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EE calculations, some not obvious (reformatted)

Posted by Thomas Klemm on 26 May 2011, 5:14 p.m.

This is a reformatted copy of the article EE calculations, some not obvious.


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.

Resistive Dividers

When calculating from known resistors, enter the lower resistor repeatedly.

To design such a network, key in the output resistance desired, and enter repeatedly.

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.


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.

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.

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 0-5 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: 1-1/6-1/4.8, 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 2 Pi F R C = 1.
The known values are multiplied, and the product inverted to give the missing element.

For constant frequency solutions enter 2 Pi F repeatedly.

Tuned circuits

Here intermediate results in the solution give other useful information.

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.

This procedure gives quick trials of various circuit values, give impedance values for filter and by-pass 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 I R product.

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


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