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16. Equations

NB Equations apply to all units regardless of quantity expressed, e.g. milli: m, micro: µ, etc.

Var. = variable; SI = International System of Units

SYMBOLS = Lucida Console font; supported symbols from MS Windows character map:

⌠ ⌡ ∫ │ ─ √ φ θ Θ ∂ δ ζ ξ ς λ ψ ω  τ µ  Ω ∆ Δ ∑ ∏ π ∞ ∝ Ξ ○ ≠ ³ ² ± ​≤ 

ELECTRONICS

Q1 OHMS LAW

Q1a   V = IR

Q1b   I = V/R

Q1c   R = V/I

Q1d   P = IV

Q1e   I = P/V

Q1f   V = P/I

Q1g   P = V²/R

Q1h   V = PR
Q1i   R = V
²/P

Q2 OSCILLOSCOPE BANDWIDTH & RISE TIME

Q2a BANDWIDTH wrt RISE TIME

http://www.amplifier.cd/Test_Equipment/Tektronix/Tektronix_7000_series_special/calibration-fixture.htm

A close approximation of oscilloscope bandwidth can be determined from:

 

Bandwidth =    0.35   

             Rise time

Q2b RISE TIME wrt BANDWIDTH

Rise time  =   0.35      

             Bandwidth

Where

Variable    Unit        SI

Bandwidth = frequency   Hz

Rise time = seconds     s

Example:

The Tek 7104 has a nominal bandwidth of 1GHz, so its rise time = 0.35 / 1GHz = 350ps

https://w140.com/tekwiki/wiki/7104

The actual value will differ slightly from this due to manufacturing tolerances of the CRT. The original specification for the 7104 with a 7A29 states it is <350ps. Here are older 1GHz 519s rated at 300ps:

https://vintagetek.org/wp-content/uploads/2017/09/519-screen.jpg

https://vintagetek.org/519-1-ghz-oscilloscope/

https://w140.com/tekwiki/wiki/519

Q3  OSCILLOSCOPE MAINFRAME RISE TIME wrt SIGNAL RISE TIME

Q3a MAINFRAME RISE TIME

http://www.amplifier.cd/Test_Equipment/Tektronix/Tektronix_7000_series_special/calibration-fixture.htm

This webpage explains in detail how to calculate oscilloscope rise time from the following equation:

t²rise measurement = t²mainframe + t²signal

                                                  ________________________________

Oscilloscope mainframe rise time = ​t mainframe = (t²rise measurement - t²signal)

This can be determined using a signal source with a known, calibrated rise time, such as Tek's own 7k series Calibration Fixture plugins, e.g. 067-0587-01, or the faster 067-0587-02 needed for the 7104.

Example:

Tek 1GHz 7104 + 067-0587-02 Calibration Fixture with nominal 150ps max rise time. The actual value will differ slightly from this due to manufacturing tolerances of the fixture. The above webpage assumes their fixture's rise time is 20% faster at 120ps, but I have used its specified nominal value of 150ps in the equation below.

On the above webpage, the rise time of this signal measured on the 7104 = 300ps.

The nominal rise time of the Calibration Fixture = 150ps

               ________________________________

t mainframe = √ (t²rise measurement - t²signal)

               __________________

t 7104      = √ (300ps² - 150ps²) = 260ps.

So based on a 150ps Cal Fixture rise time, the 7104 mainframe rise time is 260ps.

Applying [Q2], Bandwidth =     1     x 0.35  =   1    x 0.35  =  1.346GHz  

                           Rise time            260ps

Q3b TRUE SIGNAL RISE TIME

http://www.amplifier.cd/Test_Equipment/Tektronix/Tektronix_7000_series_special/calibration-fixture.htm

This same equation can be used to determine true signal rise time from the displayed value:

 

t²rise measurement = t²mainframe + t²signal

   

                                      ___________________________________

Actual signal rise time = t signal = (t²rise measurement - t²mainframe)

Example:

Tek 7104 1GHz + 067-0587-02 calibration fixture with nominal 150ps max rise time. The actual value will differ slightly from this due to manufacturing tolerances of the fixture. The above webpage assumes their fixture's rise time is 20% faster at 120ps but I have used its specified nominal value of 150ps in the equation below.

From equation [Q2] above, the calculated mainframe rise time = 260ps.

From the webpage above, the rise time of the Cal Fixture measured on the CRT = 300ps 

               __________________________________

t signal    = √ (t²rise measurement - t²mainramw)

               __________________

t 7104      = √ (300ps² - 260ps²) = 150ps.

So the actual rise time of the Cal Fixture signal = 150ps.

Q4 NYQUIST SAMPLING RATE
 

Harry Nyquist determined the optimum sampling rate that can be applied to DSOs.

​If the signal consists of L discrete levels, Nyquist's theorem states:

Q4a    BitRate = 2 x Bandwidth x log2(L) bits/sec.

Where

       bandwidth is the bandwidth of the channel

       L is the number of signal levels used to represent data,

       BitRate is the bit rate in bits per second.

https://en.wikipedia.org/wiki/Nyquist_rate

This is referred to as the Nyquist sampling rate, and is usually simplified to x3 to x5:

[E58]: 'To capture the true shape of the signal, you need a scope with a bandwidth large enough to capture several of the signals harmonics, so ideally use a scope with 3x to 5x the bandwidth you calculated for your signal.'

https://ibex.tech/resources/geek-area/electronics/test-and-measurement/oscilloscopes/bandwidth-vs-sample-rate

Q4b MAXIMUM BANDWIDTH RATE FOR A PULSE (based on [4d] below)

Maximum signal frequency = 0.4 / rise time (20-80%)

Q4c PULSE RISE TIME FOR MAXIMUM BANDWIDTH (based on [4d] below)

 

rise time =  0.4 / Maximum signal frequency

Q4d MINIMUM BANDWIDTH SAMPLE RATE FOR A DSO TO CAPTURE A PULSE

Absolute minimum: maximum signal frequency x 3

Or better:        maximum signal frequency x 3.5, for example:

From [I24], HP54854A 4GHz 20GS/s DSO datasheet, page 4: 

Bandwidth required to measure rise time with 3% error. Example: 100ps rise time (20-80%)


Maximum signal frequency content   = 0.4/rise time (20-80%) Maximum signal frequency  = 4GHz

 

N.B. HP complicates this issue as their constant 0.4 differs from [Q2a,Q2b]: 0.35.

I suspect either is usable, and HP is increasing the signal quality by rising to 0.4.

I've also seen it relaxed to 0.3/Tr. [E57, ETS Current Probes Page 42].

Until I know better, I will leave both [Q2] and [Q4] here.


Scope bandwidth required           = 1.4 x maximum frequency Required scope bandwidth = 5.6GHz
Minimum scope sample rate required = 2.5 x bandwidth Required scope sample rate       = 14GS/s

IOW maximum signal frequency x3.5  (3.5 = 1.4 x 2.5)

Q4e OPTIMUM BANDWIDTH SAMPLE RATE FOR FOR A DSO

 

Maximum signal frequency x 5

PHYSICS

Q5: ENERGY IS POWER EXPENDED OVER TIME 

E = P x T = energy in Joules, J

Where

SU     Description   Unit     SI

E      Energy        Joules   J

P      Power         Watts    W

T      Time          Seconds  s

Worked examples given E = 24mJ, P = 6MW, T = 4ns:

Q5a      E = P x T      =  6MW x 4ns  = 24mJ

    

Q5b      P = E          = 24mJ        = 6MW
             T             4ns

Q5c      T = E          = 24mJ        = 4ns 
             P             6MW

Q6: CAPACITIVE ENERGY

E = 0.5 x C x V² = energy in Joules, J

Where

SU     Description   Unit     SI

C      Capacitance   Farads   F

V      Voltage       Volts    V

Worked example for MK367 Nd:YAG, given C = 20µF. V = 673V:

Q6a E =    0.5 x C x V²  =   0.5 x 20µF x 673V² = 4.53J

           ________       ___________

Q6b V =   /    E     =   /  4.53J      = 673V

         √  0.5 x C     √  0.5 x 20µF

Q6c C =        E     =      4.53J      = 20µF

           0.5 x V²       0.5 x 673²

Q7: FLASHLAMP PFN PULSE ENERGY

THIS NEEDS CHECKING - CHICKEN & EGG & NOT SURE IF IT'S CORRECT:

 

From [O19] CORD Module 3-2: Pulsed Laser Flashlamps and Power Supplies

 

Where

  

Var    Description                                    Unit     SI

r    = load resistance (lamp resistance)              Ohms     Ω

L    = total network inductance                       Henries  H

C    = total network capacitance                      Farads   F

T    = pulse width at 70% of average peak amplitude   seconds  s

Worked examples given L = 22µH, C = 20µF:

Q7a SIMPLE #1 MESH LCR LOAD RESISTANCE

                            ____

L  =  r²C  therefore r =   / 4L 

       4                  √   C

        ___        _________

r  =   / 4L   =   / 4 x 22µH          = 2.097Ω

      √   C      √    20µF

 

Q7b CR CHARGING TIME

C  =  T   therefore T  = C x r

      r

T  = C x r  = 20µF x 2.097Ω = 41.94µs

Q7c #5 OR MORE MESH PFN

 

       ___        _____

r  =  / L    =   / 22µH               = 1.049Ω

     √  C       √  20µF

        ___       ____________ 

T  =  2√ LC  =  2√ 22µH x 20µF        = 41.952µs

   

Or,

L  =  T x r =   41.952µs x 1.049Ω     = 22µH

       2                2     

Q10 GAS LAWS

https://en.wikipedia.org/wiki/Gas_laws

[Q10a] GAY-LASSAC'S LAW

 

'For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature'.

P ∝ T   or   P/T = k   or   P1/T1 = P2/T2

where

 

P = pressure

T = absolute temperature

k = a proportionality constant.

Q10b BOYLE'S LAW 

https://en.wikipedia.org/wiki/Boyle%27s_law

'The absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system.

The volume of a given mass of a gas is inversely related to pressure when the temperature is constant.'



 

PV= k1

where

 

P  = pressure of the system

V  = volume of the gas

k1 = constant value representative of the temperature and volume of the system.

Q10c CHARLES'S LAW

https://en.wikipedia.org/wiki/Charles%27s_law

'for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature, assuming in a closed system.

 

The volume (V) of a given mass of a gas, at constant pressure (P), is directly proportional to its temperature (T)'

V ∝ T   or   V/T = k2   or   V1/T1 = V2/T2

here

 

P  = pressure of the system

V  = volume of the gas

k2 = proportionality constant
     (k2 is not the same as the proportionality constants in the other equations in this section)

Q20 LASER EXTRA-CAVITY POWER

 

Power of beam area = watts / π x radius²

Example: 0.5J @ 10ns pulse = 50MW output pulse

For a 2mm dia 50MW beam = 50MW/(π x 0.10cm²) = 1.59GW/cm² = 1.60GW

For a 4mm dia 50MW beam = 50MW/(π x 0.20cm²) =  398MW/cm² = 0.40GW

For a 5mm dia 50MW beam = 50MW/(π x 0.25cm²) =  255MW/cm² = 0.26GW

For a 6mm dia 50MW beam = 50MW/(π x 0.30cm²) =  177MW/cm² = 0.18GW

Q21 OPO NLOs

 

This webpage is an excellent primer for NLOs in general:

http://www.repairfaq.org/sam/laserssl.htm#sslsrgs

'Optical Parametric Oscillation (OPO) is a nonlinear process in which a single input laser beam or "pump" beam is converted into two lower-energy beams known as the "signal" beam and the "idler" beam.

This nonlinear process enables a fixed wavelength laser beam to be converted into other wavelengths. The wavelengths (λ) / frequencies (f) of the three beams must satisfy:

 

                       1                  1                 1      
                     λ(Pump)     =    λ(Signal)     +    λ(Idler)

or equivalently:     f(Pump)     =    f(Signal)     +    f(Idler)

Q30 CONVERT ELECTRON-VOLT (eV) TO WAVELENGTH & VICE-VERSA

From Wikipedia (https://en.wikipedia.org/wiki/Electronvolt):

Were

Var  Description                                  Unit           SI

E =  energy                                       electron-volt  eV

F =  frequency                                    Hertz          Hz

λ =  wavelength of a photon                       Distance       nm

h =  Planck constant, 4.135667516                 Speed          m/s

c =  speed of light constant,  299792458m/s       Speed          m/s

E = hF = hc   (4.135667516 x 10^-15eV)(299792458 m/s)
         λ                  λ (nm)

This reduces to:


E (eV) = 4.135667516 feVs x F (PHz) = 1239.84193 eV-nm
                                         λ (nm)

Or,

λ (nm) = 1239.84193 eV-nm

              E (eV)

MATHEMATICS

Q50 CIRCUMFERENCE OF A CIRCLE

distance = π x diameter

 

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