# Ecalc, some calculators for basic electronic

Program under GNU GPL License V3. No warranty is given by the author on the correction of the results, nor anything.
This page is not a professionnal page; read it with a critical mind. Any problems, bug or error can be reported here. - CHANGES
Help me to fix the problems, help the community, and finaly help you. Howto help

## Calculators

Display / Hide all schemas. | Power = watts | U = volts | R = $$\Omega$$ | I = amperes | C = F (or µF (noted)) | f = Hz | T = tolerance
Resistors: Calculation with the lowest tolerance $$R-=R-\left(\frac{R \times T}{100}\right)$$ | Calculation with the highest tolerance $$R+=R+\left(\frac{R \times T}{100}\right)$$
NB: a numeric value >= 0 is required in the input fields.
$$R$$
$$I$$ -
$$U$$
$$I=\frac{U}{R}\ \text{or}\ R=\frac{U}{I}$$
NaN or Infinity in I or R? Check if $$R\ \text{or}\ I>0$$
- Top
$$U$$
$$R\ \text{or}\ I$$ -
$$I\ \text{or}\ R$$
$$U$$
$$I$$ -
$$P$$
$$U=\frac{P}{I}\ \text{or}\ I=\frac{P}{U}$$
NaN or Infinity in U or I? Check if $$I\ \text{or}\ U>0$$
- Top
$$P$$
$$I\ \text{or}\ U$$ -
$$U\ \text{or}\ I$$
$$U$$
$$P\ \text{or}\ R$$ -
$$R\ \text{or}\ P$$
$$20\%\ T\ =\$$No tolerance color
$$R$$

$$T$$ -
$$R-$$ $$R+$$
Resistors in serie (Value 0 = no resistor).
$$eqR_s = \sum_{n=1}^{n=5}R_n = R_1 + R_2 + R_3 + R_4 + R_5$$ - Top
$$R_1$$

$$T$$
$$R_2$$

$$T$$
$$R_3$$

$$T$$
$$R_4$$

$$T$$
$$R_5$$ -

$$T$$
$$eqR_s$$
$$eqR-$$ $$eqR+$$
Resistors in parallel (Value 0 = no resistor). The division by 0 is handled.
$$\frac{1}{eqR_p} = \sum_{n=1}^{n=5}\frac{1}{R_n} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$$ - Top
$$R_1$$

$$T$$
$$R_2$$

$$T$$
$$R_3$$

$$T$$
$$R_4$$

$$T$$
$$R_5$$ -

$$T$$
$$\sum_{n=1}^{n=5}\frac{1}{R_n}$$
$$eqR_p$$
$$eqR-$$ $$eqR+$$
$$\Delta;U > 0$$ and $$f > 0$$ are required, the division by 0 is handled.
$$C_{\mu F}=\frac{I}{\Delta U \times f} \times 10^6$$ - Top
$$I$$
$$\Delta U$$
$$f$$ -
$$C_{\mu F}$$
Capacitors in serie (Value 0 = no capacitor). The division by 0 is handled.
$$\frac{1}{eqC_s} = \sum_{n=1}^{n=5}\frac{1}{C_n} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4} + \frac{1}{C_5}$$ - Top
$$C_1$$
$$C_2$$
$$C_3$$
$$C_4$$
$$C_5$$ -
$$\sum_{n=1}^{n=5}\frac{1}{C_n}$$
$$eqC_s$$
Capacitors in parallel (Value 0 = no capacitor).
$$eqC_p = \sum_{n=1}^{n=5}C_n = C_1 + C_2 + C_3 + C_4 + C_5$$ - Top
$$C_1$$
$$C_2$$
$$C_3$$
$$C_4$$
$$C_5$$ -
$$eqC_p$$
Voltage divider. The division by 0 is handled.
$$U_2 = U\times \left(\frac{R_2}{R_2+R_1}\right)$$ - Top
$$U$$
$$R_1$$

$$T$$
$$R_2$$ -

$$T$$
$$U_2$$
$$U_2=f(R1+, R2-)$$ $$U_2=f(R1-, R2+)$$
Note: With the formula, when we take the lowest tolerance for R1 AND R2, this will tend to annul the total effect of the two tolerances. Ditto for the highest tolerance. For an equal tolerance, the total effect is completely canceled. So, to get the most significant differences, we first compute with the highest tolerance for R1 and the lowest it for R2, then the opposite.
Current divider. The division by 0 is handled.
$$I_1 = I\times \left(\frac{R_2}{R_2+R_1} \right)$$ | $$I_2 = I\times \left(\frac{R_1}{R_2+R_1} \right)$$ - Top
$$I$$
$$R_1$$

$$T$$
$$R_2$$ -

$$T$$
$$I_1$$
$$I_1=f(R1+, R2-)$$ $$I_1=f(R1-, R2+)$$
$$I_2$$
$$I_2=f(R1+, R2-)$$ $$I_2=f(R1-, R2+)$$
Note: With the formula, when we take the lowest tolerance for R1 AND R2, this will tend to annul the total effect of the two tolerances. Ditto for the highest tolerance. For an equal tolerance, the total effect is completely canceled. So, to get the most significant differences, we first compute with the highest tolerance for R1 and the lowest it for R2, then the opposite.

## Howto help

Top
• English is not my mother tongue. Correct my grammar and spelling, or the way I write English.
• Use a calculator to check if the calculations of Ecalc are correct.
• Send us suggestions or a good idea. Etc. ;)