WizEIA

1. Introduction

During the design of a circuit for a PCB, the problem of selecting appropriate resistance, capacitance, or inductance values often arises. This issue is further complicated by the fact that not all values are available in the catalog. For instance, in an inverting operational amplifier, the ratio between two resistances determines the gain, which is expressed by the equation: K = R1/R2.

If the amplifier is required to have a gain of exactly 8, it is necessary to find two resistors that possess this precise ratio. One possible approach would be to fix R2, for example, at 1k, and select R1 that is closest to 8k. However, 8 is not a standard value and cannot be found even in the most comprehensive E192 series.

It is important to note that the E series were originally developed based on a logarithmic progression to accommodate manufacturing tolerances. The values in each series were spaced to ensure that the tolerance bands of adjacent values did not overlap.

The first common series were:

E6: 20% tolerance (6 values per decade)
E12: 10% tolerance (12 values per decade)
E24: 5% tolerance (24 values per decade)

Manufacturing processes improved and nowadays exists the E48 series (2%), the E96 series (1%) and the E192 (0.5%).

WizEIA solves the problem of finding the right resistor values. Using the example above, introducing the R1/R2=8 equation on the solver, it will find that the values 120k and 15k that exist on the E12 series have exactly 8 times ratio between them.

Reducing component diversity

Instead of adding a new resistor value, it is often more advantageous to combine existing values to obtain the new required value. The combination can be done by placing resistors in series or in parallel. There are a few advantages to this approach:

Logistics: It's easier to find stock of resistances on lower series. It is not so unusual to have more than 30 weeks of procurement time for some resistor values on the EIA192 series.
Manufacturing: Reducing the number of different resistance values on a board will reduce it's cost. On the production floor, the machines that place components on the PCB (pick and place) have limitations on the number of reels that they can hold. A design with many different denomination will have to be reloaded more times. The reloading procedure is manually operated and often takes more time than the component placement.
Design: Ironically the highest E-series (48, 96 and 192) don't have the most common values used on the market. Often better results are obtained using lower series because the original series were defined very intelligently, so to achieve precise ratios between values. For example on the summing circuit, better results are obtained with EIA24 than with EIA48 despite the EIA48 having more values.

Tolerances: There is often the misconception that combining resistor values leads to a higher tolerance errors. This is however not correct, as is demonstrated by the calculator below. Independently if you are making a series or parallel combinations, the resulting value doesn't have a worst tolerance. In fact, it improves in terms of distribution, regardless if you start with a uniform or gaussian distribution, especially if the two values are close.

Resistor 1: 1000 Ω Tolerance (%):	Resistor 2: 2000 Ω Tolerance (%):
Distribution:
Series: Min: Ω, Max: Ω, -Tol: %, +Tol: % Parallel: Min: Ω, Max: Ω, -Tol: %, +Tol: %
Series Combination
Parallel Combination

Derating:When combining resistors we are also dividing their load. This is helpful when resistors are subject to considerable power dissipation.

The disadvantage of combining resistor values is the amount of real estate that such solution takes. In the event that your design is space-constrained, it is advisable to select the precise value required from a higher series.

2. Simple Circuits using Resistors

2.1 Nearest Resistor value

Ohm's Law: Fundamental relationship between voltage, current, and resistance.

Key Formula: V = I · R

To find the closest resistor value on an EIA series use the operator = and a resistor identifier starting with the letter R. Example : R=8k

WizEIA accepts all mathematical operators when solving a resistor value.
The following equation is valid: R = 8V/1mA

Also, it should be noted that these decade quantifiers are also recognized.

Letter	Decade
T	Tera (1E+12)
G	Giga (1E+9)
M	Mega (1E+6)
k or K	Kilo (1000)
m	mili (0.001)
u or µ	micro (1E-6)
n	nano (1E-9)
p	pico (1E-12)
f	femto (1E-15)

Note that the qualifiers are case sensitive. 'm' means mili and 'M' means Mega

All units are just decorative and are ignored by the solver.

Examples of Use: Basic circuit design, power calculation.

2.2 Resistors in Series

When two or more resistors are placed in series, the equivalent resistance corresponds to the sum of all resistor values

Series Resistor: R_eq = R₁ + R₂ + ... + R_n

In WizEIA the + operator between two resistor identifiers corresponds to placing them in series. Example: R1+R2=1234

Using the Ohm's Law, the voltage in a given resistor is given by the following equation:

Voltage Divider: V_x = V_in · (R_x / (R₁ + ... + R_n))

Example: Vdd*R1/(R1+R2)=Vout

Voltage divider Uses: Sensor voltage scaling, reference voltage generation (comparators).

2.3 Resistors in Parallel

When two resistors or more resistors are in parallel, they form an equivalent resistor which value is given by:

Parallel Resistor: R_eq = 1/(1/R₁ + 1/R₂ + ... + 1/R_n)

In WizEIA, either the | or the // operator represents the parallel between resistor identifiers. Example: R1|R2=40.5k

The current in a given resistor is given by the following equation:

Current Divider: I_x = I_total · R_eq /R_x

Example: I*R1/(R1+R2)=Ix

Examples of Use: Current sensing, load sharing.

2.4 Mixed Networks

When evaluating formulas, the parallel operator "|" has precedence over the "+" operator.

Examples:
a) R1|R2+R3=1234
b) (R1+R2)|R3=1234

2.5 Setting Boundaries

Often resistor values must fall within a specific interval. Consider the following voltage divider where Vout must be 1.225V and the current on resistor R3 must be between 100uA and 1mA.

In this situation the following equations must be set:
(R2|R3) / (R1 + (R2|R3)) = 1.225V
1.225 / R3 < 1mA
1.225 / R3 >= 100uA

Applying > and < operators can be a bit cumbersome, so, this form is also accepted:
(R2|R3) / (R1 + (R2|R3)) = 1.225V
R3:[1.225k 12.250k]

All the following is accepted:

Rname:E48[1k..20k]
Cname:E12
Cname:[10n..100n] # default is 12
Cname:E24 < 1u
Cname:EIA24[..1u] # EIA and E are equivalent
Lname:E6[1m 10m] # spaces, : or - can also be used

2.6 RC Networks

WizEIA can be used to find capacitor values. The capacitors are identified with the prefix C.
The series used for capacitors is EIA12 being the minimum 1pF and maximum 1mF.

RC Time constant

Determines the time it takes for a capacitor to charge or discharge through a resistor.

Key Formula: T = R · C

The capacitor charging time function is given by v(t) = V - V · exp(-1 · t/T). Quick engineering:

At t=0 the slope of the time function is given by V/T.
At t=0.22·T the capacitor voltage reached roughly 20% of the final value
At t=0.7·T the capacitor voltage reached roughly 50% of the final value
At t=1.6·T the capacitor voltage reached roughly 80% of the final value
At t=2.3·T the capacitor voltage reached roughly 90% of the final value
At t=3·T the capacitor voltage reached roughly 95% of the final value.

The capacitor discharging time function is given by v(t) = V · exp(-1 · t/T). Quick engineering:

At t=0 the slope of the time function is given by -V/T.
At t=0.22·T the capacitor voltage is roughly 80% of the finale value
At t=0.7·T the capacitor voltage reached roughly 50% of the final value
At t=1.6·T the capacitor voltage reached roughly 20% of the final value
At t=2.3·T the capacitor voltage reached roughly 10% of the final value
At t=3·T the capacitor voltage reached roughly 5% of the final value.

Example: R*C=Tau

Examples of Use: Timing circuits, switch deglitching.

RC Filters

Low-pass or high-pass filters to allow specific frequencies through.

Key Formula: f_c = 1/(2πRC)

Example: f=1/(2*pi*R*C)

Examples of Use: frequency filters, noise reduction.

LC Resonator

An LC network is prone to resonate at a given frequency. The generic resonance frequency of a lossless LC network is given by:

Key Formula: f_c = 1/(2π · sqrt(L·C))

Example: freq=1/(2*pi*sqrt(L*C))

Examples of Use: frequency filters, notch filters

2.7 LED Polarization

Protects LEDs by limiting current.

Key Formula: R = (V_supply - V_LED) / I_LED

Example: R=(VCC-Vf)/I

Examples of Use: LED indicators, lighting circuits.

3. Operational Amplifier Circuits

3.1 Inverting Amplifier

The input signal is applied to the inverting input through a resistor, while the non-inverting input is grounded. A feedback resistor provides negative feedback.

Key Formula: V_out = -(R_f / R_s) · V_in

Example: gain=Rf/Rs

Examples of Use: Audio amplifiers, signal processing, and waveform inversion.

3.2 Non-Inverting Amplifier

The input signal is applied to the non-inverting input, with feedback provided to the inverting input.

Key Formula: V_out = V_in · (1 + R_f / R_g)

Example: gain=1+Rf/Rs

Examples of Use: Buffer circuits, high-impedance sensors, and general-purpose amplification.

3.3 Summing Amplifier

Combines multiple input signals into a single output with adjustable weights.

Key Formula: V_out = -[(R_f/R₁)V₁ + (R_f/R₂)V₂ + ...]

Example: Rf/R_i=weight_i

Examples of Use: Audio mixing, weighted signal addition, and DACs (digital-to-analog converters).

3.4 Differential Input Amplifier

Amplifies the difference between two input signals.

(V_out - V_-)/Rf = (V_- - V_IN-)/Rn) and (V_IN+ - V₊)/Rp = V₊/Rg

Isolating V₊ it reduces to: V₊ = Rg / (Rg + Rp) · V_IN+

Isolating V_out it takes the form: V_out = (Rn + Rf)/Rn · V_- - Rf/Rn · V_IN-

Since the operational amplifier gain is high it can be assumed that V₊ ≈ V_-, then:
V_out = ((Rn + Rf)/Rn) · (Rg / (Rg + Rp)) · V_IN+ - Rf/Rn · V_IN-

When making Rn=Rp and Rg=Rf it becomes:
V_out = (~~(Rn + Rf)~~/Rn) · (Rf / ~~(Rf + Rn)~~) · V_IN+ - Rf/Rn · V_IN-
This simplifies to:

Key Formula: V_out = Rf/Rn · (V_IN+ - V_IN-)

Examples:
Symmetrical gain: gain=Rf/Rs
Asymmetrical gain: gainn=Rf/Rn and gainp=((Rn+Rf)/Rn)*(Rg/(Rg+Rp))

Examples of Use: Instrumentation, sensor signal processing, and noise rejection.

3.9 Howland Current Source

A precise current source circuit based on an op-amp, capable of driving loads with constant current regardless of voltage variations.

Key Formula: I_out = (V_in / R₁)

Example : R1=Vin/Iout

Examples of Use: Precision current drive for LEDs, motor control, and sensor excitation.

4. Comparators

4.1 Comparator with hysteresis

Analogue Comparators need a reference compare voltage to operate. In some situations it is recommended to add an hysteresis to the comparison. For example, if a comparator is used for a Power On Reset (POR) circuit, when the voltage surpasses the defined threshold and the reset is released, it may require an additional current which in turn will make the voltage input drop. In this case, to avoid an oscillation it is better to have an histersys built in.

Many comparators have an open collector output so that the user can select the output comparison voltage level. It is opted to write the full equation for this case, and let the user make the necessary simplifications for a push-pull comparator.

Note: It is good practice to choose Rup << Rh so that the V_OH of the comparator is not compromised.
It is assumed that the comparator collector output resistance is close to 0.

Key Formula:
V_th+ = (Vcc/R₁ + Vlogic/(R_h+R_up)) · (R₁|R₂|(R_h+R_up)),
V_th- = ((R₂|R_h)/(R₁ + (R₂|R_h))) · Vcc

Example : Comparator equations

Examples of Use: Debouncing switches, waveform shaping, and signal conditioning.

5. 555 Timer

The 555 timer is a relic from the past but, it's simplicity and flexibility make it still widely used still today. It has basically two configurations: Monostable and astable.

5.1 Monostable Configuration

When a falling edge is detected by the Pin 3 (trigger), a pulse is produced with a Time depending on the R C elements. This cycle is generated whenever the trigger pin is stimulated.

In this configuration the capacitor is quickly discharged every time the trigger pin is activated. The charging of the capacitor from 1/3 to 2/3 of the supply voltage determines the output pulse time. Given that the timing is calculated as a ratio of the supply voltage, the resulting time is not dependent of the VCC.

Key Formula: T = 1.1 · R · C

Example : 1.1*R*C=T

Examples of Use: Pulse detectors.

5.2 Astable Configuration

In astable configuration the capacitor C1 is charged and discharged between 1/3 and 2/3 of the voltage supply. In the traditional configuration as depicted below the charge is done by R1+R2 to VCC, whilst the discharge to ground is done only by R2.

The charging of C1 is given by the following equation:
v_C1(t) = VCC - (VCC - 1/3·VCC)·e^{-t/((R1+R2)·C)}

The t_high is reached when v_C1(t) reaches 2/3·VCC. Hence:
VCC - (VCC - 1/3·VCC)·e^{-t/((R1+R2)·C)} = 2/3·VCC <=>
1 - (1 - 1/3)·e^{-t/((R1+R2)·C)} = 2/3 <=>
(2/3)·e^{-t/((R1+R2)·C)} = 1/3<=>
e^{-t/((R1+R2)·C)} = 1/2 <=>
e^{t/((R1+R2)·C)} = 2.

In this case the charge time is given by: t_high = ln(2) · (R1 + R2) · C

The same rationale can be done for the discharging.
In this case the discharge time is given by: t_low = ln(2) · R2 · C

The frequency is then given by: freq=1/(t_high + t_low)=1/(ln(2) · (R1 + 2·R2) · C)
The duty-cycle is then given by: D=t_high/(t_high + t_low) · 100% = (R1 + R2)/(R1 + 2·R2) · 100%

Key Formula: Freq = 1.4427/((R1 + 2·R2) · C)

Example : ln(2)*(R1+2*R2)*C=period

From the formulas above it can be seen that with this configuration the duty-cycle is always higher than 50%. There are some solutions to this one, is depicted in the schematic below.

Same as before, the charging of C1 is given by R1 alone and can be calculated using the formula:
t_high = ln(2) · R1 · C

The discharging time equation is however a bit more complicated due to the fact that both R1 and R2 set both the time constant and the t=∞ value. The discharge of C1 follows this function:
v_C1(t) = VCC·R2/(R1+R2) - (VCC·R2/(R1+R2) - 2/3·VCC)·e^{-t/((R1|R2)·C)}.

It can be verified that at v_C1(t=0) = 2/3·VCC, and v_C1(t=∞) = VCC·R2/(R1+R2).

The t_low is reached when v_C1(t) reaches 1/3·VCC. This means that the condition VCC·R2/(R1+R2) < 1/3·VCC must be verified for the circuit to work properly. This implies that R1 > 2·R2 is a precondition for this circuit to work.

The t_low can be calculated from the following equation:
VCC·R2/(R1+R2) - (VCC·R2/(R1+R2) - 2/3·VCC)·e^{-t/((R1|R2)·C)} = 1/3·VCC <=>
R2/(R1+R2) - (R2/(R1+R2) - 2/3)·e^{-t/((R1|R2)·C)} = 1/3 <=>
e^{-t/((R1|R2)·C)} = (R2/(R1+R2) - 1/3)/(R2/(R1+R2) - 2/3) <=>
e^{t/((R1|R2)·C)} = (R2/(R1+R2) - 2/3)/(R2/(R1+R2) - 1/3) <=>
t_low = ln((R2/(R1+R2) - 2/3)/(R2/(R1+R2) - 1/3)) · ((R1|R2)·C) <=>

This seems a very complicated formula to resolve, but, WizEIA can digest this and solve it for you.

Key Formula:t_high ln(2) · R1 · C and t_low = ln((R2/(R1+R2) - 2/3)/(R2/(R1+R2) - 1/3)) · ((R1|R2)·C)

Example : Astable with DC close to 50%

With this configuration however, the duty cycle can be 50%, but, realistically, it can't be less than 40%. The discharge of the capacitor begins to be too slow close to the lower threshold leading to circuit being too sensitive to noise. This can be solved by adding a diode to the original configuration, as is shown here.

In this configuration the charge of the capacitor is done with only R1 since the diode shorts the R2. When 2/3 of VCC is reached, DIS pin shorts to ground and the capacitors is discharged by R2. Considering an ideal diode and using the equations we have already found for the astable operation we have:

t_high = ln(2) · R1 · C and t_low = ln(2) · R2 · C

This yields and aproximated value for freq = 1/(ln(2)· (R1+R2) · C) and duty = R1/(R1+R2) · 100%. More accurate results must take into account the diode characteristics. It is overly complicated to derive a formula that could take into account the diode, especially if it is operating close to its cutoff area. In this case the best is to use a spice simulator.

Examples of Use: Low frequency clocks, pulse generators

6. Voltage Regulators

Adjustable voltage regulators often provide a reference voltage and require a resistor network to set the desired output voltage. Normally the formulae that dictates the operation of the device is given in its datasheet. Follows a few examples of well-known voltage regulators.

LM1117 LDO - Adjustable

Key Formula:

Example : 1.25*(R1+R2)/R2=vout

LM2576 DCDC - Adjustable