Thought I would devote a single page to a bunch of electronics circuits for those who don’t want to read a lot of text.
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Simple circuits with basic components:
Lone resistor/resistive component:
- I = V/R Ohms law formula for current: Current (I) in amps, though a resistor is equal to the voltage (V) in volts across the resistor, divided by the resistance (R) of the resistor in Ohms.
- P = VI formula for power: Power (P) in Watts (W) of a component is the Voltage (V) in volts across it, times the current (I) in amps (A) through it.
- Most resistors are rated for 1/4W (0.25W) maximum. But it is still highly recommended to stay below 1/8W (0.125W).
Measuring simple circuit current with a multimeter:
- Current is the same through all series components (power supply, conductive connectors, load component(s), etc.).
- Circuit must be opened, the meter much be set to measure more current than will be put through it, and then the meter is connected in series to measure current. More current through a meter than it is set to measure will damage the meter (probably blow a fuse).
- Measuring current ads almost no extra resistance to a circuit. An impossibly perfect (ideal) ammeter/multimeter set to measure current, would have literally no resistance at all.
Current has to flow through all of the circuit resistance of series components. Therefore, all of the resistance simply adds up to the total resistance as far as the current is concerned.
- 2 series resistors with 1,000 ohms (1kΩ) each, will provide a total of 2,000 ohms (2k) of resistance to the circuit.
- 12 volts across 2,000 ohms of resistance is 12V/2,000Ω = 0.006A. Therefore 6 milliamps of current will flow through that circuit.
- The resistors divide up the voltage (covered in more detail under voltage dividers) based on their percentage of the total resistance. Since the current is 0.006A, we can use Ohms law for calculating voltage (V= IR) to find that 0.006A x 1,000Ω = 6V is across each resistor.
- 6V x 0.006A = 0.036W shows that 1/4W (0.25W max) resistors don’t generate (and have to dissipate) much heat in this circuit.
Connecting resistors in parallel results in an equivalent of less resistance. Two equal value resistors is easy to calculate and analyze.
- Each 1,000Ω resistor with 12V across it in the diagram allows 12mA of current to flow through it.
- Each resistors also needs to dissipate 12V x 0.012A = 0.144W of power (heat).
- The power supply needs to provide a total of 24mA of current (2 x 12mA) to power both resistors. The current of all parallel components add up to the total circuit current.
- A lone 500 ohm resistor would also provide 12V/500 = 0.024A of current. So, 2 parallel 1,000 ohm resistors have the equivalent resistance of 500 ohms. However, the lone 500Ω resistor would need to dissipate 12V x 0.024A = 0.288W of power. That is too much for the typical 0.25W maximum (less than 0.125W recommended) resistor.
LED protected by a resistor circuit:
- LEDs (and other diodes) do not limit current on their own and must be protected. However they do drop some voltage, so there is less voltage across the resistor that then sets the current. Red/orange LEDs commonly drop about 2V while forward biased, whereas blue/green LEDs commonly drop approximately 3V.
- A series resistor limits the circuit current based on the voltage across it (supply voltage minus LED forward voltage drop) divided by the resistor’s resistance.
- In the schematic, I showed the voltages across the components when the LED has a forward voltage (Vf) of 2V.
- LED Cathode lead (pronounced like leed) is shorter than the Anode, if the Anode hasn’t been trimmed. Often the Cathode side of the LED also has a flat edge along the rim.
- Anode needs to be more positive than Cathode (forward biased) to conduct and light up.
Forward and reverse biased rectifier diode current measurements with a multimeter:
Always keep in mind, that when a circuit with 1 or more diodes/LEDs looks like it is wired properly but no current is flowing. A diode/LED could have accidently been wired backwards (reversed biased when it was meant to be forward biased).
This is more of a bonus circuit for better familiarizing yourself with electronics than a practical circuit.
- Most FB (forward biased) diodes drop about 0.7V.
- a 9 volt battery, with a 0.7V diode drop will result in 8.3V across the current setting resistor.
- A 1,000Ω resistor with 8.3V across it will pass 0.0083A of current. 8.3V/1000Ω = 0.0083. Remember that 0.0083A is the same as 8.3mA.
When measuring current, you need to open the circuit, and you need to make sure that the meter is set to be able to measure more current than you will put through it (otherwise it my be damaged). Then you complete the circuit by connecting the multimeter in series with the components so that the current also flows through the meter.
A forward biased diode (Anode more positive and cathode more negative) will conduct fairly easily, but most FB diodes will drop about 0.7V from the current setting resistor.
A band is painted on the cathode side of the diode.
A reverse biased rectifier diode blocks current up to the breakdown voltage of that particular diode. Do not exceed that breakdown voltage. Zero amps/milliamps will show up on the meter. If the meter is sensitive enough though, you may be able to measure the leakage current.
No diode ever completely eliminates current flow, but while a rectifier diode is reverse biased, it is usually such a small amount of current leaking through that it doesn’t matter to almost any circuit. Some diodes do have more leakage than others, so that is something to be aware of if you have a circuit where leakage may be a problem.
The 1N4001 is a rectifier diode that is usually included in beginner electronic component kits. It is rated to block up to 50V while reverse biased.
Alternating LEDs by sinking or sourcing current circuit:
When you see LEDs flashing back and forth, it is likely that the output of something (in this case a single pole single throw switch) is alternating between sinking and sourcing current.
- Schematic diagram on the left has the switch set so that it is sinking current, leading to the top LED being lit up. Current is imagined as flowing from +5V, through the 1k (one thousand ohm) resistor, the LED, and then through the switch to ground.
- Pictorial circuit on the right has the switch sourcing current. It is imagined that the current is flowing from +5V, out of the switch (output), through the red LED and 220 ohm resistor to ground.
- Blue LEDs tend to be a lot brighter than red LEDs when they have the same current through them. So I used a much higher value resistor to limit the current trough the blue LED. Blue LEDs usually have a forward voltage of about 3V, wheras red LEDs usually have a forward voltage of about 2V.
Mechanical switch – High side versus low side:
- High side switching is more common. The switch is place closer to the positive side of the power supply while the load is closer to ground.
- Low side switching is also possible. The switch is closer to ground while the load is closer to the positive side of the supply.
- An open switch is “off”. No power will be applied to a load in series with the switch.
- A closed switch is “on”. Power will be applied to a load in series with the switch.
- Schematic shown here displays a normally open push button switch. It is open/off until you press and hold the button to close/turn it on.
Series components share the same current. This is good for LEDs and other low power components for a number of reasons.
- Relatively High voltage can be used to power LEDs that are connected in series because their voltage drops add up. You can add them in series as needed to drop the supply voltage away from the current limiting resistor until you get an acceptable voltage. Keep in mind that you can get electrocuted somewhere above about 40 volts. Stay away from high voltage!
- Voltage drops of series LEDs add up. Explained more in the following listed item.
- 220Ω resistor protecting a single LED from 12V will result in too much current for the LED and too much heat for the resistor. But 4 series red LEDs will usually drop about 8V from the current setting resistor. 4 series blue or green LEDs will usually drop a total of almost 12V from the resistor. You may need more voltage, even though those LEDs are usually brighter than red LEDs at low current.
- Four series LEDs that are red with the typical 2V forward voltage, will drop a total of 4 x 2V = 8V from a 12V power supply. That leaves 4 volts across the current setting resistor, which is also in series with the LEDs. Giving us 4V/220Ω = 0.018A of current through the entire circuit, and 0.018A x 4V = 0.072W of power (heat) that must be dissipated by the resistor. Remember that most resistors are 1/4W resistors rated for a maximum 0.25W, recommended 0.125W of power.
Parallel LEDs with a single protective resistor:
It is best to give each parallel LED it’s own protective resistor. Parallel LEDs of the same forward voltage, with a single resistor protecting them, will split up the current, but not necessarily evenly. If one has a higher forward voltage, then it will not pass any of the current, whereas the lower forward voltage LED will pass all the current.
A slightly higher voltage does build up across a FB (forward biased) diode as current though it goes up. So current will still start flowing through a parallel diode with a slightly higher Vf (forward voltage) that is in parallel with it. They must be within about 0.1V to 0.2V Vf of each other for most diodes though. Many circuits are sensitive to the exact voltage across a diode as current changes. So it is good to know the voltage across it over a wide range of currents through it.
Zener diode reference voltage:
- Zener diodes are specially made to conduct while reverse biased (RB).
- Zener diodes come in a lot of values where they are rated to conduct while RB. The voltage that a specific zener diode is rated to conduct while RB is called it’s zener voltage. Their value is usually written on them, but it’s very small writing that needs magnification to read.
- When a RB zener diode is conducting current, the zener voltage builds up across it. It does change slightly based on how much current is flowing through it. Upwards at higher current and downwards at lower current. The closer that you get to using a 6V zener diode, the closer it will probably stay nears it’s zener voltage.
- The voltage across the zener diode can be used as a relatively steady reference voltage, even as the power supply voltage changes. Always remember that current through a zener diode still needs to be limited.
- Current must be limited through the zener diode, usually with a resistor. The power supply voltage most be high enough to push current through the RB zener diode and the series current limiting resistor. A supply voltage of 2V higher than the zener diode, or more, will probably be enough.
Light Dependent Resistor LDR voltage divider circuit fragment:
Wired as a voltage divider…
- High side (closer to positive supply) LDR will output close to the supply voltage (5V in this circuit) under bright light, and close to 0V when it is dark.
- Low side (closer to ground) LDR will output close to ground voltage (0V in this circuit) under bright light, and close to 5V when it is dark.
- An in-between light level will output an in-between amount of voltage.
- The value of the fixed (not adjustable) resistor will help determine light sensitivity in relationship to output voltage.
Capacitor storing some charge (Smoothing voltage):
Applying a voltage to a capacitor, charges the capacitor until it is storing enough charge to have the same voltage applied to it.
Power sources instantly charge capacitors if connected directly to them (unless they are specially made not to). Therefore it is important to only instantly charge relative low value capacitors. I limit it to about 1,000µF, (one thousand micrfarads) which is the same as 0.0001F.
Also putting an LED protected by a resistor in parallel with the capacitor, allows the capacitor to discharge once the power source is removed. The LED stays lit up for a bit to give you an indication of how much capacity (energy that can be stored) a capacitor has.
Most power sources can easily charge a capacitor instantly, and keep power a load at the same time.
Larger value capacitors keep the LED lit for longer after the supply voltage is removed, in relation to smaller value capacitors. Even a 1,000µF capacitor won’t keep an LED lit for very long though.
The LED also fades off. Quickly at first and then slower as the capacitor discharges through a load. It’s voltage drops based on how much current it has had to provide.
Circuits that will be added later:
None at the moment.
This is a new page that will be updated!
- Information on this site is not guaranteed to be accurate. Always consult the manufacturer info/datasheet of parts you use. Research the proper safety precautions for everything you do.
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