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Eval » Historie » Version 19

Maximilian Seesslen, 20.07.2023 16:51

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h1. Eval
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h1. Overall design
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* 4 x red+green LEDs for condition
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* Restart button
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* Iterate over the ADC channels and get Voltage level
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* Self-Calibrating via MOSFET (shortcuit cable interface)
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* draw 100mA. Why? Sonst habe ich keinen Spannungsteiler. Je mehr Strom desto hoeher der Spannungsabfall; optional via mosfet.; 3/0,1= 30 Ohm
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* ADC loesst theoretisch auf "3 / 4096 = approx. 0,0007" Volt auf.
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* Die Kabel haben AWG28, 212.9 Ohm/km, 1.4 A nach erde,	0.23 A versorgung
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* 0,2129 Ohm/m; 1,0645 Ohm/5m; 2,129 Ohm/10m;
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* https://de.wikipedia.org/wiki/E-Reihe#/media/Datei:E12_values_graph.svg
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* -Buzzer would be nice for indication; durchklingeln- Da werde ich lieber Multimeter nehmen.
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* -PWM Output-
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* Pinsocket connected to wires to measure resistance/Voltage directly
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* https://www.aeq-web.com/spannungsteiler-microcontroller-berechnen-und-dimensionieren/
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* Spannungen einzeln schaltbar damit auch kurzschluesse detektiert werden koennen
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h1. Schema
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       3V
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        |
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       | |  Cable
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        |-------------->ADC
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       | | 30Ohm
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        |
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       ===
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h1. Theoretische Spannungen bei AWG28
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theoretically an device that draws 0,1A, 5V and 5m AWG28 cable:
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U2=((5*50)/ (50+1,0645) ) = 4,8957691; that should work
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theoretically an device that draws 0,5A, 5V and 10m AWG28 cable:
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U2 = ( (5*10) / (10+2,129) ) = 4,1223514 V
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theoretically an device that draws 0,1A, 3V and 5m AWG28 cable:
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U2=((3*30)/ (30+1,0645) ) = 2,8971978; that should work
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theoretically an device that draws 0,1A, 3V and 10m AWG28 cable:
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U2=((3*30)/ (30+2,129) ) = 2,8012076
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theoretically an device that draws 0,25A, 3V and 10m AWG28 cable:
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U2= (3 * 12) / (12 + 2,129) = approx. 2,547951
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theoretically an device that draws 0,5A, 3V and 10m AWG28 cable:
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U2= (3 * 6) / (6 + 2,129) = approx. 2,2142945
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U2= (3 * 6,8) / (6,8 + 2,129) = approx. 2,2846903
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h1. Calculations I
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U2=((U*R2)/R_GES)
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U
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R*I
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5V/0,1A= 50 Ohm
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5V*0,1A= 0,5 W
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5V/0,5A= 10 Ohm
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RGes = R1+R2 = 50
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3V=(5*R2)/50
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*R2 = (3*50)/5 = 30 -> 27
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R1 = 22*
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RGES = 49
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U2 = 5*27/49 = 2,755102
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h1. Calculations II
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* Ein Kabel >= 10m soll durchfallen. Koennen trotzdem groessere Wiederstaende verwendet werden?
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Ohne Widerstand:
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U2 = 2,755102
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Obige werte 1fach, 10m Kabel:
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R2 = 27
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R1 = 22
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mit 10m 
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U2 = (5*27)/(49+2,129) = 2,6403802
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2,755102-2,6403802=0,1147218
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0,1147218/0,00073242188 = 156,6335
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Obige werte 10fach, 10m Kabel:
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R2 = 270
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R1 = 220
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mit 10m 
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U2 = (5*270)/(490+2,129) = 2,7431832
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Bei 12Bit ADC: 3 / 4096 = approx. 0,00073242188 V pro ADC-Wert.
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2,755102-2,7431832=0,0119188
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0,0119188 / 0,00073242188 = approx. 16,273135. Thats not super much
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h1. Calculations III
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h1. Keyfeatures
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* Cable checker
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* Buzzer Durchgangspruefer
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* PWM Output
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h1. MCU
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* "stm32f051c4":https://www.mouser.de/datasheet/2/389/stm32f051c4-1851079.pdf
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** I have 17; 
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** 16 ADC channels
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* STM32G030C8T6
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** nearly same specs but smaller footprint; LQFP 48
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** 2,68€ inc. Mwst.
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** ADC faster
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** Mentions an 16Bit ADC value via oversampling, but thats complicated: adding white noise in order to calculate further 2 bits by software. Every MCU can do that.