Repeating t register usage Message #1 Posted by Sam Levy on 10 Apr 2007, 9:00 a.m.
HEWLETT PACKARD CALCULATIONS
I used HP RPN pocket calculators for electronic solutions, and I wanted to find a fast way to solve common problems using as few keystrokes as possible, and getting maximum information during the process. This method uses the repeating t register. Formulas now in use were derived for pen and pencil mathematics, totally outmoded by the calculator. This is the keystroke procedure for quick solutions. With programmable calculators I added a small program to enter 3 times to fill the stack or clear it using zero entry.
Resistive Dividers; when calculating from known resistors, enter the lower resistor repeatedly. Key in the top resistor, and add, the result is the total resistance, then divide, the result is the fractional output voltage of the divider. Then multiply and subtract to see the output resistance from the divider tap.
To design such a network, key in the output resistance desired, and enter repeatedly. Key in the division fraction, and divide to see the upper resistance. Divide again to restore the fraction, key 1, subtract, change sign, divide, and see the lower resistor. The relation between the upper resistor and the output resistance is the fraction of the divider. 1 minus the fraction is the relation between the output resistance and the lower resistance. Given the fraction and either resistor, the other value may be found.
I have used this method to find the added end resistors for setting particular potentiometer control ranges. The difference between the two potentiometer fractions desired at the potentiometer ends, relates the potentiometer resistance to the total resistance. Thus a particular value of potentiometer may be used to obtain the desired range of fractions.
EXAMPLE: It was desired to use a 1K potentiometer to adjust a power supply to vary between 24 and 16 volts. The IC regulated at 4.75 volts from the potentiometer. It is desired to calculate the end resistors needed to set the correct range. 4.75/24 = 0.2 store 1. 4.75/16 = 0.3 store 2. Recall 1, subtract, see 0.1, key 1000 swap X&Y, divide, see 10105, the total resistance. Enter several times. Recall 1, multiply, see 2000, the resistance when the pot is at the lower extreme, add 2000 ohms to the lower end of the pot. Cl X, recall 2, multiply, see 3000, this is the sum of the pot resistance and the pot lower end resistor. The upper end resistor must be 10205 minus 3000 = 7105. Some compromise values may be used.
Transistor bias; I was solving for a transistor bias divider, and wondered how to account for the base current drop in the divider. I thought to divide the divider output resistance by Beta and add it to the actual emitter resistor. To do this procedure, solve the bias divider resistors, storing the divider fraction. With R0 in the register key Beta, divide, and see the emitter resistor equivalent to be added. Key the actual emitter resistor and add, see the total emitter resistor, accounting for the bias source drop. Key the source voltage; recall the fraction, multiply and see the divider voltage. Key 0.7, subtract, see the emitter voltage. Interchange X&Y and divide, to see the emitter current: check by multiplying by the collector resistor, to see the collector resistor drop.
Summing network; The resistive divider can be extended to a more general case. I needed to design a telemetry output that combined several signals, to a common output, having a 10 K output resistance. The signal of interest was a +/ 12 volt VCO control voltage. The telemetry output was to be 05 volts, using the range 0.5 to 4.5 volts normally. The centering voltage was 2.5 volts from a regulated 12 volts supply. A third resistor to ground gave the desired output resistance. I found a general solution. Each resistor is determined by the fraction of its input voltage contributed to the output sum, thus the telemetry signal was to output 1/6th of its signal. The resistor was then 10K times 6 or 60K. The centering voltage was 2.5 volts derived from +12, a ratio of 4.8 times, 48K. The third resistor was found by subtracting the other two fractions from 1: 11/61/4.8, a fraction of 0.625 and dividing the output resistance to get a resistance of 16K. Some compromise values were used, which proved satisfactory.
RC frequency corners; the process may be simplified as 2PiFRC=1. The known values are multiplied, and the product inverted to give the missing element. For constant frequency solutions enter 2PiF repeatedly.
Tuned circuits; here intermediate results in the solution give other useful information. Key in L, then C, and store C in a register, divide and take the square root; this is the impedance of the resonant elements. Recall the capacitance value and multiply, then multiply by 2Pi and invert to see the frequency in Hertz. This is the frequency corner formula using R0 as the R value. Q if known may be used to compute the resonant impedance, or the Q may be found by dividing the parallel circuit damping resistance by the impedance, or the impedance by the series circuit resistance. The square root of L/C is also the output resistance of a simple LC filter, such as a power supply, and can be used to estimate transient voltages from current changes.
Resonant values; When using a constant frequency, resonant values may be calculated quickly. Key in 2PiF and enter it repeatedly. Then key in either L or C , multiply twice, and invert to see the other resonant circuit element. When 2PiFL is shown, it is the reactance of the circuit values. This procedure gives quick trials of various circuit values, give impedance values for filter and bypass elements, and value choices for a given Q.
Ohms law; this may seem simple, but can be used as a rapid verification of dissipation and the voltage and current relation. Remember the little circles of voltage E divided by the IR product. Key in the voltage repeatedly, and divide by either I or R to see the other value, as in E=IR. The power in a resistor can be found by keying in the resistor and dividing to get I, then multiplying to get EI. It can be used to solve W=EI, by keying in power repeatedly, and dividing to get either E or I.
These methods allow such rapid solutions; they will increase your ability to investigate possibilities in your designs. Practice will allow fast and sure answers to a wide range of problems. Familiarity will allow you to extend these methods to your particular problems. I trained one tech to calculate time payments faster than they could be found in a book.
E. Samuel Levy, 754 Temple St. San Diego 92106, 6192236292 designnut@cox.net
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