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 This tutorial is dedicated to use a 8051 based Microcontroller core and program it using keil's C51 tools on the Proteus Source code editor.  1- Brief History of 8051 one of the first microprocessors 4004 was invented by Intel Corporation as well as  8085 and 8086 microprocessors back in 1981,shortly after Intel introduced an 8-bit microcontroller called the 8051 . It was referred to as system on a chip because it had 128 bytes of RAM, 4K byte of on-chip ROM, two timers, one serial port, and 4 ports (8-bit wide), all on a single chip. When it became so widespread, Intel allowed other manufacturers to make and market different flavors of 8051 with its code compatible with 8051. It means that if you write your program for one flavor of 8051, it will run on other flavors too, regardless of the manufacturer. This has led to several versions with different speeds and amounts of on-chip RAM. 2- Features of AT892051 Microcontroller Compatible with MCS®-51Products 2K ...
  Direct Memory Access (DMA) is a feature of computer systems that allows hardware subsystems to access main system memory (RAM) with minimal intervention from the Central Processing Unit (CPU). The Direct Memory Access (DMA) hardware peripheral on PIC ® microcontrollers (MCUs) allows for direct memory-to-memory and peripheral-to-memory transfers and concurrent operation of the DMA and the CPU (while sharing system bus resources). This enables more flexible usage of eXtreme Low Power (XLP) technologies through core-independent movement of data in low-power modes and improvements in data throughput and latency.     Why Is DMA Used? Most CPU operations in MCU-based systems involve the movement of data. In many cases, this can significantly limit the throughput and responsiveness of a system. For example, an interrupt can cause the CPU to stop what it is currently doing to service the interrupt. In low-power applications, this can significant...
You want to heat your small garage using a couple of electric radiators. The power and voltage requirements for each radiator are 1200 W, 240 V. But you are not sure how to wire the radiators to the power supplied to the garage. Should you use the wiring diagram on the left or the one on the right? Does it make any difference?
Assume that the voltage at the terminals of the element in Fig. 1.5, whose current was defined in Assessment Problem 1.3, is v=0t<0;v=10e−5000t kV, t≥0. 1. Calculate the power supplied to the element at 1 ms. 2. Calculate the total energy (in joules) delivered to the circuit element. Solution 1. Since the current is entering the + terminal of the voltage drop defined for the element in Fig. 1.5, we use a “+” sign in the power equation. p = vi = (10,000e−5000t)(20e−5000t) = 200,000e −10,000t W.p(0.001)= 200,000e − 10,000(0.0001) = 200,000e −10= 200,000(45.4×10−6) = 9.08 W. 2. From the definition of power given in Eq. 1.3, the expression for energy is w(t)=∫0tp(x)dx. To find the total energy delivered, integrate the expresssion for power from zero to infinity. Therefore, wtotal=∫0∞200,000e−10,000x dx=200,000e−10,000x−10,000|0∞ =−20e−∞−(−20e−0 )=0+20=20 J. Thus, the total energy supplied to the circuit element is 20 J. Assessment Problems Objective 3—Know and u...
Whenever the reference direction for the current in an element is in the direction of the reference voltage drop across the element (as in Fig. 1.5 ), use a positive sign in any expression that relates the voltage to the current. Otherwise, use a negative sign. We apply this sign convention in all the analyses that follow. Our purpose for
p=vi,(1.4) where p=the power in watts,v=the voltage in volts,i=the current in amperes. Equation 1.4 shows that the power associated with a basic circuit element is the product of the current in the element and the voltage across the element. Therefore, power is a quantity associated with a circuit element, and we have to determine from our calculation whether power is being delivered to the circuit element or extracted from it. This information comes from correctly applying and interpreting the passive sign convention . If we use the passive sign convention, Eq. 1.4 is correct if the reference direction for the current is in the direction of the reference voltage drop across the terminals. Otherwise, Eq. 1.4 must be written with a minus sign. In other words, if the current reference is in the direction of a reference voltage rise across the terminals, the expression for the power is p=−vi. The algebraic sign of power is based on charge movement through voltage drops an...
A diagram shows a rectangle with two terminals marked as 1 and 2 at its ends. Positive voltage is indicated as v, with polarities at 1 and 2 marked as plus and minus, respectively. A right-facing arrow pointing at terminal 1, and labeled as i, represents positive current in the circuit. Power in the circuit is represented as "p = vi". A diagram shows a rectangle with two terminals marked as 1 and 2 at its ends. Positive voltage is indicated as v, with polarities at 1 and 2 marked as plus and minus, respectively. A left facing arrow pointing at terminal 2, and labeled as i, represents negative current in circuit. Power in the circuit is represented as "p = minus vi". A diagram shows a rectangle with two terminals marked as 1 and 2 at its ends. Negative voltage is indicated as v, with polarities at 1 and 2 marked as minus and plus, respectively. A right-facing arrow pointing at terminal 1, and labeled as i, represents positive curre...
p=dwdt, (1.3) where p=the power in watts,w=the energy in joules,t=the time in seconds. Thus, 1 W is equivalent to 1 J/s. The power associated with the flow of charge follows directly from the definition of voltage and current in Eqs. 1.1 and 1.2, or p=dwdt=(dwdq)(dqdt),
Power and energy calculations are important in circuit analysis. Although voltage and current are useful variables in the analysis and design of electrically based systems, the useful output of the system often is