======================RF_ELECTRONICS====================== ______________ | |__| | ___| |___ |___| 1 VCC 14 |___|______ | | | .22uF ___| |___ _|_ |___| 2 13 |___| ___ | | | ___| |___ _|_ |___| 3 12 |___| /// | | ___| |___ |___| 4 11 |___| | | ___| |___ Typical Digital |___| 5 10 |___| Supply Bypass | | ___| |___ |___| 6 9 |___| | | ___| |___ ___|___| 7 GND 8 |___| | | | _|_ |______________| /// For low frequencies the general rules are... 1) Supply bypass capacitors 2) Short leads 3) The greater the distance, the less the stray C 4) Watch for cross talk...etc __\_ H | / | _______________|_____________ ()_____________\|/____________) R_cm --> I | V L_cm |_/_ \ SELF INDUCTANCE (uH) for ROUND WIRE (L_cm,R_cm) L_uH=.002*L_cm*( ln(2*L_cm/R_cm) -.75 ) When entering the RF world, you begin to see that having short leads on your supply bypass capacitor is not telling you the whole story. A good way to see this is to measure the impedance of a 0.22uf bypass capacitor over frequency. IMPEDANCE OF 0.1uF CAPACITOR 10 |..C............................................. | C . . . L . | C . . L . | C . L . . | C. . L . . | C . L . . 1 |..............C...........L..................... | . C .L . . | . C L . . | . C L. . . | . R . . . | . . . . .1 |................................................ | . . . | . . . . | . . . . | . . . . | . . . . |______________________________________________ . 100KHz 1MHz 10MHz 100MHz 1GHz above 10MHz, you really don't have capacitor. self inductance takes over, and digital supply current spikes really seeing a "bypass inductor" The RF guys have long known this. they never use things like 0.22uf capacitors any where critical like to use those silver-mica capacitors in picoFarads. ^ VCC 0.1uF /_\ Internal prop delays for a CS80CBI _______|______ inverter typ below 100pS. | => I | _____ _|_ _||<- | ___ __ | || /|\ Vout | |_ | ||__ ___| | ___| |__ =>I | _|_ | __| |______ ... | /Vin\ | || _|_ | \___/ |_|| ___ | | ||-> ______|__ ... |_____|_________| I<= <= I _|_ /// current always flows in loops. Say you have an inverter swinging from low to high. Current coming out of inverter to charge the stray C. inverter get this current from the supply bypass. current will form the loop shown above. _______\_____ I <--- | / | ____________|____ | | \ | _____ /|\ \ \|/ H Field \ | __\__ | \ | _____ | Magnetic Radiation \ | \ | \ ---> I \ | \____________\ | | | |______/______| \ in terms of how much magnetic field does current spike loop generate to understand size of the "bypass inductor" . inductance of loop wire can show the inductance is related to loop area. _ _ Dia_inch | | | | _| | | |_ / __| |__ \ / / ^ \ \ R_inch/Dia_inch > 2.5 / / /|\ \ \ | | | | | | | R_inch | | \ \ / / \ \__ __/ / \_ --- _/ ------- SELF INDUCTANCE (uH) for a RING OF WIRE (R_cm, Dia_inch) L_uH (R_inch/100)*( 7.353*log(16*R_inch/Dia_inch) -6.386) area of the loop which defines the size of the inductance. when you make the loop area small? ___\__ --> I ______|___/__|________________ ->()___________\|/______________ | |_/____| _/___ | | ________\_______|_\___|_____| | <-()_____________________|_______| <-- I |___\_| __ / | |__ _____ |__| | | | |*() () | | \_/\_/\_/ L1 | Lb M _ _ _ | Lb = L1 + L2 -2*M /*\/ \/ \ L2 | if L1 = L2 =M __ | () () | | | |__| |_____| then Lb=0 |__| When return path flow right next to each other, mutual inductance cancel out self inductance. Remember the short lead rule? Suppose don't make bypass capacitor leads short, rather just make return current flow next to any signal current. ( make Loop area is small) _____________________________ ->()____________________________) <- d_cm | D_cm Z_ohms sqrt(L/C) | _____________________________ ->()____________________________) L_cm C_pf/meter = 12.06/log(2*D_cm/d_cm) L_uH/meter = 0.92*log(2*D_cm/d_cm) Z_ohms = 276*log(2*D_cm/d_cm) now a transmission line. current loop will still generate a magnetic field. magnetic field is completely confined in space between conductors. A transmission line just like microwave wave guide. have current, voltage, E fields and H fields. are happening all together in a small area of space Voltage dropping current \ -----> \ \____\ \______\ \ \ \ \ \ \ _ \ _ ->E \| \ \ _ \| \ \|/ _ /_ \ ^ _ /_ v H |_/ \ _V Voltage \ |_/ \ __/_ \ Pulse \ __/_ |_/ \ \ \ |_/ \ | _V | voltage \ \ - + \ \____\ \______\ \ \ \ \ \ ---> \ _ \ _ | E \| \ \ _ \| \ \|/ _ /_ \ ^ _ /_ v H |_/ \ _V \ |_/ \ __/_ \ __/_ ->E |_/ \ \ |_/ \ \|/ | | v H \ \ \ \____\ \______\ \ \ \ \ current <------ Voltage increasing Between conductors, capacitance. a fast voltage pulse, transmission line form current loops within itself. confine both the E field and H field completely between conductors. So now signal really both voltage and current and electromagnetic wave, in same place. if signal current is differential (twisted pair lines), transmission line will look like no signal at short distance away. If only air bewteen conductors, this wave will travel at speed of light. ( 1ns =about 1foot) _____________________________ ->()____________________________) <- d_cm | D_cm Z_ohms sqrt(L/C) | _____________________________ ->()____________________________) L_cm C_pf/meter = 12.06/log(2*D_cm/d_cm) L_uH/meter = 0.92*log(2*D_cm/d_cm) Z_ohms = 276*log(2*D_cm/d_cm) Z_air =377 ohms Rabbit ears 300 ohm Transmission line of rabbit ears follow equation. Transmission lines have resistance impedances. violate short lead supply bypass capacitor rule and connect 5 ohm transmission line between IC and supply bypass as shown below,now be bypassing your supply with 5 ohm resistor all frequencies 10MHz ___\__ --> I ______|___/__|______________ 5 Ohm ->()___________\|/_____________|__/\ __ |_/____| _/___ \/ | ________\_______|_\___|_____ _|_ <-()_____________________|_____|_ ___ Bypass <-- I |___\_| |______| 0.22uF In terms of a PC board layout is is common to have a signal wire over ground plane with small enough dielectric between. recommended spacings result in 50ohm trans line _ I signal current /| / / / / /_____/ / / / |_____|/ / / / /_________________/ / |/_ I return path |_________________|/ simply laying out PC boardthat all signal current runs next to return path, easy to have all high frequency circuitry see nothing but transmission lines. ^ VCC 0.1uF /_\ Internal prop delaysCS80CBI _______|______ inverter typically below 100pS. | => I | _____ _|_ _||<- | ___ __ | || /|\ Vout | |_ | ||__ ___| Signal Current Sees | ___| |__ Minimum Inductance | _|_ | __| | | /Vin\ | || | =>I ___________... | \___/ |_|| |_____| | | ||-> ________| Any Type of tran line |_____|_________| I<= |___________... <= I _|_ /// THIS IS WELL WORTH THE EFFORT. 1) High frequency energy completely confined a) This greatly reduces crosstalk Low RF radiation both in and out. b) Signals travel faster (No Inductors) speed light Efficient use of signal energy 2) termination ( Impedance at end =same as transmission Z) a) Keeps the ringing down. costto know where all signal current flows and simply provide a path of least inductance (Path of least indcutance = min loop area) Supply bypass ____________ 0.1uF | | _______| |_________________ _|_ |___| 1 6 |___| __________ 50_Ohms ___ VCC | SOT6 | CS | ______|_______| |_______\__________ | _|_ |___| 2 5 |___| /__________ 50_Ohms _|_ ___ GND | | DOUT | \ / |_________| |_______\__________ V Local | |___| 3 4 |___| /__________ 50_Ohms Gnd | VIN | | CLK | _ | |____________| _|_ | |_/\ ___| adcv0831 \ / Local |_| \/ V Gnd use of this principle involved the building of first SOT23-6 tester for an ADC as is shown. In this case, the challenges were.. A) no separate analog and digital ground.ran out of pins B) 3 foot cables connect the Eagle tester to the SOT23-6 handler. A standard cable size is 50 ohms change thinking for TD using outer shield connector as current return paths, In the RF world, if you can't transmit, you also can't receive. tansmission line mentality introduces Electromagnetic shielding as well as Electrostatic shielding. Travelling wave string on tension tension s mass_per_unit u velocity sqrt(tension/mass_per_unit) ------------------------------------------------------------- H ^ /|\- | - |______\ E \ /- \ - \ - - \ - \ E /______\ -\ |\ - | \ -\|/ \ V H \- _\| S Electrical c =sqrt(1/(e0*u0)) phi=u0*H eo*E Z = E/H = sqrt(u0/e0) = 377ohms ------------------------------------------------------------- Maxwell's Equations __ \/ dot J = -delta_rho/dt __ \/ cross E = -delta_B/dt V=delta_Phi/dt __ \/ cross H = J+ delta_D/dt H =I*N __ \/ dot D = rho __ \/ dot B = 0 -----------------------SUPPLY_BY_PASS----------------------- ______________ | |__| | ___| |___ |___| 1 VCC 14 |___|______ | | | .1uF ___| |___ _|_ |___| 2 13 |___| ___ | | | ___| |___ _|_ |___| 3 12 |___| /// | | ___| |___ |___| 4 11 |___| | | ___| |___ |___| 5 10 |___| | | ___| |___ |___| 6 9 |___| | | ___| |___ ___|___| 7 GND 8 |___| | | | _|_ |______________| /// world of supply supply bypass decoupling has been changing due to the ever higher and higher speeds at which digital circuit can now operate. A standard off_the_shelf digital circuit with its supply by_pass is shown above. At speeds which digital circuits using CS80CBI process at, this 0.1uF capacitor really is not a capacitor. SELF INDUCTANCE (uH) for ROUND WIRE (L_cm,R_cm) __\_ H | / | _______________|_____________ ()_____________\|/____________) R_cm --> I | V L_cm |_/_ \ L_uH=.002*L_cm*( ln(2*L_cm/R_cm) -.75 ) Inductances in the nanoHenry range are not hard to come by given the package lead sizes. equation for single wire inductance is give above. IMPEDANCE OF 0.1uF CAPACITOR 10 |..C............................................. | C . . . L . | C . . L . | C . L . . | C. . L . . | C . L . . 1 |..............C...........L..................... | . C .L . . | . C L . . | . C L. . . | . R . . . | . . . . .1 |................................................ | . . . | . . . . | . . . . | . . . . | . . . . |______________________________________________ . 100KHz 1MHz 10MHz 100MHz 1GHz The 0.1uF supply bypass capacitance may have the impedance as is shown above. Above a certain frequency, the size of the capacitor no longer matters. Rather, it is how the capacitor is connected that defines the impedances. _______\_____ I <--- | / | ____________|____ | | \ | _____ /|\ \ \|/ H Field \ | __\__ | \ | _____ | Magnetic Radiation \ | \ | \ ---> I \ | \____________\ | | | |______/______| \ Current really always flows in loops. Wher current flows, a magnetic field. However it is the Area which current loop encircles determins how much magnetic field gets radiated. To have inductor, you need space for magnetic field. Little space, little magnetic field and inductance. SELF INDUCTANCE (uH) for a RING OF WIRE (R_cm, Dia_inch) _ _ Dia_inch | | | | _| | | |_ / __| |__ \ / / ^ \ \ R_inch/Dia_inch > 2.5 / / /|\ \ \ | | | | | | | R_inch | | \ \ / / \ \__ __/ / \_ --- _/ ------- L_uH (R_inch/100)*( 7.353*log(16*R_inch/Dia_inch) -6.386) equations of inductance for a ring of wire is given above. If current loop is made to runthat currents flowing in opposite directions are next to each other, you get a transmission line. ___\__ --> I _______ ______|___/__|______________| | ->()___________\|/_____________| | |_/____| _/___ |50Ohms | ________\_______|_\___|_____| | <-()_____________________|_____| | <-- I |___\_| |_______| / From outside world, transmission lines have magnetic fields cancelling out. In terms loop current, impedances look like a resistor. Depending on spacing of the wires,it is common to set this resistance value to be typically 50 ohms. Without inductance, the moment you apply signal voltage, you instantly get signal current. Things happen close to speed of light. Voltage dropping current \ -----> \ \____\ \______\ \ \ \ \ \ \ _ \ _ ->E \| \ \ _ \| \ \|/ _ /_ \ ^ _ /_ v H |_/ \ _V Voltage \ |_/ \ __/_ \ Pulse \ __/_ |_/ \ \ \ |_/ \ | _V | voltage \ \ - + \ \____\ \______\ \ \ \ \ \ ---> \ _ \ _ | E \| \ \ _ \| \ \|/ _ /_ \ ^ _ /_ v H |_/ \ _V \ |_/ \ __/_ \ __/_ ->E |_/ \ \ |_/ \ \|/ | | v H \ \ \ \____\ \______\ \ \ \ \ current <------ Voltage increasing w0=sqrt(L*C) = sqrt(dL*dC*X^2) =X*sqrt(dL*dC) velocity=(2*pi()/sqrt(dL*dC) transmission linethought of as a "wave_guide". E and H field travel down the such the E field is set by signal voltage and H field is set by signal current.Given transmission line typically 50 ohms, power, voltage, current, etc are all defined. Using a transmission line really win/win situation. you remove inductance from equation which greatly speeds things up. also prevent electromagnetic radiation leaving or entering circuit. you confine the signal energy to flow where you want, everything gets better. ^ VCC 0.1uF /_\ Internal prop delays for a CS80CBI _______|______ inverter typically typically 100pS | => I | _____ _|_ _||<- | ___ __ | || /|\ Vout | |_ | ||__ ___| Signal Current Sees | ___| |__ Minimum Inductance | _|_ | __| | | /Vin\ | || | =>I ___________... | \___/ |_|| |_____| | | ||-> ________| Any Type of trans line |_____|_________| I<= |___________... <= I _|_ /// Inside chip, things can now happen very fast. prop delays for inverter CS80CBI typically under 100 picoseconds! 10GHz bandwidth. If digital circuit is connected to outside world shown above, current loop can be confined in space consisting of transmission line and supply bypass capacitor. to make this whole path look real in impedances by having currents flowing in opposite directions next to each other so that their magnetic fields cancel out. Another way say it to make lead lengths of bypass capacitor as short as possible. Supply bypass ____________ 0.1uF | | _______| |_________________ _|_ |___| 1 6 |___| __________ 50_Ohms ___ VCC | SOT6 | CS | ______|_______| |_______\__________ | _|_ |___| 2 5 |___| /__________ 50_Ohms _|_ ___ GND | | DOUT | \ / |_________| |_______\__________ V Local | |___| 3 4 |___| /__________ 50_Ohms Gnd | VIN | | CLK | _ | |____________| _|_ | |_/\ ___| adcv0831 \ / Local |_| \/ V Gnd This relationship ofsupply bypass to signal path ofADC seen in the lab using the test circuit shown Because of ESD requirements, all pad are going to have around 5pF of capacitance. every digital input or output pin going to cause some supply current to flow through 0.1uF capacitor and ADCV0831's ground lead. , sample and hold only has access to internal ground node to which references its Analog input voltage. When change lead length of supply bypass capacitor, can modulate analog input voltage which is what graph shows below. First Code Transition ^ /_\ | 2.2_lsb _| * * | * 2.1_lsb _| * | * * 2.0_lsb _| | * * 1.9_lsb _| | * 1.8_lsb _| * * | |______________________________________|\ | | | | | | | |/ 2mm 3mm 4mm 5mm 6mm Supply bypass Lead length This type input modulation consistant. should not effect linearity of the ADC. present test system may have linearity modulating error. major code transition on test is coming close to having missing code while the lab set up does not correlate This error only possible if the Data output signal from DUT is getting back into the analog signal line. ____ Sample and Hold 10nsec after this edge _|_ \ / V ____ CHIP SELEC ___ |_______________________________________________| _ __ __ __ __ __ __ __ __ __ __ | | | | | | | | | | | | | | | | | | | | | | | V | V | V | V | V | V | V | V | V | CLOCK |_| |_| |_| |_| |_| |_| |_| |_| |_| |_| |__ ________________________________________ | | | | | | | | | ............|Bit7|Bit6|Bit5|Bit4|Bit3|Bit2|Bit1|Bit0 |... TRI STATE |____|____|____|____|____|____|____|_____| TRI STATE Data Output signal comes well after external analog voltages capture, reducting feedback from output to input should not be that hard. input sample and hold circuit need an external digital signal to tell when to capture analog signal. Care taken in design to have few of things happening possible inside die untilsample and hold process is done. still youhave a digital input signal and how much time to wait until you sample? Supply bypass ____________ 0.1uF | | _______| |_________________ _|_ |___| 1 6 |___| __________ 50_Ohms ___ VCC | SOT6 | CS | ______|_______| |_______\__________ | _|_ |___| 2 5 |___| /__________ 50_Ohms _|_ ___ GND | | DOUT | \ / |_________| |_______\__________ V Local | |___| 3 4 |___| /__________ 50_Ohms Gnd | VIN | | CLK | _ | |____________| _|_ | |_/\ ___| adcv0831 \ / Local |_| \/ V Gnd offical end point offset will probably be defined as what part measures in a common lab set up. On SOT production tester, things more challenging in that nothing can come up close to DUT. all cable are at least 2 feet long. have to generate a local ground close to DUT such that really apply an accurate Analog input voltage. capacitor form local ground to input appears to common mode out most of trouble input pin draws no current and some resistors can be added if they help. Hopefullycan geterrors small or at least consistant. best to spec end point offset numbers separately on the spec so they can be handled independently.