APPLIED ANALOGUE ELECTRONICS
OSCILLATORS
Questions & Solutions
Question 1
Consider the oscillator circuit below. The component values are 𝐶 = 0.01 𝜇𝐹, 𝑅 = 8.2 𝑘Ω, 𝑅1 𝑎𝑛𝑑 𝑅2 .
You are given two resistors (360 𝑘Ω 𝑎𝑛𝑑 4.7 𝑘Ω) that can be used as 𝑅1 . The supply voltage 𝑉𝑐𝑐 = 12 𝑉.
The output voltage is 𝑉0 (𝑡).
a) Sketch the output voltage, 𝑣𝑜 (𝑡) versus time.
b) Derive an expression for the feedback factor 𝛽(𝜔) where 𝜔 is the radian frequency.
Using one of the RC stages;
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑎𝑛 𝑖𝑑𝑒𝑎𝑙 𝑜𝑝 𝑎𝑚𝑝 𝑎𝑛𝑑 𝑎𝑝𝑝𝑙𝑦𝑖𝑛𝑔 𝐾𝐶𝐿 𝑎𝑡 𝑛𝑜𝑑𝑒 𝑥;
𝑉1 − 𝑉2 𝑉2 1
= 𝑏𝑢𝑡; 𝑋𝑐 = 𝑤ℎ𝑒𝑟𝑒 𝑆 = 𝑗𝜔
𝑅 𝑋𝑐 𝑆𝐶
𝑉2 (𝑠) 1 1 1
= = 𝑙𝑒𝑡 𝑅𝐶 =
𝑉1 (𝑠) 𝑅𝑆𝐶 + 1 𝑗𝜔𝑅𝐶 + 1 𝜔1
𝐷𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑟𝑒𝑠𝑒𝑛𝑐𝑒 𝑜𝑓 𝑏𝑢𝑓𝑓𝑒𝑟 𝑎𝑚𝑝𝑙𝑖𝑓𝑖𝑒𝑟𝑠 𝑎𝑛𝑑 𝑣𝑖𝑟𝑡𝑢𝑎𝑙 𝑔𝑟𝑜𝑢𝑛𝑑 𝑎𝑡 𝑡ℎ𝐸 𝑖𝑛𝑝𝑢𝑡 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒
𝑖𝑛𝑣𝑒𝑟𝑡𝑖𝑛𝑔 𝑜𝑝 𝑎𝑚𝑝;
𝑉𝑖 (𝑠)
𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑓𝑎𝑐𝑡𝑜𝑟, 𝛽(𝑗𝜔) =
𝑉𝑜 (𝑠)
𝑉𝑖 (𝑠) 1
𝑇ℎ𝑒 𝑓𝑒𝑒𝑑𝑏𝑎𝑐𝑘 𝑓𝑎𝑐𝑡𝑜𝑟; 𝛽(𝑗𝜔) = =
𝑉𝑜 (𝑠) 𝜔 3
[1 + 𝑗 (𝜔 )]
1
1
𝛽(𝑗𝜔) =
𝜔 𝜔 2 𝜔 3
1 + 3𝑗 ( ) − 3 ( ) − 𝑗 ( )
𝜔1 𝜔1 𝜔1
1
𝛽(𝑗𝜔) = 2
𝜔 𝜔 𝜔 2
[1 − 3 (𝜔 ) ] + 𝑗 (𝜔 ) [3 − (𝜔 ) ]
1 1 1
But at the oscillating frequency, 𝛽(𝑗𝜔0 )𝑖𝑚 = 0 𝑚𝑒𝑎𝑛𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑖𝑡 𝑖𝑠 𝑝𝑢𝑟𝑒𝑙𝑦 𝑟𝑒𝑎𝑙.
𝜔𝑜 2 1
𝐻𝑒𝑛𝑐𝑒; 3 − ( ) = 0 𝑎𝑛𝑑 𝑠𝑖𝑛𝑐𝑒 𝜔1 =
𝜔1 𝑅𝐶
√3
𝜔𝑜 = √3𝜔1 =
𝑅𝐶
√𝟑
𝒇𝒐 =
𝟐𝝅𝑹𝑪
1 1 1
𝛽(𝑗𝜔)𝑟𝑒𝑎𝑙 = 2 = 2 =−
𝜔 8
[1 − 3 (𝜔0 ) ] √3𝜔
[1 − 3 ( 𝜔 1 ) ]
1
1
𝟏
𝜷(𝝎) = −
𝟖
c) What is the value of the oscillating frequency, 𝑓0 and the closed loop gain 𝐴𝑐𝑙 ?
√3 √3
𝑓0 = = = 𝟑. 𝟑𝟔𝟐 𝒌𝑯𝒛
2𝜋𝑅𝐶 2𝜋(8.2𝑘)(0.01𝜇)
𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝐵𝑎𝑟𝑘ℎ𝑎𝑢𝑠𝑒𝑛 𝑐𝑟𝑖𝑡𝑒𝑟𝑖𝑎, 𝐴𝑐𝑙 (𝑓𝑜 )𝛽(𝑓𝑜 ) = 1
𝐻𝑒𝑛𝑐𝑒; 𝑨𝒄𝒍 = 𝟖
d) What are the requirements for sustained oscillations in the circuit?
For sustained oscillations, |𝐴𝑐𝑙 (𝑓𝑜 )𝛽(𝑓𝑜 )| = 1 𝑎𝑛𝑑
< 𝐴𝑐𝑙 (𝑓𝑜 )𝛽(𝑓𝑜 ) = ±360𝑛 𝑤ℎ𝑒𝑟𝑒, 𝑛 = 0, 1, 2 …
e) What is the condition for the oscillator to start up?
The total phase shift should be 3600 𝑜𝑟 00
The |𝐴𝑐𝑙 (𝑓𝑜 )𝛽(𝑓𝑜 )| ≥ 1
f) Which one of the two resistors should be selected as 𝑅1 and why?
The 4.7 𝑘𝛺 resistor should be selected because it results into a higher closed loop gain of the op-amp
if the 𝑅2 is held constant thus better sustained oscillations.
g) Calculate the value of 𝑅2 given your choice of 𝑅1 in part f) above.
𝑅1 = 8(4.7𝑘) = 𝟑𝟕. 𝟔 𝒌𝛀
h) What is the minimum slew rate required for the op-amps above if you limit peak value 𝑣0 (𝑡) a
peak value of 5 𝑣𝑜𝑙𝑡𝑠?
𝑆𝑅 ≥ 2𝜋𝑓0 𝑉𝑝𝑒𝑎𝑘
𝑆𝑅 = 2𝜋(3362)(5) = 𝟎. 𝟏𝟎𝟔 𝑽/𝝁𝒔
Question 2
Consider the oscillator circuit below that uses a dual power supply at ±5𝑉 and outputs a
voltage 𝑉𝑜 (𝑡) = 2.5 sin 2𝜋𝑓0 𝑡. the circuit parameters are 𝐶1 = 1 𝑛𝐹, 𝐶2 = 2 𝑛𝐹, 𝑅1 = 1.5 𝑘Ω, 𝑅3 =
1.5 𝑘Ω 𝑎𝑛𝑑 𝑅4 = 3.3 𝑘Ω.
a) Derive an expression for the oscillating frequency, 𝑓𝑜 and the feedback factor 𝛽(𝜔) where 𝜔 is
the radian frequency?
Redrawing the circuit in a more familiar format;
Assuming that the op-amp is ideal, with infinite input impedance and infinite open loop gain.
Hence no current flows into the op-amp.
𝑅1 𝑋𝐶1
𝐿𝑒𝑡; 𝑍𝑎 = 𝑎𝑛𝑑 𝑍𝑏 = 𝑅4 + 𝑋𝐶2
𝑅1 + 𝑋𝐶1
By voltage division at node x, since no current flows into the op-amp, the feedback factor will be given
as;
𝑅1 𝑋𝐶1
𝑉𝑓 𝑍𝑎 𝑅1 + 𝑋𝐶1
𝛽= = =
𝑉0 𝑍𝑎 + 𝑍𝑏 𝑅1 𝑋𝐶1
𝑅1 + 𝑋𝐶1 + 𝑅4 + 𝑋𝐶2
𝑅1 𝑋𝐶1 1 1
𝛽= 𝐵𝑢𝑡 𝑋𝐶1 = 𝑎𝑛𝑑 𝑋𝐶2 =
(𝑅4 + 𝑋𝐶2 )(𝑅1 + 𝑋𝐶1 ) + 𝑅1 𝑋𝐶1 𝑗𝜔𝐶1 𝑗𝜔𝐶2
1
𝑅1
𝑗𝜔𝐶1
𝛽(𝑗𝜔) =
1 1 𝑅
(𝑅4 + ) (𝑅1 + )+ 1
𝑗𝜔𝐶2 𝑗𝜔𝐶1 𝑗𝜔𝐶1
1
𝑅1 𝑗𝜔𝐶
1
𝛽(𝑗𝜔) =
1 1 1 𝑅1
(𝑅4 𝑅1 + 𝑅4 𝑗𝜔𝐶 + 𝑗𝜔𝐶 𝑅1 − 2 ) + 𝑗𝜔𝐶
1 2 𝜔 𝐶 𝐶
1 2 1
1
𝑅1 𝜔𝐶
1
𝛽(𝑗𝜔) =
1 1 1 𝑅
(𝑗𝑅4 𝑅1 + 𝑅4 𝜔𝐶 + 𝜔𝐶 𝑅1 − 𝑗 2 ) + 𝜔𝐶1
1 2 𝜔 𝐶1 𝐶2 1
1
𝑅1 𝜔𝐶
1
𝛽(𝑗𝜔) =
𝑅 1 1 1
([𝜔𝐶1 +𝑅4 𝜔𝐶 + 𝜔𝐶 𝑅1 ] + 𝑗 [𝑅4 𝑅1 − 2 ])
1 1 2 𝜔 𝐶1 𝐶2
𝛽(𝑗𝜔) = 𝛽(𝑗𝜔)𝑟𝑒𝑎𝑙 + 𝛽(𝑗𝜔)𝑖𝑚
Since at the oscillating frequency, 𝛽(𝜔0 )𝐴𝑐𝑙 (𝜔0 ) = 1 (according to Barkhausen criterion), hence
𝛽(𝑗𝜔)𝑖𝑚 = 0
1
𝑅4 𝑅1 − =0
𝜔𝑜2 𝐶1 𝐶2
1
𝜔𝑜 =
√𝑅1 𝑅4 𝐶1 𝐶2
𝟏
𝒇𝒐 = 𝒊𝒔 𝒕𝒉𝒆 𝒐𝒔𝒄𝒊𝒍𝒍𝒂𝒕𝒊𝒏𝒈 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 𝒆𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏.
𝟐𝝅√𝑹𝟏 𝑹𝟒 𝑪𝟏 𝑪𝟐
Since 𝛽(𝑗𝜔𝑜 )𝑖𝑚 = 0, it follows that,
1
𝑅1 𝜔𝐶
1
𝛽(𝑗𝜔)𝑟𝑒𝑎𝑙 = 𝛽(𝜔𝑜 ) =
𝑅 1 1
([𝜔𝐶1 +𝑅4 𝜔𝐶 + 𝜔𝐶 𝑅1 ])
1 1 2
𝑹𝟏 𝑪𝟐
𝜷(𝝎𝒐 ) =
𝑹𝟏 𝑪𝟐 + 𝑹𝟒 𝑪𝟐 + 𝑹𝟏 𝑪𝟏
b) Find the value of the closed loop gain 𝐴𝑐𝑙
−1 −1
𝑅1 𝐶2 1.5𝑘 × 2𝑛
𝐴𝑐𝑙 = [ ] =[ ] = 𝟑. 𝟕
𝑅1 𝐶2 + 𝑅4 𝐶2 + 𝑅1 𝐶1 1.5𝑘 × 2𝑛 + 3.3𝑘 × 2𝑛 + 1.5𝑘 × 1𝑛
c) Find the value of 𝑅2
𝑅2
+ 1 = 3.7
𝑅3
𝑅2 = 1.5𝑘(2.7) = 𝟒. 𝟎𝟓 𝒌𝛀
d) What value of 𝑅2 would you pick to enable the oscillator start up?
I would pick any value greater than 4.05 𝑘Ω like 4.7 𝒌𝛀 such that the closed loop gain of the amplifier
is always large enough for sustained oscillations.
e) What is the minimum slew rate of the op-amp?
1 1
𝑓𝑜 = = = 50.583 𝑘𝐻𝑧
2𝜋√𝑅1 𝑅4 𝐶1 𝐶2 2𝜋√3.3𝑘(1.5𝑘)(1𝑛)(2𝑛)
𝑆. 𝑅𝑚𝑖𝑛 = 2𝜋𝑓0 𝑉𝑝
𝑆𝑙𝑒𝑤 𝑟𝑎𝑡𝑒 = 2𝜋(50.583𝑘)(2.5) = 𝟎. 𝟕𝟗𝟓 𝑽/𝝁𝒔
Question 3
Consider the oscillator circuit below. The circuit parameters are 𝑉𝑐𝑐 = 5𝑉, 𝑉𝑜 (𝑡) = 3𝑆𝑖𝑛(2𝜋𝑓𝑜 𝑡), 𝐶𝑛 =
5𝑝𝐹, 𝐶𝑜 = 50𝑝𝐹, 𝐶𝑝 = 0.02𝜇𝐹, 𝐶𝑞 = 100𝑝𝐹, 𝐶𝑟 = 100𝑝𝐹, 𝐿1 = 3𝜇𝐻, 𝑅1 = 1𝑀𝛺.
a) Derive an expression for the oscillating frequency, fo and the feedback factor, β(ω) where ω is
the radian frequency?
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑟𝑘, 𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑘 𝑚𝑢𝑠𝑡 𝑏𝑒
𝑧𝑒𝑟𝑜,𝑜𝑛 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝐿𝐶 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑝𝑢𝑟𝑒𝑙𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑤𝑖𝑡ℎ 𝑛𝑜 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠;
𝑋𝐶2 (𝜔0 ) + 𝑋𝐶𝑟 (𝜔𝑜 ) + 𝑋𝐿1 (𝜔𝑂 ) = 0 𝑤ℎ𝑒𝑟𝑒 𝑋𝐶2 (𝜔0 ) = 𝑋𝐶𝑜 (𝜔𝑜 )||𝑋𝐶𝑞 (𝜔𝑜 )
𝑋𝐶2 (𝜔0 ) + 𝑋𝐿1 (𝜔𝑂 ) = −𝑋𝐶𝑟 (𝜔𝑜 ) … … … … … … 1)
1 1
+ + 𝑗𝜔𝑜 𝐿1 = 0
𝑗𝜔𝑜 𝐶2 𝑗𝜔𝑜 𝐶𝑟
1 1 1
[ + ] = −𝑗𝜔𝑜 𝐿1
𝑗𝜔𝑜 𝐶2 𝐶𝑟
𝐶2 +𝐶𝑟 𝐶2 +𝐶𝑟 𝐶𝑜 + 𝐶𝑞 +𝐶𝑟 1
[ ] = 𝜔𝑜2 𝐿1 𝑏𝑢𝑡 = =
𝐶2 𝐶𝑟 𝐶2 𝐶𝑟 (𝐶𝑜 + 𝐶𝑞 )𝐶𝑟 𝐶𝑇
1 1
𝜔𝑜2 𝐿1 = 𝑡ℎ𝑢𝑠 𝜔 =
𝐶𝑇 √𝐿1 𝐶𝑇
𝟏
𝒇𝒐 =
𝟐𝝅√𝑳𝟏 𝑪𝑻
𝐼𝑔𝑛𝑜𝑟𝑖𝑛𝑔 𝑅2 𝑏𝑦 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑓𝑟𝑞𝑢𝑒𝑛𝑐𝑦, 𝑅2 ≫ 𝑋𝐶 2
𝑟𝑒𝑐𝑎𝑙𝑙 𝐶2 = (𝐶0 + 𝐶𝑞 )
𝑉𝑖𝑛 (𝜔𝑜 ) 𝑋𝐶2 (𝜔𝑜 )
𝑏𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛, 𝛽(𝜔𝑜 ) = =
𝑉𝑜 (𝜔𝑜 ) 𝑋𝐿1 (𝜔𝑜 ) + 𝑋𝐶2 (𝜔𝑜 )
𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1, 𝑋𝐶2 (𝜔0 ) + 𝑋𝐿1 (𝜔𝑂 ) = −𝑋𝐶𝑟 (𝜔𝑜 )
1
𝑗𝜔𝑜 𝐶2 𝐶𝑟
𝛽(𝜔𝑜 ) = =−
−1 𝐶2
𝑗𝜔𝑜 𝐶𝑟
𝑪𝒓
𝜷(𝝎𝒐 ) = −
𝑪𝒐 + 𝑪𝒒
b) Find the value of the closed loop gain, 𝐴𝑐𝑙 .
−1
1 100𝑝
𝐴𝑐𝑙 = = −[ ] = −𝟏. 𝟓
𝛽 50𝑝 + 100𝑝
c) What is the minimum slew rate of the op-amplifier, and find the values of 𝑅2 𝑎𝑛𝑑 𝑅3 ?
1
𝐺𝑒𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐶𝑇 = 0.06 𝑛𝐹, 𝑓𝑜 = = 11.863 𝑀𝐻𝑧
2𝜋√3𝜇 × 0.06𝑛
𝑆. 𝑅𝑚𝑖𝑛 = 2𝜋𝑓0 𝑉𝑝 = 2𝜋(11.863𝑀)(3) = 𝟐𝟐𝟑. 𝟔𝟏 𝑽/𝝁𝒔
𝑅1
𝑅2 =
𝐴𝑐𝑙
𝑅3 = 𝑅1 ||𝑅2
Question 4
Consider the oscillator circuit below, with the following component values
𝐶1 = 𝐶2 = 64 𝑝𝐹, 𝐿𝑠 = 529 𝑚𝐻. The output voltage is 𝑣𝑜 (𝑡)
a) Derive an expression for the feedback factor 𝛽(𝜔) where 𝜔 is the radian frequency.
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑟𝑘, 𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑘 𝑚𝑢𝑠𝑡 𝑏𝑒
𝑧𝑒𝑟𝑜,𝑜𝑛 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝐿𝐶 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑝𝑢𝑟𝑒𝑙𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑤𝑖𝑡ℎ 𝑛𝑜 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠;
𝑋𝐶1 (𝜔0 ) + 𝑋𝐶2 (𝜔𝑜 ) + 𝑋𝐿1 (𝜔𝑂 ) = 0
𝑋𝐶1 (𝜔0 ) + 𝑋𝐿𝑠 (𝜔𝑂 ) = −𝑋𝐶2 (𝜔𝑜 ) … … … … … … 1)
1 1
+ + 𝑗𝜔𝑜 𝐿𝑠 = 0
𝑗𝜔𝑜 𝐶1 𝑗𝜔𝑜 𝐶2
1 1 1
[ + ] = −𝑗𝜔𝑜 𝐿𝑠
𝑗𝜔𝑜 𝐶2 𝐶1
𝐶2 +𝐶1 𝐶2 +𝐶1 1
[ ] = 𝜔𝑜2 𝐿𝑠 𝑏𝑢𝑡 =
𝐶2 𝐶1 𝐶2 𝐶1 𝐶𝑇
1 1
𝜔𝑜2 𝐿𝑠 = 𝑡ℎ𝑢𝑠 𝜔𝑜 =
𝐶𝑇 √𝐿𝑠 𝐶𝑇
𝟏
𝒇𝒐 =
𝟐𝝅√𝑳𝒔 𝑪𝑻
𝑉𝑖𝑛 (𝜔𝑜 ) 𝑋𝐶1 (𝜔𝑜 )
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑎𝑛 𝑖𝑑𝑒𝑎𝑙 𝑜𝑝 𝑎𝑚𝑝, 𝑏𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛, 𝛽(𝜔𝑜 ) = =
𝑉𝑜 (𝜔𝑜 ) 𝑋𝐿𝑠 (𝜔𝑜 ) + 𝑋𝐶1 (𝜔𝑜 )
𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1,
1
𝑗𝜔𝑜 𝐶1 𝐶2
𝛽(𝜔𝑜 ) = =−
−1 𝐶1
𝑗𝜔𝑜 𝐶2
𝑪𝟐
𝜷(𝝎𝒐 ) = −
𝑪𝟏
b) What is the value of the oscillating frequency, 𝑓0 and the closed loop gain 𝐴𝑐𝑙
𝐺𝑒𝑡 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐶𝑇 = 32 𝑝𝐹
1
𝑓𝑜 = = 𝟑𝟖. 𝟔𝟖 𝑯𝒛
2𝜋√529𝑚 × 32𝑝
𝐶1
𝐴𝑐𝑙 = − = −𝟏
𝐶2
Question 5 (Hartley oscillator)
Consider the oscillator circuit below. The circuit parameters are 𝑉𝑐𝑐 = 5 𝑉, 𝐶𝑛 = 5 𝑝𝐹, 𝑉𝑜 (𝑡) =
3 sin 2𝜋𝑓𝑡 , 𝐶𝑜 = 100 𝑝𝐹, 𝐶𝑝 = 0.02 𝜇𝐹, 𝑡𝑎𝑝𝑝𝑒𝑑 𝑖𝑛𝑑𝑢𝑐𝑡𝑜𝑟, 𝐿1 = 90 𝜇𝐻. The inductor tap is 30% of the total
number of turns, 𝑅1 = 10 𝑘Ω. Assume mutual inductance is zero.
a) Derive an expression for the oscillating frequency, 𝑓𝑜 and the feedback factor 𝛽(𝜔) where 𝜔 is
the radian frequency.
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑟𝑘, 𝑎𝑡 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑡ℎ𝑒 𝑛𝑒𝑡 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝐿𝐶 𝑛𝑒𝑡𝑤𝑜𝑘 𝑚𝑢𝑠𝑡 𝑏𝑒
𝑧𝑒𝑟𝑜,𝑜𝑛 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝐿𝐶 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑝𝑢𝑟𝑒𝑙𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑣𝑒 𝑤𝑖𝑡ℎ 𝑛𝑜 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒
𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠;
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒; 𝐿 𝑇 = 𝐿′1 + 𝐿′′1 + 2𝑀 𝑤ℎ𝑒𝑟𝑒 𝑀 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑢𝑡𝑢𝑎𝑙 𝑖𝑛𝑑𝑢𝑐𝑡𝑎𝑛𝑐𝑒,
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑀 = 0, 𝐿 𝑇 = 𝐿′1 + 𝐿′′1
′ (𝜔 ) ′′
𝑋𝐶0 (𝜔0 ) + 𝑋𝐿1 𝑂 + 𝑋𝐿1 (𝜔𝑂 ) = 0
′ (𝜔 ) ′′
𝑤ℎ𝑒𝑟𝑒 𝑋𝐿1 (𝜔0 ) = 𝑋𝐿1 𝑜 + 𝑋𝐿1 (𝜔𝑜 ) 𝑎𝑠 𝑠ℎ𝑜𝑤𝑛 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑘𝑒𝑡𝑐ℎ.
𝑋𝐶0 (𝜔0 ) = −𝑋𝐿1 (𝜔𝑂 ) … … … … … … 1)
1
= −𝑗𝜔𝑜 𝐿1
𝑗𝜔𝑜 𝐶0
1 1
𝜔𝑜2 𝐿1 = 𝑡ℎ𝑢𝑠 𝜔0 =
𝐶0 √𝐿1 𝐶0
𝟏
𝒇𝒐 =
𝟐𝝅√𝑳𝟏 𝑪𝟎
𝐴𝑠𝑠𝑢𝑚𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑡ℎ𝑒 𝑟𝑒𝑠𝑜𝑛𝑎𝑛𝑡 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑅2 ≫ 𝑋𝐿′1
𝑉𝑖𝑛 (𝜔𝑜 ) 𝑋𝐿′1 (𝜔𝑜 )
𝐵𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛, 𝛽(𝜔𝑜 ) = =
𝑉𝑜 (𝜔𝑜 ) 𝑋𝐿′1 (𝜔𝑜 ) + 𝑋𝐶0 (𝜔𝑜 )
𝐹𝑟𝑜𝑚 𝑒𝑞𝑛 1, 𝑋𝐶0 (𝜔0 ) + 𝑋𝐿′1 (𝜔𝑂 ) = −𝑋𝐿′′1 (𝜔𝑜 )
𝑋𝐿′1 (𝜔𝑜 )
𝛽(𝜔𝑜 ) =
−𝑋𝐿′′1 (𝜔𝑜 )
𝑗𝜔𝐿′1
𝛽(𝜔𝑜 ) =
𝑗𝜔𝐿′′1
𝑳′𝟏
𝜷(𝝎𝒐 ) = −
𝑳′′
𝟏
b) Find the value of 𝑓𝑜 and the closed loop gain 𝐴𝑐𝑙
1 1
𝑓𝑜 = = = 1.6776 𝑀𝐻𝑧
2𝜋√𝐿1 𝐶0 2𝜋√90𝜇 × 100𝑝
𝐿′′1 90𝜇(70%) 7
𝐴𝑐𝑙 = − ′ =− = − = −𝟐. 𝟑𝟑𝟑
𝐿1 90𝜇(30%) 3
c) What is the minimum slew rate of the amplifier?
𝑆. 𝑅𝑚𝑖𝑛 = 2𝜋𝑓0 𝑉𝑝 = 2𝜋(1.6776𝑀)(3) = 𝟑𝟏. 𝟔𝟐 𝑽/𝝁𝒔
d) Find the values of 𝑅2 𝑎𝑛𝑑 𝑅3
𝑅1 7 3
= 𝐻𝑒𝑛𝑐𝑒 𝑅2 = (10𝑘) = 𝟒. 𝟑 𝒌𝛀
𝑅2 3 7
(4.3𝑘 × 10𝑘)
𝑅3 = = 𝟑 𝒌𝛀
4.3𝑘 + 10𝑘
Question 6
a) Give the functions of a piezoelectric crystal in LC oscillators.
It is a frequency selective element.
Improves quality factor and frequency stability in parallel resonant circuits of oscillators.
b) What is the Barkhausen criterion and why is it important?
It states that for a circuit with open loop gain A and feedback factor β to oscillate, the total loop
gain Aβ must be unity and the total phase shift must be ±360𝑛 where n is an integer including
zero at the frequency of oscillation
It is used to determine whether the linear electronic circuit will oscillate or not.
c) Give one reason why inductors are not used to build audio frequency range oscillators.
At low frequencies, in audio range, inductors are large, heavy and expensive to design
d) .Give one advantage of using positive feedback in an application.
Positive feedback is applied in oscillators since loop gain gradually builds up oscillations.
e) Give reasons why it is impossible to build an ideal (brick wall) analog filter in practice.
Because filters have exactly flat bands, zero gain and zero width transitional bands hence very
hard to design such filters using existing technologies.
f) What is the function of LC oscillator in a clapp oscillator circuit?
It provides necessary feedback for oscillations to occur.
It is responsible for determining the frequency of oscillations in the clapp oscillator.
