The multivibratores are used to produce two states circuits such as bisatle, monostable and astable.

1. Multivibrator
BY DR MOHAMMAD RASHTIAN
CIVIL AVIATION TECHNOLOGY COLLEGE
6:34 PM 1

2. K2A1 K1 A2
Positive Feedback:
If K1 and K2 are:
1- Simple Resistor Networks  Bi-stable
2- One RC network and the other
one a resistor network  Mono-stable ( Timer )
3- Both RC Networks  A-stable ( Pulse generator )
A1A2K1K2 > 1 A1,A2 < 0 A1 and A2 : Amplifies
6:34 PM 2

3. Bi-stable
Q1 Q2
RC1
RB1
RB2
RC2
VCC
6:34 PM 3

4. Q2Q1
RC1=2k
RB2=20k
RC2=2k
RB1=20k
C1 C2
VCC=12V
VO1 VO2
VBE(ON) = 0.6V
VCESat = 0.1V
βmin = ?
Bi-stable
6:34 PM 4

5. Q2Q1
RC1=2k
RB2=20k
RC2=2k
RB1=20k
C1 C2
VCC=12V
VO1 VO2
Assumption : Q1 : Off , Q2: On
VCC
RB2=20k
RC1=2k
RB1=20k
Rc2=2k
c1
e1
b2
VCC
b1
off
e2
VCESAT=0.1V
IB2
IC2
IC2 =
VCC − VCEsat
RC
=
12 − 0.1
2k
= 5.95mA IB2 =
VCC−VBE(ON)
RC+RB
=
12−0.6
2k+20k
= 0.51mA
βmin =
IC
IB
=
5.95
0.51
= 11.66 VC1 = Vcc
RB2
RB2 + RC1
+ VD
RC1
RB2 + RC1
= 10.96
6:34 PM 5

6. VBE(ON) = 0.6V
VCESat = 0.1V
βmin = ?
VCC=12V
Q2Q1
RC1=2k
RB2=20k
RC2=2k
RB1=20k
C1 C2
VO1 VO2
RB3=100k
RB4=100k
V=-2
`
6:34 PM 6

7. RB2=20k
RC1=2k
RB1=20k
Rc2=2k
VCC
c1
e1
b2
VCC
b1
off
e2
R=100k
V=-2
VCESAT=0.1V
R=100k
V=-2
Assumption : Q1 : Off , Q2: On
VB1 = −2 ∗
20
120
+ 0.1 ∗
100
120
= −0.25V
IB2 =
12 − 0.6
20k + 2k
−
0.6 − −2
100k
= 0.518mA − 0.026mA = 0.492mA
IC2 =
12 − 0.1
2k
−
0.1 − −2
100k + 20k
= 5.93mA
5.93
0.492
= 12.05βmin =
VCC=12V
Q2Q1
RC1=2k
RB2=20k
RC2=2k
RB1=20k
C1 C2
VO1 VO2
RB3=100k
RB4=100k
V=-2
`
6:34 PM 7

8. SQ2Q1
RC1
RB2
RC2
RB1
C1 C2
VCC
Q3 Q4
Q Q
R
Bi-Stable Usages: 1- RS Flip Flop
6:34 PM 8

9. Bi-Stable Usages: 2- T Flip Flop
Q2Q1
RC1
RB2
RC2
RB1
V=0
VCC
V 0
R*
VCC VCC
0
First Assumption : Q1: on , Q2: off
Vcc
Vcc – 0.7
- +
Before
Trigger t
V 0
V = 0.7 V 0
R* has a small value
6:34 PM 9

10. VCC
Q2Q1
RC1
RB2
RC2
RB1
V=0 - +
V 0
R*
VCC VCC
0
V=0
Vcc – 0.7
t
During
Trigger
V=0
V 0V -Vcc + 0.7
Bi-Stable Usages: 2- T Flip Flop
6:34 PM 10

11. Bi-Stable Usages: 2- T Flip Flop
VCC
Q2Q1
RC1
RB2
RC2
RB1
V=0
V 0
R*
VCC VCC
0
V 0
t
After
Trigger
Vcc
V = 0.7V 0
Vcc – 0.7
+ -
Trigger result : Q1: off , Q2: on
6:34 PM 11

12. Bi-Stable Usages: 2- T Flip Flop 2
Q2Q1
RC1
RB2
RC2
RB1
V=0
VCC
D1 D2
D3
RsVCC VCC
0
First Assumption : Q1: on , Q2: off
Before
Trigger t
V 0 Vcc
Vcc – 0.7
- +
V = 0.7 V 0
V 0
6:34 PM 12

13. Bi-Stable Usages: 2- T Flip Flop 2
VCC VCC
0
During
Trigger
t
Q2Q1
RC1
RB2
RC2
RB1
V=0
VCC
D1 D2
D3
Rs
V 0 0.7
Vcc – 0.7
- +
V 1.4 - Vcc V 0
V 0
6:34 PM 13

14. VCC VCC
0
After
Trigger
t
Q2Q1
RC1 RB2
RC2
RB1
V=0
VCC
D1 D2
D3
Rs
V = 0.7
V 0
Vcc
+ -
V 0
V 0
Vcc
Trigger result : Q1: off , Q2: on
Bi-Stable Usages: 2- T Flip Flop 2
6:34 PM 14

15. Bi-Stable Usages: 2- T Flip Flop 3 ( Base Triggered )
Q2Q1
RC1
RB
RC2
RB
V 0
RB2
RB2
-VB
Vcc
RS
RS
D1 D2
CS CS
VCC VCC
0
Before
Trigger
t
First Assumption : Q1: on , Q2: off
_
Vcc
+
V 0
V 0 Vcc
Vcc-0.7
- +
0.7V V 0
6:34 PM 15

16. Bi-Stable Usages: 2- T Flip Flop 3 ( Base Triggered )
Q2Q1
RC1
RB
RC2
RB
RB2
RB2
-VB
Vcc
RS
RS
D1 D2
CS CS
VCC VCC
0
After
Trigger
t
V 0
V 0
V 0
V 0Vcc
0.7V
Vcc-0.7
+ -
_
Vcc
+
6:34 PM 16

17. Schmitt trigger
5m
Unideal
Water Level
Ideal Water
Level = 5m
5.05m
4.95m
Unideal
Water Level
6:34 PM 17

18. Q1 Q2
RC1=2k RC2=2k
RB2=10k
RB1=10k
C=50PF
2mA
Vcc = 15V
Vin
Vout
VBE
IC
0.5 0.7
6:34 PM 18

19. Assumption : Q1 : Off , Q2: On
Q1 Q2
RC1=2k RC2=2k
RB2=10k
RB1=10k
C=50PF
2mA
Vcc = 15V
Vin
Vout
Q2
RC1=2k
RB2=10k
RB1=10k
RC2=2k
2mA
Q1: off
VE2
Vout
VC1
VE1=VE2
+15+15
VE1 = VE2 Vout = VCC − 2 ∗ IEE = 11V
if Vin = 0.5 + VE1 ⇒ condition change if Vin > 6.61V ⇒ Q1 ∶ On , Q2 ∶ Off
UTP : Upper Threshold Point = 6.61V
⇒ VE2 = 6.81 − 0.7 = 6.11V = VE1VB2 = VCC ∗
RB1
RB1 + RB2 + RC1
= 15 ∗
10
10 + 10 + 2
= 6.81V
6:34 PM 19

20. New condition : Q1 : On , Q2: Off
2mA
RC1=2k
RB2=10k
RB1=10k
+15
VC1
VB2
15 − VC1
2k
= 2mA +
VC1
20k
⇒ VC1 = 10V
VE2 = VE1 = Vin − 0.7
Question: For which Input Voltage (Vin) Q2 starts to turn on?
Vin = 5 + 0.7 − 0.5 = 5.2 = LTP: Lower Threshold Point
Q1 Q2
RC1=2k RC2=2k
RB2=10k
RB1=10k
C=50PF
2mA
Vcc = 15V
Vin
Vout
VB2 =
𝑅 𝐵1
𝑅 𝐵1 + 𝑅 𝐵2
∗ 𝑉𝐶1
IC2 = 0 ⇒ Vout = 15V = VCC
VB2 = VC1 ∗
RB1
RB1 + RB2
= 10 ∗
1
2
= 5V VE2 = Vin − 0.7
VBE2 = VB2 − VE2 = 5 − Vin − 0.7 if VBE2 = 0.5, Q2 is on threshold
𝑉𝐵2 = 5V & VE2 = 6.11 ⇒ Q2 ∶ Off
6:34 PM 20

21. 11
15
5.2 6.61
Vout
Vin
LTP = 5.2 & UTP = 6.61
6:34 PM 21

22. Trimming LTP and UTP
2mA
RC1=2k
RB2=10k
RB1=10k
+15
VB2
Q1
RE1
e2
-RE1IEE+
Vin
Q1 Q2
RC1=2k RC2=2k
RB2=10k
RB1=10k
C=50PF
2mA
Vcc = 15V
Vin
Vout
RE1
Vin
Q1 Q2
RC1=2k RC2=2k
RB2=10k
RB1=10k
C=50PF
2mA
Vcc = 15V
Vout
RE2
RE1
6:34 PM 22

23. Trimming LTP and UTP
2mA
RC1=2k
RB2=10k
RB1=10k
+15
VB2
Q1
RE1
e2
-RE1IEE+
Vin
By Adjusting RE1 , LTP could be trimmed accurately
2mA
RC2=2k
+15
Q2
RE2
e1
-RE2IEE+
VBE1 = 0.5 ⇒ state change
Vin − VB2 − 0.7 − RE2IE2 = 0.5
LTP UTPLTP UTP
VB2 − VE2 = 0.5 ⇒ VB2 − LTP′
− 0.7 − RE1IE1 = 0.5
LTP′ = VB2 + 0.7 − 0.5 + RE1IE1 = LTP + RE1IE1
UTP′ = UTP − RE2IE2
UTP = VB2 − 0.7 + 0.5 = VB2 − 0.2
UTP′
= VB2 − 0.2 − RE2IE2
6:34 PM 23

24. Using REE instead of Current Source
Q1: off ⇒ VB2 = 15 ∗
10
10 + 10 + 2
= 6.81
⇒ VE2 = 6.11 and IE2 =
6.11
3
= 2.03mA
⇒ UTP = 0.5 + 6.11 = 6.61V
VOUT = Vcc − RC2
Vb2 − 0.7
RE
= 10.94
VE2 = VE1 = Vin − 0.7
𝑉𝑐𝑐 − 𝑉𝐶1
𝑅 𝐶1
= 𝛼
𝑉𝑖𝑛 − 0.7
𝑅 𝐸𝐸
+
𝑉𝐶1
𝑅 𝐵1 + 𝑅 𝐵2
& VB2 = VC1 ∗
10
10 + 10 + 2
LTP: VE2 = 𝑉𝐸1 = 𝑉𝑖𝑛 − 0.7; 𝑉𝐵𝐸2 = 0.5 = 𝑉𝐵2 − 𝑉in − 0.7 ⇒ Vin = LTP
Q1
REE=3k
RC1=2k
RB2=10k
RB1=10k
Vcc = 15V
Vin
Now inspecting the case of Q1: On , Q2: Off
VC1=14.6-0.606Vin; VB2=7.3-0.303Vin;=> 7.3-0.303Vin-(Vin-0.7)=0.5
=> LTP=5.76V
6:34 PM 24

25. Triangle Pulse Generator
6:34 PM 25

26. 6:34 PM 26
Triangle Wave Generator

27. Mono-Stable (Timer)
Q2Q1
RC1=2k
RB2=10k
RC2=2k
10uF
VCC=12V
10k
15k
-5
VCC VCC
0
6:34 PM 27

28. Mono-Stable (Timer)
Q2Q1
RC1=2k
RB2=10k
RC2=2k
10uF
VCC=12V
10k
15k
-5
12V
10k
2k
c1
e1
15k
-5
ic1
B2
2k
12V
10k
12V
- Vc +
iB1
B1
E1
VB2 = −5 ∗
10
10 + 15
= −2V IC1 =
12
2k
−
5
25k
= 5.8mA & IB1 =
12 − 0.7
10k
IB1 = 1.13mA βmin =
5.8
1.13
= 5.13 VC 0−
= 12 − 0.7 = 11.3
6:34 PM 28
In Stable Mode, Q1 Is saturated & Q2 Is Off
Pre requistion for saturating the Q1

29. Q2Q1
RC1=2k
RB2=10k
RC2=2k
10uF
VCC=12V
10k
15k
-5
Equivalant Circuit just After Trigger:
IC2MAX =
12 − (−11.3)
10k
+
12
2k
= 8.33mA
βmin =
8.33
0.56
= 14.87
VC1 = 12 − 2 ∗
12 − 0.7
12k
= 10.1V
IB2 =
12 − 0.7
12
−
5.7
15
= 0.56mA
10k
2k
2k
12
c1
e1
b2
12
e2
15k
-5
10k
10uF c2
e2
12
iB2
VB1
6:34 PM 29

30. T
12
0.7
-11.3
-2
0.7
12
0.7
-2
10.1
0.5
0.9
0.4
0.4
RCCB
Recovery
time
VC1
VC2
VB1
VB2
6:34 PM 30
Q2Q1
RC1=2k
RB2=10k
RC2=2k
10uF
VCC=12V
10k
15k
-5

31. VB1 = ? & VB1 0+
= −VCC + VBE(ON)
VB1 ∞ = VCC
VB1 = VCC + −VCC + VBE ON − VCC e
−
t
RBCB
VB1|t=T = 0.5V = Vγ
−2VCC + VBE ON e
−
T
RBCB = −VCC + Vγ
Calculating period time:
10k
2k
2k
12
c1
e1
b2
12
e2
15k
-5
10k
10uF c2
e2
12
iB1
VB1
e−
T
τ =
−VCC+Vγ
−2VCC+VBE(ON)
=
VCC−Vγ
2VCC−VBE(ON)
⇒ T = −τ ln
VCC−Vγ
2VCC−VBE(ON)
≅ −τ ln
1
2
= τ ln 2
T = 100ms ∗ ln
24 − 0.7
12 − 0.5
≅ 70ms
* Max Reverse VBE is practically about less than 7 Volts
6:34 PM 31

32. Emitter Coupled Mono-Stable (Timer)
VBE2 ON = 0.7 V , VBE1 ON = 0.6 V , Vγ = 0.5 V , VCEsat = 0.1 V , β = 100
6:34 PM 32

33. Assuming Q1: Off , Q2:On
−12 + 50IB2 + 0.7 + β + 1 IB2 ∗ 1k = 0
151IB2 = 11.3 ⇒ IB2 = 0.075mA
IC2 = 7.5mA ⇒ VB2 = 12 − 50 ∗ 0.075 = 8.25V
VE = 7.55V; VC = 12 − 1k ∗ 7.5mA = 4.5V
Since VCE < 0 then Q2 must be saturated and we must recalculate
6:34 PM 33

34. Saturation model for Q2:
1212
0.1V0.7V
1k
1k
50k
VC2
VE2
VE2 = 12 ∗
1
1k + 1k ∥ 50k
− 0.7 ∗
1k ∥ 1k
1k ∥ 1k + 50k
− 0.1 ∗
1k ∥ 50k
1k ∥ 50k + 1k
≅ 6𝑉
VB1 = 12 ∗
4
4 + 8
= +4 ⇒ VBE1 = −2V ⇒ Q1: Off and Q2 ∶ Saturated
, VCap = VC1 − VB2 , VC1 = 12V ; VC2 = 6.1V
VCap = 12 − 6.7 = 5.3V ,
6:34 PM 34

35. After Trigger: Q1: On , Q2: Off
VB1 = 4V ⇒ VE = 3.4V ⇒ IE1 = 3.4mA ≅ IC1
𝑉𝐶1 0+
=?
Right After Trigger:
12 − VC1
2
+
12 − (VC1 − 5.3)
50k
= 3.4mA ⇒ VC1 0+ = 5.66V
⇒ VB2 0+
= 5.66 − 5.3 = 0.36V ⇒ Q2: Off
VC1SS = 12 − 2 ∗ 3.4 = 5.2V , VB2 ∞ = 12
VC1 t = 5.2 + 5.66 − 5.2 e−
t
τ = 5.2 + 0.46e−
t
τ
VB2 t = 12 + 0.36 − 12 e−
t
τ = 12 − 11.64e−
t
τ
τ = (RB + RC1)∗C = 52k∗1uF = 52mS , VE2 = VE1 = 3.4
if VB2 = 0.5 + VE2 => Due to the positive feedback => Q2: On
VB2|t=t1 = 3.9 , 12 − 11.64e
−
t1
52mS = 3.9 ⇒ t1 = 18.85mS
𝑉𝐶1 𝑡 = 18.85𝑚𝑆−
= 5.2 + 0.46e
−
t1
52 = 5.52𝑉 , 𝑉𝐶𝑎𝑝 𝑡 = 18.85𝑚𝑆−
= 5.52 − 3.9 = 1.62
6:34 PM 35

36. Recovery phase : t > 18.85mS t = 18.85mS+ Q2: On , Q1: Off
1212
0.1V
0.7V
1k
1k
50k
VC2
VE2
1.62
2k
12
VC1
VB2
12 − 0.1 − VE2
1k
+
12 − 0.7 − VE2
50k
+
12 − 1.62 − 0.7 − VE2
2k
=
VE2
1k
VE2 = 6.73V ⇒ VC2 = 6.83V VB2 = 7.43V , VC1 = 9.05V
τRecovery = C RC1 + (RC2 ∥ RE ∥ RB)
6:34 PM 36

37. VC1
VB2
VC2
τ2
τ2
τ2
τ1
12
6.7
6.1
12
0.36
5.66 5.52
9.05
12
6.7
6.1
3.9
7.43
6.83
t=0
18.85mS
6:34 PM 37

38. Trimming the circuit for Decreasing Recovery Time
6:34 PM 38

39. A-Stable
Such Circuit Would Produce a symmetrical periodic output
6:34 PM 39

40. e1
12k 1k
- C2 +
+6+6
b1
e2
c2
+6
1k
c1
e1
12k
+6
ic1
B2
+ C1 -
Assumption: Q1: Sat , Q2: Off
VCap1 0+ = VCap2 ∞ = 5.3V
VB2 0+
= −VCap1 0+
= −5.3V
VB2SS = VCC = 6
VB2 t = VCC + −VCC + 0.7 − VCC e−
t
τ
VB2 t = t1 = 0.5 ⇒
VCC − 0.5
2VCC − 0.7
= e−
t
τ
t1 = τ ln
2VCC − 0.7
VCC − 0.5
t1 ≅ τ ln 2 , τ = CRB
τ =12mS , t1=12ms ∗ ln
12 − 0.7
6 − 0.5
= 8.65mS
τ2 = C∗RC = 1uF ∗ 1k = 1ms , VCap2 0+ = Last Voltage of previous state
VCap2 0+
= −0.5V , VC2 = VB1 0+
− 0.5 , VB1 0+
= 0.9 , VC2 0+
= 0.4V
6:34 PM 40

41. VC1
VC2
VB1
VB2
6 6
0.7 0.7
0.70.7
0.9
0.9
0.9
0.4
0.4
0.4
0.5
0.5
-5.3
-5.3-5.3
6:34 PM 41

42. What is minimum β?
5.6mA
0.44mA
6.94mA
6.45mA
ic1
ib1
IB1 =
6 − VB1
12k
+ IC
variable
IBmin =
6 − 0.7
12 𝑘
= 0.44 mA
IBMax =
6 − 0.9
12
+
6 − −0.5 − 0.9
1k
= 6.11mA
IC1 0+
=
6 − 0
1k
+
6 − (−5.3)
12k
= 6.94mA
IC1 8.04mS =
6
1k
+
6 − 0.5
12k
= 6.45mA
βmin =
ICMAX
IBmin
=
6.94
0.44
= 15.77
ib1
e1
12k 1k
- C2 +
+6+6
b1
e2
c2
+6
1k
c1
e1
12k
+6
ic1
B2
+ C1 -
6:34 PM 42

43. Solving the output exponential form
Q2Q1
1k
1k
VCC=6V
12k12k1k 1k
6:34 PM 43

44. Emitter Coupled Multi-vibrator
VBE ON = 0.7 V , Vγ = 0.5 V , β = 200
Obviously Q1 & Q2 are Current Sources
IC1&C2 =
2.7 − 0.7
4k
= 0.5mA
*Positive Feedback is in place, so Q3 & Q4
cannot be on simultaneously.
6:34 PM 44

45. Q2Q1
Q4
Q3
10nF
4k 4k
1.2k4k
+12V+10V
2.7V
Vout
+5V
A B
- +
Assumption: Q4: Off , Q3: On
IE4 = 1mA ⇒ VC4 = 12 − 1.2 = 10.8V
IB4 = 50uA ⇒ VB4 = 10 − 4k ∗ 50u ≅ 10V ⇒ VE4 = 9.3V = VB
in this state voltage of node A is linearly decreasing because of the current source Q1, Once VA
reaches 4.5 Volts, Q3 will be at threshold to turn on. Now we consider the state change: Q3:
On , Q4: Off VC3 = 10 − 4 ∗ 0.5 + 0.5 ≅ 6V
in this state voltage of node B will fall until VB Equals 5.5 then Q4 will be at threshold to turn on:
VCap t1−
= VB t1−
− VA t1−
= 5.5 − 4.3 = 1.2
VA t1+
= VB t1+
− VCap t1+
, VCap t1+
= VCap t1−
VA t1+
= 9.3 − 1.2 = 8.1V
VB4 = VC3 = 10 − 4 0.5 + 0.5 = 6V
Q3: Off , Q4: On
+10
Q3
4k
+5
VB4
A B
0.5mA0.5mA
Vout
Q4
1.2k
+10
4k
+10
A B
0.5mA
0.5mA
6:34 PM 45

46. VCap
4.8V
1.2V
T
C∆V = I∆T ⇒ 10 ∗ 10−9
∗ 4.8 − 1.2 = 0.5 ∗ 10−3
∗
T
2
⇒
T
2
= 72uS ⇒ f =
1
2 ∗ 72uS
f =
1000kHz
2 ∗ 72
= 6.94kHz
6:34 PM 46

47. Q2Q1
RR
A AC1
I2I1
IXIX
Q4Q3
DD
VCC
VOB VOB
Emitter Coupled Multi-vibrator
Q1 and Q2 cant be turned on
Simultaneously.
Suppose that Q1:off Q2:on
VB3 ≅ VCC
VOB = Ve3 = VCC − VBE on
VE2 = VCC − 2 ∗ VBE(on)
VB4 = VCC − VBE(on)
VOB′ = VCC − 2 ∗ VBE on ⇒
if VE1 == VCC − 3 ∗ VBE on
then the state will change
φ ≡ VBE on
6:34 PM 47

48. VA -VA
VA
VA
VB
VB
Q2:ON
Q1:ON
VCC-Ψ
VCC-2Ψ
VCC-Ψ
VCC-2Ψ
VCC-3Ψ
VCC-2Ψ
VCC-Ψ
VCC-3Ψ
VCC-2Ψ
VCC-Ψ
-Ψ
Ψ
Q2Q1
RR
A AC1
I2I1
IXIX
Q4Q3
DD
VCC
VOB VOB
Q1 and Q2 cant be turned on
Simultaneously.
VB3 ≅ VCC
VOB = Ve3 = VCC − VBE on
VE2 = VCC − 2 ∗ VBE(on)
VB4 = VCC − VBE(on)
t1 = t2 =
C∆V
I1
=
2CVBE(on)
I1
⇒ f0 =
1
t
=
1
2 ∗ t1
=
I1
4C1VBE(on)
VOB′ = VCC − 2 ∗ VBE on ⇒ if VE1 == VCC − 3 ∗ VBE on
then the state will change
φ ≡ VBE on
The frequency is a function of VBE which itself is
temperature dependent .
6:34 PM 48

49. 𝐾𝑉𝐿: −𝑉𝐶 + 𝑉𝐵𝐸5 + 𝑅1 𝐼1 − 𝑅1 𝐼2 − 𝑉𝐵𝐸7 = 0
∆𝐼 = 𝐼1 − 𝐼2 ≅
𝑉𝐶 − ∆𝑉𝐵𝐸
𝑅1
≅
𝑉𝐶
𝑅1
①
2𝐼1 + 2𝐼2 = 𝐼0 ⇒ 𝐼1 + 𝐼2 =
𝐼0
2
②
①&② ⇒ 𝐼1 =
𝐼0
4
+
𝑉𝐶
2𝑅1
𝐼0 =
𝑉𝐵𝐸10
𝑅2
⇒ 𝐼1 =
𝑉𝐵𝐸10
4𝑅2
+
𝑉𝐶
2𝑅1
𝑓0 =
𝐼1
4𝑉𝐵𝐸 𝐶
=
𝑉𝐵𝐸
4𝑅2
+
𝑉𝐶
2𝑅1
4𝑉𝐵𝐸 𝐶
𝑓0 =
1
16𝐶1 𝑅2
1 +
2𝑅2
𝑅1 𝑉𝐵𝐸(𝑜𝑛)
𝑉𝐶
The circuit’s frequency will
be completely temperature
independent only at VC=0
6:34 PM 49

50. 6:34 PM 50