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Understanding Gate Delay: Voltage Rise, Capacitance, and Resistance

Learn about gate delay in electric circuits, calculating delays using RC analysis, electric charge, current-voltage relationships, current density, and conceptual issues related to current flow. Develop a solid foundation in electric circuits.

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Understanding Gate Delay: Voltage Rise, Capacitance, and Resistance

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  1. Lecture 4 • Last time • Gate delay: why it happens (Capacitance! Resistance!) • Voltage rise and fall through logic circuits is gradual • t (delay between input and output at 50% of final value) found graphically • Used clock signal to prevent false output (due to signals delayed different amounts) • Now we will • Get to calculating delays for circuits (RC analysis) • Review fundamental knowledge of electric circuits

  2. CHARGE Most matter is macroscopically electrically neutral most of the time. Exceptions: clouds in thunderstorm, people on carpets in dry weather, plates of a charged capacitor… Microscopically, of course, matter is full of charges. The application of an electric field causes charges to drift, or move. Electrons will naturally move from lower electric potential to higher potential. The rate at which the charges move depends on the magnitude of the potential difference and the properties of the matter. CURRENT – VOLTAGE RELATIONSHIP

  3. = 1 Ampere flow of 1 Coulomb per second æ ö dq C = ç ÷ i (A) dt S è ø MOVING CHARGE Charge flow Current Charge storage  Energy Definition of Current where q is the charge in Coulombs and t is the time in seconds

  4. Wire attached to ends CURRENT & CURRENT DENSITY Semiconductor with free electrons Define Current Density by Example: • Suppose we force a + 1A current from C1 to C2. Then • Current of 1A flows in the semiconductor (in +X direction) • Current Density = Current / (Cross-Sectional area current goes thru) • Current density of above = 1 A / (2 cm • 1 cm) = 0.5 A/cm2 • Electron flow (in negative X direction) consists of

  5. Note: Typical dimensions of integrated circuit components are in the range of . What is the current density in a component with [1m]² area carrying 5A? Answer CURRENT & CURRENT DENSITY (cont.) • A wire carrying 1 amp has a cross-section, then the current density is

  6. POSSIBLE CONCEPTUAL ISSUES Remember a wire has a huge number of free carriers moving very fast but randomly (because of thermal energy) Drift concept: Now add even a modest electric field Carriers “feel” an electric field along the wire and tend to drift with it (+ sign charge) or against it ( charge carrier). This drift is still small compared to the random motion. • How does charge move through the wire? • Sign of the charge carriers: It is often negative (for metals); in silicon, it can be either negative or positive…we don’t care for circuit analysis.

  7. POSSIBLE CONCEPTUAL ISSUES (con’t) 3. Electric field is in the direction positive charge carriers move. Thus if we have an electrical field in the Z direction, positive charges (ions, positrons, whatever) will experience a force in the positive Z direction. Negatively charged particles will experience the force in the –Z direction. Thus the free carriers we will be concerned with (electrons with negative charge and “holes” with positive charge) will move against and with the electric field respectively. 4. Field or Current can have positive or negative sign. Examples: I = 0.2 mA same as I = 0.2 mA   going left to right going right to left

  8. V I + | REFERENCE DIRECTIONS A question like “Find the current” or “Find the voltage” is always accompanied by a definition of the direction: In this example if the current turned out to be 1mA, but flowing to the left we would merely say I = -1mA. To solve circuits, you may need to specify reference directions for currents. But there is no need to guess the reference direction so that the answer comes out positive….Your guess won’t affect what the charge carriers are doing!

  9. Vab | + VOLTAGE “Vab” means the potential at a minus the potential at b. Voltage is the difference in electric potential between two points. Potential is always defined by two points. We can use the subscript convention above to define a voltage between two labeled points (e. g. a and b above), Vx or draw a + and – indicating polarity.

  10. Answer: terminal b Now you make a change. What would this circuit measure? • Answer: +1.401V Note that we have used “ground” symbol ( ) for reference node on DVM. Often it is labeled “C” or “common.” SIGN CONVENTIONS Suppose you have an unlabelled battery and you measure its voltage with a digital voltmeter. It will tell you magnitude and sign of the voltage. With this circuit, you are measuring (or ). DVM indicates 1.4, so by 1.4 V. Which is the positive battery terminal?

  11. Example 1 B C A D + + + + + + 9V 1.5V 1.5V How are single-subscript voltages related to double-subscript voltages? Example 2 B C A D 9V 1.5V 1.5V - + V1 - + VX Lets put a bunch of batteries, say 1.5V and 9V in series to see what we already know about sign conventions: SIGN CONVENTIONS (cont.) How is VAD related to VAB, VBC etc ? VAB = -1.5, VBC = -1.5, and VAC = -3 so clearlyVAC =VAB+VBC etc. Answer: Clearly A is above D by 9V – 3V or 6V. That is consistent with VAB+VBC + VCD= -1.5 –1.5 + 9 = 6V What is VAD ? Answer: Clearly V1 = 1.5 = - VAB = VBA VX = -6 = VDA

  12. Example 2 Solve for ix • Note that in this case, the current flows counterclockwise, i.e., opposite to arrow defining • Here Just as in the physical world, we do not know a priori the sign (or magnitude) of voltages or currents; we don’t know them in the theoretical world until solving for them, so we just define voltages and currents and accept that half of the time they will be negative (quite analogously to probing with a DC meter). SIGN CONVENTIONS (cont.) Solve for Example 1 • Remembering Ohm’s law, • Here

  13. - + 2 Using sign conventions: V a c + + v 1 V cd - - b d - + v bd KEEPING THE VOLTAGE SIGNS STRAIGHT Labeling Conventions • Indicate + and  terminals clearly; or label terminals with letters • The + sign corresponds to the first subscript; the  sign to the second subscript. Therefore, Vab = - Vba Note: The labeling convention has nothing to do with whether or not v > 0 or v < 0 Obviously Vcd +Vdb = Vcb too. Then if Vbd = 5V, what is Vcd? Answer: Vcd -5 = -1 so Vcd = 4V

  14. POWER IN ELECTRIC CIRCUITS Power: Transfer of energy per unit time (Joules per second = Watts) Concept: in falling through a positive potential drop V, a positive charge q gains energy • potential energy change = qV for each charge q • Rate is given by # charges/sec Power = P = V (dq/dt) = VI P = V  I Volt  Amps = Volts  Coulombs/sec = Joules/sec = Watts Circuit elements can absorb or release power (i.e., from or to the rest of the circuit); power can be a function of time. How to keep the signs straight for absorbing and releasing power? + Power  absorbed into element  Power  delivered from element

  15.  i “ASSOCIATED REFERENCE DIRECTIONS” It is often convenient to define the current through a circuit element as positive when entering the terminal associated with the + reference for voltage Here, I and V are “associated” + - Circuit element • For positive current and positive voltage, positive charge “falls down” a potential “drop” in moving through the circuit element: it absorbs power • P = VI > 0 corresponds to the element absorbing power if the definitions of I and V are associated. How can a circuit element absorb power? By converting electrical energy into heat (resistors in toasters); light (light bulbs); acoustic energy (speakers); by storing energy (charging a battery). Negative power  releasing power to rest of circuit

  16. So, (delivered out of battery) “ASSOCIATED REFERENCE DIRECTIONS” (cont.) We know i = +1 mA A) For Resistor: P = i vR (because we are using “associated reference directions”) 1.5K Hence, for the resistor, P = i vR = +1.5mW (absorbed) B) For battery, current i is opposite to associated

  17. “ASSOCIATED REFERENCE DIRECTIONS” (cont.) Again, iR = 1mA therefore iB = -1mA B) Battery: iB and vB are associated, therefore P= iB vB . Thus A) Resistor: P = iB vB = +1.5mW

  18.  2 - V +  c a  + + + 3 mA 2.5 mA 3 1 V V 1 V 0.5 mA - - - b flip current direction: EXAMPLES OF CALCULATING POWER Find the power absorbed by each element Element  : Element  : Element  : Element  :

  19. Voltage Source Current Source Resistor Capacitor Inductor (like ideal battery) (always supplies some constant given current) (Ohm’s law) (capacitor law – based on energy storage in electric field of a dielectric) (inductor law – based on energy storage in magnetic field produced by a current) BASIC CIRCUIT ELEMENTS

  20.  Current through voltage source can take on any value (positive or negative) but not infinite DEFINITION OF IDEAL VOLTAGE SOURCE Symbol Note: Reference direction for voltage source is unassociated (by convention) Examples: 1) V = 3V 2) v = v(t) = 160 cos 377t • Special cases: upper case V  constant voltage … called “DC” lower case v  general voltage, may vary with time

  21. IDEAL CURRENT SOURCE “Complement” or “dual” of the voltage source: Current though branch is independent of the voltage across the branch + note unassociated v direction  Actual current source examples – hard to find except in electronics (transistors, etc.), as we will see upper-case I  DC (constant) value lower-case implies current could be time-varying i(t)

  22. Ideal voltage source Assume unassociated signs   If V is positive and I is only positive  But this is arbitrary; i might be negative so we extend into 2nd quadrant i absorbing power releasing power V  But this is still arbitrary, V could be negative; all four quadrants are possible releasing power absorbing power (charging) CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES Describe a two-terminal circuit element by plotting current vs. voltage

  23. Ideal current source + v    releasing power absorbing power i V CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES (con’t) If i is positive then we are confined to quadrants 4 and 1: Remember the voltage across the current source can be any finite value (not just zero) And do not forget i can be positive or negative. Thus we can be in any quadrant.

  24. Answer: V = 0  I = 0 V = 1V  I = 1 mA V = 2V  I = 2 mA etc Slope = 1/R RESISTOR If we use associated current and voltage (i.e., i is defined as into + terminal), then v = iR (Ohm’s law) Question: What is the I-V characteristic for a 1K resistor? Draw on axis below. Note that all wires and circuit connections have resistance, though we will most often approximate it to be zero. But we can (and do) deliberately construct circuit elements with some desired resistance, even very large values such as 10M.

  25. We learned about the parallel-plate capacitor in physics. If the area of the plate is A, the separation d, and the dielectric constant of the insulator is , the capacitance equals C = A /d. Symbol Constitutive relationship: Q = C (Va Vb). (Q is positive on plate a if Va > Vb) But so where we use the associated reference directions. CAPACITOR Any two conductors a and b separated by an insulator with a difference in voltage Vab will have an equal and opposite charge on their surfaces whose value is given by Q = CVab, where C is the capacitance of the structure, and the + charge is on the more positive electrode.

  26. Thus, energy is . ENERGY STORED IN A CAPACITOR You might think the energy (in Joules) is QV, which has the dimension of joules. But during charging the average voltage was only half the final value of V.

  27. ENERGY STORED IN A CAPACITOR (cont.) More rigorous derivation: During charging, the power flow is v  i into the capacitor, where i is into + terminal. We integrate the power from t = 0 (v = 0) to t = end (v = V). The integrated power is the energy but dq = C dv. (We are using small q instead of Q to remind us that it is time varying . Most texts use Q.)

  28. Inductors are the dual of capacitors – they store energy in magnetic fields that are proportional to current. INDUCTORS Switching properties: Just as capacitors demand v be continuous (no jumps in V), inductors demand i be continuous (no jumps in i ). Reason? In both cases the continuity follows from non-infinite, i.e., finite, power flow.

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