Child-Langmuir Law
edit
I
a
{\displaystyle I_{a}}
S
{\displaystyle S}
d
{\displaystyle d}
V
a
{\displaystyle V_{a}}
potential difference from anode to cathode.
P
{\displaystyle P}
perveance of the device.
I
a
=
(
4
ϵ
0
9
2
e
m
e
)
(
S
d
2
)
V
a
3
2
{\displaystyle I_{a}=\left({\frac {4\epsilon _{0}}{9}}{\sqrt {\frac {2e}{m_{e}}}}\right)\left({\frac {S}{d^{2}}}\right)V_{a}^{\frac {3}{2}}}
I
a
=
k
1
(
S
d
2
)
V
a
3
2
{\displaystyle I_{a}=k_{1}\left({\frac {S}{d^{2}}}\right)V_{a}^{\frac {3}{2}}}
I
a
=
P
V
a
3
2
{\displaystyle I_{a}=PV_{a}^{\frac {3}{2}}}
I
a
∝
V
a
3
2
{\displaystyle I_{a}\propto V_{a}^{\frac {3}{2}}}
P
∝
S
d
2
{\displaystyle P\propto {\frac {S}{d^{2}}}}
Triode equation (and rearrangements thereof)
edit
μ
{\displaystyle \mu }
I
a
=
P
(
V
a
μ
+
V
g
)
3
2
{\displaystyle I_{a}=P\left({\frac {V_{a}}{\mu }}+V_{g}\right)^{\frac {3}{2}}}
V
a
=
μ
(
(
I
a
P
)
2
3
−
V
g
)
{\displaystyle V_{a}=\mu \left(\left({\frac {I_{a}}{P}}\right)^{\frac {2}{3}}-V_{g}\right)}
V
g
=
(
I
a
P
)
2
3
−
V
a
μ
{\displaystyle V_{g}=\left({\frac {I_{a}}{P}}\right)^{\frac {2}{3}}-{\frac {V_{a}}{\mu }}}
Derivatives of triode equation
edit
r
a
{\displaystyle r_{a}}
g
m
{\displaystyle g_{m}}
r
a
=
μ
g
m
{\displaystyle r_{a}={\frac {\mu }{g_{m}}}}
μ
=
d
V
a
d
V
g
{\displaystyle \mu ={dV_{a} \over dV_{g}}}
g
m
=
d
I
a
d
V
g
=
3
2
P
2
I
a
3
{\displaystyle g_{m}={dI_{a} \over dV_{g}}={\frac {3}{2}}\ {\sqrt[{3}]{P^{2}I_{a}}}}
r
a
=
d
V
a
d
I
a
=
2
3
μ
P
2
I
a
3
{\displaystyle r_{a}={dV_{a} \over dI_{a}}={\frac {2}{3}}\ {\frac {\mu }{\sqrt[{3}]{P^{2}I_{a}}}}}
DC to DC converters - conversion ratios
edit
Buck
Boost
Buck-boost (inverting)
Buck-boost (non-inverting)
D
=
V
V
g
{\displaystyle D={\frac {V}{V_{g}}}}
D
=
V
−
V
g
V
{\displaystyle D={\frac {V-V_{g}}{V}}}
D
=
−
V
V
g
−
V
{\displaystyle D={\frac {-V}{V_{g}-V}}}
D
=
V
V
g
+
V
{\displaystyle D={\frac {V}{V_{g}+V}}}
M
(
D
)
=
V
V
g
=
D
{\displaystyle M(D)={\frac {V}{V_{g}}}=D}
M
(
D
)
=
V
V
g
=
1
1
−
D
{\displaystyle M(D)={\frac {V}{V_{g}}}={\frac {1}{1-D}}}
M
(
D
)
=
V
V
g
=
−
D
1
−
D
{\displaystyle M(D)={\frac {V}{V_{g}}}={\frac {-D}{1-D}}}
M
(
D
)
=
V
V
g
=
D
1
−
D
{\displaystyle M(D)={\frac {V}{V_{g}}}={\frac {D}{1-D}}}
d
d
D
M
(
D
)
=
1
{\displaystyle {\frac {d}{dD}}M(D)=1}
d
d
D
M
(
D
)
=
1
(
1
−
D
)
2
{\displaystyle {\frac {d}{dD}}M(D)={\frac {1}{(1-D)^{2}}}}
d
d
D
M
(
D
)
=
−
1
(
D
−
1
)
2
{\displaystyle {\frac {d}{dD}}M(D)={\frac {-1}{(D-1)^{2}}}}
d
d
D
M
(
D
)
=
1
(
1
−
D
)
2
{\displaystyle {\frac {d}{dD}}M(D)={\frac {1}{(1-D)^{2}}}}
Type II Compensator transfer function
edit
H
(
s
)
=
V
e
V
(
s
)
=
−
R
3
C
1
s
+
1
[
R
1
(
C
1
+
C
2
)
s
]
(
R
3
C
1
C
2
C
1
+
C
2
s
+
1
)
{\displaystyle H(s)={\frac {V_{e}}{V}}(s)=-{\frac {R_{3}C_{1}s+1}{[R_{1}(C_{1}+C_{2})s](R_{3}{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}s+1)}}}
H
(
s
)
=
V
e
V
(
s
)
≈
−
R
3
C
1
s
+
1
(
R
1
C
1
s
)
(
R
3
C
2
s
+
1
)
{\displaystyle H(s)={\frac {V_{e}}{V}}(s)\approx -{\frac {R_{3}C_{1}s+1}{(R_{1}C_{1}s)(R_{3}C_{2}s+1)}}}
C
2
≪
C
1
{\displaystyle C_{2}\ll C_{1}}
Type III Compensator transfer function
edit
H
(
s
)
=
V
e
V
(
s
)
=
−
(
R
3
C
1
s
+
1
)
[
C
3
(
R
1
+
R
4
)
s
+
1
]
[
R
1
(
C
1
+
C
2
)
s
]
(
R
3
C
1
C
2
C
1
+
C
2
s
+
1
)
(
R
4
C
3
s
+
1
)
{\displaystyle H(s)={\frac {V_{e}}{V}}(s)=-{\frac {(R_{3}C_{1}s+1)[C_{3}(R_{1}+R_{4})s+1]}{[R_{1}(C_{1}+C_{2})s](R_{3}{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}s+1)(R_{4}C_{3}s+1)}}}
H
(
s
)
=
V
e
V
(
s
)
≈
−
(
R
3
C
1
s
+
1
)
[
R
1
C
3
s
+
1
]
(
R
1
C
1
s
)
(
R
3
C
2
s
+
1
)
(
R
4
C
3
s
+
1
)
{\displaystyle H(s)={\frac {V_{e}}{V}}(s)\approx -{\frac {(R_{3}C_{1}s+1)[R_{1}C_{3}s+1]}{(R_{1}C_{1}s)(R_{3}C_{2}s+1)(R_{4}C_{3}s+1)}}}
C
2
≪
C
1
{\displaystyle C_{2}\ll C_{1}}
R
4
≪
R
1
{\displaystyle R_{4}\ll R_{1}}
Formula for R3
edit
For type III compensation,
R
3
=
F
c
F
l
c
V
o
s
c
V
i
n
R
1
{\displaystyle R_{3}={\frac {F_{c}}{F_{lc}}}{\frac {V_{osc}}{V_{in}}}R_{1}}
Voltage Feedforward
edit
The following formulae are for one particular voltage feedforward circuit which uses a bias winding to charge a capacitor through a resistor to create the ramp proportional to Vbias, which is proportional to Vsec
V
o
s
c
=
N
b
2
N
s
V
s
e
c
(
1
−
e
−
1
F
s
R
C
)
{\displaystyle V_{osc}={\frac {N_{b2}}{N_{s}}}V_{sec}\left(1-e^{\frac {-1}{F_{s}RC}}\right)}
V
s
e
c
V
o
s
c
=
N
s
N
b
2
(
1
−
e
−
1
F
s
R
C
)
{\displaystyle {\frac {V_{sec}}{V_{osc}}}={\frac {N_{s}}{N_{b2}\left(1-e^{\frac {-1}{F_{s}RC}}\right)}}}
Find RC for minimum acceptable Vosc at minimum Vpri (input voltage).
R
C
=
−
1
F
s
l
n
(
1
−
N
p
N
b
2
V
o
s
c
V
p
r
i
)
{\displaystyle RC={\frac {-1}{F_{s}ln\left(1-{\frac {N_{p}}{N_{b2}}}{\frac {V_{osc}}{V_{pri}}}\right)}}}
Unfortunately any Offset voltage introduces an error:
Vosc=Nb2/Ns * Vsec * [1-e^(-1/FsRC)] + Voffset
Vsec/Vosc = Vsec/ [Nb2/Ns * Vsec * [1-e^(-1/FsRC)] + Voffset]
Vsec/Vosc = Vsec/ Vsec[Nb2/Ns * [1-e^(-1/FsRC)] + Voffset/Vsec]
Vsec/Vosc = 1/ [Nb2/Ns * [1-e^(-1/FsRC)] + Voffset/Vsec]
Offset error keeps Vsec in the equation, Vsec proportional to Vin, so PWM comparator gain is a function of input voltage.
From Summing Amplifier equation, with V2 =0V and IN+=Vref .
V
o
u
t
=
−
R
3
(
V
i
n
−
V
r
e
f
R
1
−
V
r
e
f
R
2
)
+
V
r
e
f
{\displaystyle V_{out}=-R_{3}\left({\frac {V_{in}-V_{ref}}{R_{1}}}-{\frac {V_{ref}}{R_{2}}}\right)+V_{ref}}
V
o
u
t
=
V
r
e
f
(
R
3
R
2
+
R
3
R
1
+
1
)
−
R
3
R
1
V
i
n
{\displaystyle V_{out}=V_{ref}\left({\frac {R_{3}}{R_{2}}}+{\frac {R_{3}}{R_{1}}}+1\right)-{\frac {R_{3}}{R_{1}}}V_{in}}
For the target condition, Vout = Vref (i.e. opamp output equal to non-inverting input) . When this is true, the above can be simplified to:
V
i
n
=
V
r
e
f
(
R
1
R
2
+
1
)
{\displaystyle V_{in}=V_{ref}\left({\frac {R_{1}}{R_{2}}}+1\right)}
or to find R2 for a target Vin :
R
2
=
R
1
V
i
n
V
r
e
f
−
1
{\displaystyle R_{2}={\frac {R_{1}}{{\frac {V_{in}}{V_{ref}}}-1}}}
A slightly different configuration can be found by swapping Vref and ground to make the circuit a typical summing amplifier. This may be useful when Vin is negative when referenced to ground.
V
o
u
t
=
−
R
3
(
V
i
n
R
1
+
V
r
e
f
R
2
)
{\displaystyle V_{out}=-R_{3}\left({\frac {V_{in}}{R_{1}}}+{\frac {V_{ref}}{R_{2}}}\right)}
Note, the target condition is now Vout =0V as the non-inverting input is grounded. When this is the case, above can be simplified to:
V
i
n
=
−
V
r
e
f
R
1
R
2
{\displaystyle V_{in}=-V_{ref}{\frac {R_{1}}{R_{2}}}}
K factor - phase boost
edit
Based on H. Dean Venable's well known paper which defines a pole and a zero around a centre frequency for a given phase boost at that frequency. Phase boost will be maximum at this frequency.
Below these well documented formulae are 2 more sets of formulae which I have derived for finding the phase boost (θ) at any frequency (f) between a zero-pole pair rather than just the centre frequency, and for finding the frequencies between these zero-pole pairs which give a known phase boost.
Type II
Type III
f
z
=
f
c
K
{\displaystyle f_{z}={\frac {f_{c}}{K}}}
f
z
=
f
c
K
{\displaystyle f_{z}={\frac {f_{c}}{\sqrt {K}}}}
f
p
=
f
c
K
{\displaystyle f_{p}=f_{c}K}
f
p
=
f
c
K
{\displaystyle f_{p}=f_{c}{\sqrt {K}}}
θ
=
arctan
K
−
arctan
1
K
{\displaystyle \theta =\arctan K-\arctan {\frac {1}{K}}}
θ
=
2
(
arctan
K
−
arctan
1
K
)
{\displaystyle \theta =2\left(\arctan {\sqrt {K}}-\arctan {\frac {1}{\sqrt {K}}}\right)}
K
=
tan
(
θ
2
+
45
∘
)
{\displaystyle K=\tan \left({\frac {\theta }{2}}+45^{\circ }\right)}
K
=
tan
2
(
θ
4
+
45
∘
)
{\displaystyle K=\tan ^{2}\left({\frac {\theta }{4}}+45^{\circ }\right)}
θ
=
arctan
f
f
z
−
arctan
f
f
p
{\displaystyle \theta =\arctan {\frac {f}{f_{z}}}-\arctan {\frac {f}{f_{p}}}}
θ
=
2
(
arctan
f
f
z
−
arctan
f
f
p
)
{\displaystyle \theta =2\left(\arctan {\frac {f}{f_{z}}}-\arctan {\frac {f}{f_{p}}}\right)}
f
=
f
p
−
f
z
tan
θ
±
(
f
z
−
f
p
tan
θ
)
2
−
4
f
p
f
z
2
{\displaystyle f={\frac {{\frac {f_{p}-f_{z}}{\tan \theta }}\pm {\sqrt {\left({\frac {f_{z}-f_{p}}{\tan \theta }}\right)^{2}-4f_{p}f_{z}}}}{2}}}
f
=
f
p
−
f
z
tan
θ
2
±
(
f
z
−
f
p
tan
θ
2
)
2
−
4
f
p
f
z
2
{\displaystyle f={\frac {{\frac {f_{p}-f_{z}}{\tan {\frac {\theta }{2}}}}\pm {\sqrt {\left({\frac {f_{z}-f_{p}}{\tan {\frac {\theta }{2}}}}\right)^{2}-4f_{p}f_{z}}}}{2}}}
RLC series circuits
edit
X
L
=
2
π
f
L
{\displaystyle X_{L}=2\pi fL}
X
C
=
1
2
π
f
C
{\displaystyle X_{C}={\frac {1}{2\pi fC}}}
X
=
X
L
−
X
C
{\displaystyle X=X_{L}-X_{C}}
L and XC are 180° out of phase.
Z
=
R
2
+
X
2
=
R
2
+
(
X
L
−
X
C
)
2
{\displaystyle Z={\sqrt {R^{2}+X^{2}}}={\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}}
At resonance XC =XL , therefore X=0, Z=R.
Gain of a low pass RLC filter
edit
G
a
i
n
=
X
C
Z
=
X
C
R
2
+
(
X
L
−
X
C
)
2
{\displaystyle Gain={\frac {X_{C}}{Z}}={\frac {X_{C}}{\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}}}
G
a
i
n
=
X
C
Z
=
1
2
π
f
C
R
2
+
(
2
π
f
L
−
1
2
π
f
C
)
2
{\displaystyle Gain={\frac {X_{C}}{Z}}={\frac {1}{2\pi fC{\sqrt {R^{2}+\left(2\pi fL-{\frac {1}{2\pi fC}}\right)^{2}}}}}}
At resonance, Gain = XC /R.
Gain of a high pass RLC filter
edit
G
a
i
n
=
X
L
Z
=
X
L
R
2
+
(
X
L
−
X
C
)
2
{\displaystyle Gain={\frac {X_{L}}{Z}}={\frac {X_{L}}{\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}}}
G
a
i
n
=
X
L
Z
=
2
π
f
L
R
2
+
(
2
π
f
L
−
1
2
π
f
C
)
2
{\displaystyle Gain={\frac {X_{L}}{Z}}={\frac {2\pi fL}{\sqrt {R^{2}+\left(2\pi fL-{\frac {1}{2\pi fC}}\right)^{2}}}}}
At resonance, Gain = XL /R.
Phase of a low pass RLC filter
edit
θ
=
−
arctan
[
Q
(
f
f
p
−
f
p
f
)
]
−
90
∘
{\displaystyle \theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}-90^{\circ }}
where
Q
=
1
R
L
C
{\displaystyle Q={\frac {1}{R}}{\sqrt {\frac {L}{C}}}}
f
p
=
1
2
π
L
C
{\displaystyle f_{p}={\frac {1}{2\pi {\sqrt {LC}}}}}
Therefore,
θ
=
−
arctan
[
1
R
L
C
(
2
π
f
L
C
−
1
2
π
f
L
C
)
]
−
90
∘
{\displaystyle \theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}-90^{\circ }}
Phase of a high pass RLC filter
edit
θ
=
−
arctan
[
Q
(
f
f
p
−
f
p
f
)
]
+
90
∘
{\displaystyle \theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}+90^{\circ }}
θ
=
−
arctan
[
1
R
L
C
(
2
π
f
L
C
−
1
2
π
f
L
C
)
]
+
90
∘
{\displaystyle \theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}+90^{\circ }}
Inductor gap
edit
l
g
=
μ
0
L
I
m
a
x
2
B
m
a
x
2
A
c
{\displaystyle l_{g}={\frac {\mu _{0}LI_{max}^{2}}{B_{max}^{2}A_{c}}}}
B
m
a
x
=
μ
0
L
I
m
a
x
2
l
g
A
c
{\displaystyle B_{max}={\sqrt {\frac {\mu _{0}LI_{max}^{2}}{l_{g}A_{c}}}}}
I
m
a
x
=
l
g
B
m
a
x
2
A
c
μ
0
L
{\displaystyle I_{max}={\sqrt {\frac {l_{g}B_{max}^{2}A_{c}}{\mu _{0}L}}}}
μ
0
=
4
π
⋅
10
−
7
{\displaystyle \mu _{0}=4\pi \cdot 10^{-7}}
l
g
{\displaystyle l_{g}}
L
{\displaystyle L}
I
m
a
x
{\displaystyle I_{max}}
B
m
a
x
{\displaystyle B_{max}}
A
c
{\displaystyle A_{c}}
2 )
Snubber design
edit
For an unknown inductance L resonating with an unknown capacitance C1 at a measurable frequency F, add capacitance C2 in parallel with C1 until the resonant frequency halves. From the LC frequency equation we know that for frequency to half, capacitance must be quadrupled, therefore C2 =3*C1 and we can calculate C1 from C2 . From F and C1 , we can calculate the resonant impedance (reactance of C1 at the resonant frequency) and this becomes the snubber resistance Rsnb . Power lost in the snubber is proportional to Csnb so its value should be minimised while being significantly greater than C1 . C2 may be used in the final snubber as its capacitance is arguably "significantly" greater than C1 (Usually in electronics significatly greater means 5-10x greater).
Modelling thermal system
edit
THIS IS JUST ME THINKING OUT LOUD. DON'T TAKE THIS SECTION AS ANY RELIABLE SOURCE
Discharging capacitor through a resistor
edit
A charged capacitor discharging through a resistor from voltage V0 to 0
V
(
t
)
=
V
0
e
−
t
τ
{\displaystyle V(t)=V_{0}e^{-{\frac {t}{\tau }}}}
τ
{\displaystyle \tau }
V
0
{\displaystyle V_{0}}
V
0
/
e
{\displaystyle V_{0}/e}
τ
=
R
C
{\displaystyle \tau =RC}
Transfer function:
H
C
(
s
)
=
V
C
(
s
)
V
i
n
(
s
)
=
1
1
+
R
C
s
=
1
1
+
τ
s
{\displaystyle H_{C}(s)={V_{C}(s) \over V_{in}(s)}={1 \over 1+RCs}={1 \over 1+\tau s}}
Thermal equivalent
edit
A hot object falling from starting temperature
T
0
{\displaystyle T_{0}}
T
r
t
{\displaystyle T_{rt}}
T
(
t
)
=
T
0
e
−
t
τ
{\displaystyle T(t)=T_{0}e^{-{\frac {t}{\tau }}}}
τ
{\displaystyle \tau }
1
e
(
T
0
−
T
r
t
)
{\displaystyle {\frac {1}{e}}(T_{0}-T_{rt})}
Transfer function:
H
(
s
)
=
T
(
s
)
T
i
n
(
s
)
=
1
1
+
τ
s
{\displaystyle H(s)={T(s) \over T_{in}(s)}={1 \over 1+\tau s}}
Really I need a transfer function of T(s)/Pin (s).
This is a good start because it means I only have to measure temperature and time to get Tau, and not need to know the thermal equivalents of R and C. The end result should be a way to correctly tune a PID controller for an critically damped step response in a thermal system.
![{\displaystyle g_{m}={dI_{a} \over dV_{g}}={\frac {3}{2}}\ {\sqrt[{3}]{P^{2}I_{a}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d2d78635d4175dca5a5324ac3c0ca0ffee3bc20)
![{\displaystyle r_{a}={dV_{a} \over dI_{a}}={\frac {2}{3}}\ {\frac {\mu }{\sqrt[{3}]{P^{2}I_{a}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c347fb0b827035564376f40eafa5f4324296a9d6)
}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d072ff98f3f37075d7560a7388d3da81bb011cac)
![{\displaystyle H(s)={\frac {V_{e}}{V}}(s)=-{\frac {(R_{3}C_{1}s+1)[C_{3}(R_{1}+R_{4})s+1]}{[R_{1}(C_{1}+C_{2})s](R_{3}{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}s+1)(R_{4}C_{3}s+1)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae89e42fd0488fbb36e019842bbeee3d52c082f3)
![{\displaystyle H(s)={\frac {V_{e}}{V}}(s)\approx -{\frac {(R_{3}C_{1}s+1)[R_{1}C_{3}s+1]}{(R_{1}C_{1}s)(R_{3}C_{2}s+1)(R_{4}C_{3}s+1)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7056f6889303fb920d5d0c3cd1d68a5e4b2aecb)
![{\displaystyle \theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}-90^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e53b32685a507a98e0c0e5dfdfe830ae5650bf57)
![{\displaystyle \theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}-90^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa10416d33368ea5c5fd19a4d0ec317cdb6217ae)
![{\displaystyle \theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}+90^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6593f218488a0793adab07d6a1d5443fbc9cc767)
![{\displaystyle \theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}+90^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/193b973a061edb5fd1eb59058e75613275fa6d48)
