HP EE calculations


I used HP pocket calculators for electronic calculations, and I needed 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 then in use were derived for pen and pencil mathematics, and have been totally outmoded by the calculator. These are keystroke procedures for quick solutions.
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.
To design such a divider, 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 either resistor and the fraction; the other value may be found. I have used this method to find the added end “stop” resistors for setting particular potentiometer control ranges. The difference between the end range of the potentiometer fractions relates the potentiometer resistance to the total resistance. Thus a particular value of potentiometer may be used.
An example of this solution: I chose to use a 1K potentiometer to adjust a power supply output, for use as a lamp dimmer. The control IC regulated at a voltage from the potentiometer of 4.75 volts. The voltage range was chosen as 16 to 24 volts output. Here we solve for the resistors to be added to the potentiometer to divide the output voltage to 4.75 volts which is the control voltage of an IC.
Key 4.75, enter, 16 divide, see the fraction 0.3, and store 0. Key 4.75, enter, 24 divide, see, 0.2, store 1, and subtract. See 0.10, this is the change in fraction caused by changing the potentiometer setting, store 2. Key 1000, recall 2, divide, see 10105 ohms, this is the total resistance of the circuit, enter it repeatedly. Recall 2, multiply, see 1000 as a check of the procedure. Clear X, recall 0, multiply, see 3000. Clear X, recall 1, multiply, see 2000, as a further check, it is 1000 ohms different from the previous solution, thus when the potentiometer is moved to the end, the resistance to ground changes by 1000 ohms. 2000 must be added from the lower end of the potentiometer to the ground return. To find the upper resistance, you can subtract 3000 from the total resistance, see 7105 ohms, this is the upper resistance added to the potentiometer circuit. Some compromise values may be chosen.
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, add 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; 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 volt, 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, enter repeatedly, and multiply by 6, see 60K. The centering voltage was 2.5 volts derived from +12, a ratio of 4.8 times, multiply to get 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, to get a resistance of 16K. Some compromise values were used, which proved satisfactory.
RC frequency corners; the process may be simplified in that 2PiFRC=1. The known values are multiplied together and the product inverted to give the missing element. For a constant frequency solution enter 2PiF repeatedly. For a variety of solutions 2Pi may be entered repeatedly.
Tuned circuits; intermediate results in the solution can give useful information. The square root of L/C is the impedance value of the component values alone at resonance. 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 Ro as the R value.
The square root of L/C is also the output resistance Ro, of a simple LC filter, such as a power supply filter, and can be used to estimate transient voltages caused by 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, gives impedance values for filter and by-pass elements, and value choices for a given circuit Q.
Ohms law; this may seem simple, but it 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 entering the power repeatedly, and dividing to get either E or I.
These methods allow rapid solutions; they will increase your ability to investigate possibilities in your choices. 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.
E. Samuel Levy, 754 Temple St. San Diego 92106 Tel 619-223-6292 e-mail designnut@cox.net



I read some lines of your post in a brief, but this was enough for me to understand this is not a post to be read in brief. I'm taking some notes about it, I'm reading it carefully, so I can understand your method.

Anyway, as far as I have already gone, I see that you use HP calculators the way I believe they must be used: with good sense. Knowing about their powerful resources will lead the user to easily find the answers. Also, using the basics with deep understanding will help the development of such descriptive methods.

I'm thanked for your time spent so I can see what you have been done. Sharing your method will, surely, help other users to use their calculator more efficiently.

Best regards

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