half the value found. What value of f H results? Note the tradeoff between gain and the bandwidth. If R
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L3_1 5.9.2 Part II CE High Frequency Response Exercises -...

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