User:KX36

Maths Sandbox[edit]

Child-Langmuir Law[edit]

$I_{a}$ is anode current.

$S$ is surface area of anode.

$d$ is distance between anode and cathode.

$V_{a}$ is the potential difference from anode to cathode.

$P$ is the perveance of the device.

$I_{a}=\left({\frac {4\epsilon _{0}}{9}}{\sqrt {\frac {2e}{m_{e}}}}\right)\left({\frac {S}{d^{2}}}\right)V_{a}^{\frac {3}{2}}$

$I_{a}=k_{1}\left({\frac {S}{d^{2}}}\right)V_{a}^{\frac {3}{2}}$

$I_{a}=PV_{a}^{\frac {3}{2}}$

$I_{a}\propto V_{a}^{\frac {3}{2}}$

$P\propto {\frac {S}{d^{2}}}$

Triode equation (and rearrangements thereof)[edit]

$\mu$ is the amplification factor of the triode.

$I_{a}=P\left({\frac {V_{a}}{\mu }}+V_{g}\right)^{\frac {3}{2}}$

$V_{a}=\mu \left(\left({\frac {I_{a}}{P}}\right)^{\frac {2}{3}}-V_{g}\right)$

$V_{g}=\left({\frac {I_{a}}{P}}\right)^{\frac {2}{3}}-{\frac {V_{a}}{\mu }}$

Derivatives of triode equation[edit]

$r_{a}$ is the anode resistance

$g_{m}$ is the transconductance of the device

$r_{a}={\frac {\mu }{g_{m}}}$

$\mu ={dV_{a} \over dV_{g}}$

$g_{m}={dI_{a} \over dV_{g}}={\frac {3}{2}}\ {\sqrt[{3}]{P^{2}I_{a}}}$

$r_{a}={dV_{a} \over dI_{a}}={\frac {2}{3}}\ {\frac {\mu }{\sqrt[{3}]{P^{2}I_{a}}}}$

DC to DC converters - conversion ratios[edit]

Buck	Boost	Buck-boost (inverting)	Buck-boost (non-inverting)
$D={\frac {V}{V_{g}}}$	$D={\frac {V-V_{g}}{V}}$	$D={\frac {-V}{V_{g}-V}}$	$D={\frac {V}{V_{g}+V}}$
$M(D)={\frac {V}{V_{g}}}=D$	$M(D)={\frac {V}{V_{g}}}={\frac {1}{1-D}}$	$M(D)={\frac {V}{V_{g}}}={\frac {-D}{1-D}}$	$M(D)={\frac {V}{V_{g}}}={\frac {D}{1-D}}$
${\frac {d}{dD}}M(D)=1$	${\frac {d}{dD}}M(D)={\frac {1}{(1-D)^{2}}}$	${\frac {d}{dD}}M(D)={\frac {-1}{(D-1)^{2}}}$	${\frac {d}{dD}}M(D)={\frac {1}{(1-D)^{2}}}$

Type II Compensator transfer function[edit]

$H(s)={\frac {V_{e}}{V}}(s)=-{\frac {R_{3}C_{1}s+1}{[R_{1}(C_{1}+C_{2})s](R_{3}{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}s+1)}}$

$H(s)={\frac {V_{e}}{V}}(s)\approx -{\frac {R_{3}C_{1}s+1}{(R_{1}C_{1}s)(R_{3}C_{2}s+1)}}$ when $C_{2}\ll C_{1}$

Type III Compensator transfer function[edit]

$H(s)={\frac {V_{e}}{V}}(s)=-{\frac {(R_{3}C_{1}s+1)[C_{3}(R_{1}+R_{4})s+1]}{[R_{1}(C_{1}+C_{2})s](R_{3}{\frac {C_{1}C_{2}}{C_{1}+C_{2}}}s+1)(R_{4}C_{3}s+1)}}$

$H(s)={\frac {V_{e}}{V}}(s)\approx -{\frac {(R_{3}C_{1}s+1)[R_{1}C_{3}s+1]}{(R_{1}C_{1}s)(R_{3}C_{2}s+1)(R_{4}C_{3}s+1)}}$ when $C_{2}\ll C_{1}$ and $R_{4}\ll R_{1}$

Formula for R3[edit]

For type III compensation,

$R_{3}={\frac {F_{c}}{F_{lc}}}{\frac {V_{osc}}{V_{in}}}R_{1}$

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

$V_{osc}={\frac {N_{b2}}{N_{s}}}V_{sec}\left(1-e^{\frac {-1}{F_{s}RC}}\right)$

Comparator gain [N.B. gain is independant of Vsec (proportional to input voltage). Classical voltage mode control has a comparator gain (Vsec/Vosc) which proportiobal to Vsec, which causes problems. The idea of votlage feedforward is to make Vosc proportional to Vsec so that comparator gain is constant]:

${\frac {V_{sec}}{V_{osc}}}={\frac {N_{s}}{N_{b2}\left(1-e^{\frac {-1}{F_{s}RC}}\right)}}$

Find RC for minimum acceptable Vosc at minimum Vpri (input voltage).

$RC={\frac {-1}{F_{s}ln\left(1-{\frac {N_{p}}{N_{b2}}}{\frac {V_{osc}}{V_{pri}}}\right)}}$

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.

EA gain[edit]

From Summing Amplifier equation, with V₂=0V and IN+=V_ref.

$V_{out}=-R_{3}\left({\frac {V_{in}-V_{ref}}{R_{1}}}-{\frac {V_{ref}}{R_{2}}}\right)+V_{ref}$

$V_{out}=V_{ref}\left({\frac {R_{3}}{R_{2}}}+{\frac {R_{3}}{R_{1}}}+1\right)-{\frac {R_{3}}{R_{1}}}V_{in}$

For the target condition, V_out = V_ref (i.e. opamp output equal to non-inverting input) . When this is true, the above can be simplified to:

$V_{in}=V_{ref}\left({\frac {R_{1}}{R_{2}}}+1\right)$

or to find R2 for a target V_in:

$R_{2}={\frac {R_{1}}{{\frac {V_{in}}{V_{ref}}}-1}}$

A slightly different configuration can be found by swapping V_ref and ground to make the circuit a typical summing amplifier. This may be useful when V_in is negative when referenced to ground.

$V_{out}=-R_{3}\left({\frac {V_{in}}{R_{1}}}+{\frac {V_{ref}}{R_{2}}}\right)$

Note, the target condition is now V_out=0V as the non-inverting input is grounded. When this is the case, above can be simplified to:

$V_{in}=-V_{ref}{\frac {R_{1}}{R_{2}}}$

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
$f_{z}={\frac {f_{c}}{K}}$	$f_{z}={\frac {f_{c}}{\sqrt {K}}}$
$f_{p}=f_{c}K$	$f_{p}=f_{c}{\sqrt {K}}$
$\theta =\arctan K-\arctan {\frac {1}{K}}$	$\theta =2\left(\arctan {\sqrt {K}}-\arctan {\frac {1}{\sqrt {K}}}\right)$
$K=\tan \left({\frac {\theta }{2}}+45^{\circ }\right)$	$K=\tan ^{2}\left({\frac {\theta }{4}}+45^{\circ }\right)$
$\theta =\arctan {\frac {f}{f_{z}}}-\arctan {\frac {f}{f_{p}}}$	$\theta =2\left(\arctan {\frac {f}{f_{z}}}-\arctan {\frac {f}{f_{p}}}\right)$
$f={\frac {{\frac {f_{p}-f_{z}}{\tan \theta }}\pm {\sqrt {\left({\frac {f_{z}-f_{p}}{\tan \theta }}\right)^{2}-4f_{p}f_{z}}}}{2}}$	$f={\frac {{\frac {f_{p}-f_{z}}{\tan {\frac {\theta }{2}}}}\pm {\sqrt {\left({\frac {f_{z}-f_{p}}{\tan {\frac {\theta }{2}}}}\right)^{2}-4f_{p}f_{z}}}}{2}}$

RLC series circuits[edit]

$X_{L}=2\pi fL$ $X_{C}={\frac {1}{2\pi fC}}$

Total reactance $X=X_{L}-X_{C}$ because X_L and X_C are 180° out of phase.

$Z={\sqrt {R^{2}+X^{2}}}={\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}$

At resonance X_C=X_L, therefore X=0, Z=R.

Gain of a low pass RLC filter[edit]

$Gain={\frac {X_{C}}{Z}}={\frac {X_{C}}{\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}}$

$Gain={\frac {X_{C}}{Z}}={\frac {1}{2\pi fC{\sqrt {R^{2}+\left(2\pi fL-{\frac {1}{2\pi fC}}\right)^{2}}}}}$

At resonance, Gain = X_C/R.

Gain of a high pass RLC filter[edit]

$Gain={\frac {X_{L}}{Z}}={\frac {X_{L}}{\sqrt {R^{2}+(X_{L}-X_{C})^{2}}}}$

$Gain={\frac {X_{L}}{Z}}={\frac {2\pi fL}{\sqrt {R^{2}+\left(2\pi fL-{\frac {1}{2\pi fC}}\right)^{2}}}}$

At resonance, Gain = X_L/R.

Phase of a low pass RLC filter[edit]

$\theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}-90^{\circ }$

where $Q={\frac {1}{R}}{\sqrt {\frac {L}{C}}}$ and $f_{p}={\frac {1}{2\pi {\sqrt {LC}}}}$ .

Therefore,

$\theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}-90^{\circ }$

Phase of a high pass RLC filter[edit]

$\theta =-\arctan {\left[Q\left({\frac {f}{f_{p}}}-{\frac {f_{p}}{f}}\right)\right]}+90^{\circ }$

$\theta =-\arctan {\left[{\frac {1}{R}}{\sqrt {\frac {L}{C}}}\left(2\pi f{\sqrt {LC}}-{\frac {1}{2\pi f{\sqrt {LC}}}}\right)\right]}+90^{\circ }$

Inductor gap[edit]

$l_{g}={\frac {\mu _{0}LI_{max}^{2}}{B_{max}^{2}A_{c}}}$

$B_{max}={\sqrt {\frac {\mu _{0}LI_{max}^{2}}{l_{g}A_{c}}}}$

$I_{max}={\sqrt {\frac {l_{g}B_{max}^{2}A_{c}}{\mu _{0}L}}}$

$\mu _{0}=4\pi \cdot 10^{-7}$ permeability of free space (H/m)

$l_{g}$ gap length (m)

$L$ Inductance (H)

$I_{max}$ max peak current (A)

$B_{max}$ max flux density (T)

$A_{c}$ core cross-sectional area (m²)

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 R_snb. Power lost in the snubber is proportional to C_snb 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

$V(t)=V_{0}e^{-{\frac {t}{\tau }}}$

$\tau$ is the time taken for voltage to fall from $V_{0}$ to $V_{0}/e$ when the end point is V=0.

$\tau =RC$

Transfer function:

$H_{C}(s)={V_{C}(s) \over V_{in}(s)}={1 \over 1+RCs}={1 \over 1+\tau s}$

Thermal equivalent[edit]

A hot object falling from starting temperature $T_{0}$ to room temperature $T_{rt}$ :

$T(t)=T_{0}e^{-{\frac {t}{\tau }}}$

$\tau$ is the time taken for temperature to fall to ${\frac {1}{e}}(T_{0}-T_{rt})$

Transfer function:

$H(s)={T(s) \over T_{in}(s)}={1 \over 1+\tau s}$

Really I need a transfer function of T(s)/P_in(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.