Resistance is an important concept in electrical work. By changing the amount of resistance in a circuit, it's possible to change the voltage within that circuit. This ensures that every component within the circuit gets just the right amount of electricity to power it without causing damage. Adding resistance allows you to step down the voltage of a 12V circuit to only 9V, but you have to make sure that you don't overdo it; adding too much resistance will lower the voltage too much, potentially causing problems for those components that are starved for power.
To reduce a 12V circuit to 9V, place two resistors in series within the circuit. If using multiple resistors, take that figure and compare it to the total desired output voltage 9V ; this gives you a ratio, meaning that the second resistor in the sequence should have three times as many ohms as the first. As electricity flows through a material, it encounters resistance. This means that not all of the voltage passing through the material makes it through as output voltage, as some portion of it is absorbed by the material itself and turned to heat.
This is actually how electric heaters work; high-resistance materials have electricity run through them, causing the materials to heat up and emit that heat into the surrounding air. When electrical current needs reduction in an electrical circuit, resistance is added in the form of resistors. A resistor is a high-resistance material encased in a protective coating often epoxy to prevent them from radiating heat while still providing resistance within the circuit.
Resistors are made to provide a specific amount of resistance, measured in ohms, and are color-coded for easy identification.
The color code used depends on the type of resistor you use.
Choosing the Resistor to Use With LEDs
If you need to step down a 12V current to 9V within a circuit, you'll have to first determine how many resistors you need and how many ohms of resistance they should provide. To determine how many ohms of resistance you need to step down these three volts, you'll need to know how many amps are in your circuit as well; this can vary from one circuit to another and will depend on the materials used, your power source and how you've designed the circuit.
Once you know your resistance, you can decide whether you want to use a single resistor or if you want to break it down across multiple resistors. Once you've calculated your resistor needs, it's time to install the resistors into your circuit. If using a single resistor, you simply need to install it between your power source and the device or load that requires a 9V current. If using multiple resistors, they will go in the same location between the power source and the load. Install the smaller resistor first, stepping down your voltage from 12V to 11V.
Once you've added the first resistor to your circuit, install the larger resistor to step the voltage down again. This resistor will take the remaining 11V current and reduce it to the 9V output you desire.
Once your resistors are installed in your circuit, be sure to test its voltage with a multimeter. The input voltage on the circuit should still be 12V, but the output voltage should drop to 9V as the current runs across the resistors. If the voltage drops as expected, finalize the circuit and solder everything into place. If the output voltage is wrong, however, recheck your calculations and change your resistors until you get the proper voltage shift. Holding a BS in computer science and several years of experience building, repairing and maintaining computers and electronics, Jack Gerard has had a love of science and mathematics for years.
About the Author. Photo Credits.How to use a resistor
Copyright Leaf Group Ltd.However, after a few trivial board designs I'm facing a challenge that must be very simple I'm making a mountain out of a molehill. I'm trying to perform some logic in between 2 existing components.
I want to invert the signal and perform a specified delay on the sensor. The delay is fairly straightforward using a delay line from DigiKey. The inverted signal is fairly straightforward as well. I need to provide 24VDC mA back to the circuit board. I am not sure how to get the voltage back up to 24VDC, and at the higher current. I am now considering a DC-DC converter link belowwhich is powered by a transistor whose base is connected to the delay, and emitter is connected to another 78 series voltage regulator 5VDC.
I MUST be missing something obvious. How do I get back to 24VDC output from the board?
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What component or circuit is common practice? A small amount of guidance and I will gladly spend the time researching appropriate usage.
Thanks, Dan J. User mini profile. Using a with 24v input for any amount of current will get too hot very quickly. There are some switching regulator modules that look and work like the without the heat, Do you need to delay powering up the sensor or to delay the output of the sensor?
Can you give some more info on the source of the 24v, the delay you want, and a data sheet for the sensor - that would give people something to go on. Here is a link to the sensor in use. Since the sensor is rated for mA, I expect that the board requires some amount less than mA for activation.
The sensor is active for approximately 35ms which is too short. Using a LTC delay chip, I want to delay the trailing edge of the signal, so that the active time is approximately 60ms. The current circuit board needs a longer signal in order to reliably perform its function. That signal will also need to be trailing edge delayed to function properly. There is an existing circuit board that performs logic unreliably due to the duration of the signal from a photoelectric sensor.
The sensor's active time is insufficient, so I want to add a board in between the sensor and the circuit board. I have access to the source 24VDC power supply if necessary. The board I add will increase the active time for the sensor using a delay chip by delaying the trailing edge of the signal. Eventually, I want to replace this with a Baumer XF camera which will provide a signal which is inverted from the photoelectric sensor the need for inverting the signaland will also require the delay chip for proper function.
Does this provide all the information? Thanks again, Dan. The sensor has a power input always on and a switched output connected to 24V when on, floating when off. I would tie the sensor power to your 24v rail. Take the switched output, add a pulldown resistor, and have it switch a transistor to generate a 0 or 5V signal to trigger the delay. Have the delay's output drive a transistor to generate the 0 or 24V logic output. That should point me in the correct direction. I appreciate the assistance!!!!Voltage conversion from 24VAC to 5VDC is quite useful, because a lot of home automation devices use 24VAC, including sprinkler solenoids, home surveillance systems etc.
Having a conversion module makes it easy to use a single power supply, without a separate 5V adapter for your control circuit. There are plenty of resources you can find online about it. But these resources are rather scattered. So in this blog post I will summarize and discuss the common choices. Before we begin, the first step is to have a rectifier that converts voltage from AC to DC. The common choices are half-wave rectifier which requires just one diode or full-wave rectifier which requires four diodes.
For simplicity, I will use half-wave rectifier as an example. The typical schematic of a half-wave rectifier is as follows:. As we know, diode only allows current to flow in one direction, so after the AC voltage passes through the diode, only positive voltage remains. The diode must be selected based on the maximum reverse voltage and the maximum current. So a diode with 70V peak reverse voltage is sufficient. This is typical, and the voltage is supposed to drop close to 24VAC under maximum load i.
These should work well for common scenarios. Again, when the transformer is well below maximum load, the output voltage can go as high as There are a variety of solutions:. Probably the simplest solution is to use a Zener diode.
This condition is met when the current flowing through it in reverse direction is at least a few milli-amps 5mA typical but less than the maximum current allowed e. The typical schematic is shown as follows:. Here resistor R1 is used for current limiting. Assume D2 is a 5V Zener diode, and the circuit on the right-hand side draws about mA current. R1 must be selected such that the current flowing through it is mA plus at least 5mA to keep D2 in break-down condition. Note that D2 should be rated at least 1W, because in case of open-circuit, it needs to absorb the entire mA without burning out.
Since the current flowing through it is mA, the power is 0. Well, this is the unfortunate drawback of a Zener diode based regulator, that is, it can waste a lot of power and require a bulky resistor. Fundamentally, it regulates the voltage by converting the voltage differential to heat.
In this case, the voltage differential is quite big 33V vs. Another drawback is that to increase the current draw, we must decrease R1. Otherwise, if the output circuit starts to draw, say mA, that will take D2 out of its break-down condition, and the output voltage is not regulated any more. So overall it is only suitable if the current draw is constant and small e.Transformers, which are often used to step-down voltage, work with AC alternating current voltages, not DC direct current voltages.
To step down a DC voltage we need to use some other method to accomplish this task. There are many methods to choose from but the two simplest are the "series voltage dropping resistor" and "the voltage divider network. Calculate the resistance of the "series voltage dropping resistor. Calculate the power dissipation rating of the power resistor and add a 25 percent safety factor. Select the next closest standard resistor wattage rating value, which is 10 watts. Connect one end of the voltage dropping resistor to the negative battery terminal using a jumper.
Connect the other end of the voltage dropping resistor to one of the terminals on the sealed beam lamp with a jumper. Complete the circuit by connecting the other terminal on the lamp to the positive battery terminal. The light will illuminate. Check the circuit voltages. Place the voltmeter probes across the resistor. The meter will indicate 6 volts. Place the probes across the terminals on the lamp. The meter will read 6 volts. Calculate the "bleeder current.
The rule of thumb for designing a voltage divider is to make the bleeder current 10 percent of the load current. Our load current is 1 amp, therefore our bleeder current equals 0.
Calculate the total resistance for the voltage divider network. Total resistance equals the source voltage divided by the bleeder current. Calculate the current flowing through R1. The current flowing through this resistor will be equal to the sum of the "bleeder current" plus the "load current.
Calculate the resistance value of R1.
In this case we would round down to 5 ohms, which would provide 5. That is close enough to the 6 volt rating of the load. Calculate the power rating for R1. Use the next closest value, or 10 watts. Calculate the resistance of R2. In this case we'll round up to ohms. Calculate the power dissipation rating for R2. Connect R1 and R2 in series using a jumper lead. Connect this series circuit between the positive and negative terminals of the battery. Using two more jumpers connect the seal beam lamp across R1.
Electrical Engineering Stack Exchange is a question and answer site for electronics and electrical engineering professionals, students, and enthusiasts. It only takes a minute to sign up. The source will be 24 VDC and I want the input to be manually adjustable assuming with a potentiometer. The input impedance for the VDC input is kohms and the input impedance for the mA input is ohms.
I'm not sure what the best way of going about doing this is or which input would suit my needs better. I'm assuming there are certain pros and cons for each one. Possibly one being more steady than the other?
I was originally trying to figure it out using 2 resistors as a voltage divider to get down to 5 VDC and then a potentiometer to vary the voltage on that 5V leg. Or possibly using 1 resistor in series and 1 in parallel with a potentiometer to vary the current? I'm a bit lost on exactly how to do it as well as which way is best.
If you need anymore info just let me know. The input impedance for the VDC input is kohms. We don't want the input impedance to load the pot significantly or the response will not be linear.
Now we just need to calculate the value of R1 to drop 19 V across it. The 4 - 20 mA will be read as 1 - 5 V by the analog input. Both those input types have in common that they depend on an internal to the PLC reference - they are not 'ratiometric' or proportional to the input voltage. To get a high degree of accuracy and stability you need an external reference so that your pot rotation results in a known and stable in absolute terms voltage or current.
Once you have the signal, current is more suitable than voltage for transmission over long distances in an electrically noisy environment. If the pot is in the same cabinet as the PLC it doesn't much matter for most purposes. The easiest and perhaps best solution is to buy a pot-to-current or voltage transmitter which will convert the pot rotation to voltage or current.
You can then use a pot of reasonable value such as 1KK and connect the wiper directly to the PLC input- if your PLC provides such an output 5Vthat would be even better. You can make a pot to current converter as well, but it's a bit more involved- there are so-called 2-wire transmitter chips available that do much of the work for you, check out Texas Instruments' nee Burr-Brown offerings in their XMTR transmitter series.Did you use this instructable in your classroom?
How to Step Down DC Voltage Without a Transformer
Add a Teacher Note to share how you incorporated it into your lesson. The easiest way is to use one of the online calculators provided below. Just click on one and enter the info from the previous step and you're set! You only need to go to one. Go to Evil Mad Scientist Labs web page at this link and print and make your own slide rule-like calculator.
PDF, assembly and usage instructions are all on the page linked above. It's pretty nifty and ends up being about business card size so you can keep one in that box with the rest of your LEDs.
But wait! And indeed you do. That's because its hard to buy a ohm resistor and easy to buy a ohm one. Just use the nearest one you can easily find.
To learn more about this magic formula read about Ohms Law. Tip 7 days ago. Reply 6 weeks ago. This is a late answer, but will be helpful to others. This would need a diode ideally 4 of them, connected to make a full-bridge rectifier, to avoid flickering in addition to the large resistor. A diode is like a 1-way valve--it prevents electricity from flowing "backwards", essentially making AC into DC. Reply 4 years ago on Introduction.
Reply 5 years ago on Introduction. Reply 5 years ago. Question 7 months ago on Introduction. Answer 6 weeks ago. That is literally the exact question this Instructable answers.
Read it, and you will have your answer! Question 4 months ago. Hi, I am trying to rig up a single LED for an aircraft, 24 volt system, breakout box.
It will be tested between 2 banana plug sockets. I need to have a LED light that will withstand higher current applications.
And it needs to stay small, like for on top of a mini banana plug grounding plug. Any thoughts? Obviously it's 10 years too late for this guy, but maybe this will help a future reader. Amperage doesn't work like that. As you probably know, the ampere is a unit of current.
As such, it only applies in the context of a particular device. When you say the system is amps, this is referring to the maximum amount of current the source can supply.Is it possible to take a 24vdc power source and reduce the voltage down to 10vdc? I have a vdc analog input card for a PLC with a potentiometer for a variable vdc input. I want to avoid buying a 10vdc power supply and use the 24vdc one I have in my panel.
I think it is possible with a resistor in series with the POT but I must be using ohms law wrong. Well, since you are just inputting a signal to a probably relatively high and relatively constant impedance ADC input the inefficiency of a resistor voltage divider is not too important. If you put a fixed resistor with 1. You might want to look into a zeneer regulator, as that may work better, but you asked about resistors, and resistors could be used.
LEDs are light-emitting diodes. They will pass current in only one direction. You can power them from AC just fine, but they will flicker, as they'll only conduct for half of the cycle. AC won't be good for any additional electronics, though, so if your LEDs have a "mystery circuit box" attached, you might not want to try this.
The 24V is more of a problem. If you can wire the lights in series, each will drop 12V and you will be ok. You could connect a second identical pair in parallel, in opposite polarity, which would conduct in the opposite phase, and balance the transformer output voltage, but there's no need to do this.
You can connect pairs of lights in parallel until you reach the current capacity of your transformer most indicator LEDs draw mA, and your transformer probably can handle at least a couple hundred mA.
I would use a LML voltage regulator chip with a 10K pot on the adjust output leg. On the output series in a and a ohm resistor to the adjust leg as well.
Install a IN or better diode jumpered between the input and output, with the diode reverse biased. Use another same diode inline forward biased on the input 24vdc i nput. Adjust pot as needed to actuate the device. This is a voltage output, not a current output. Answer Save. Jennifer Lv 4. How do you think about the answers? You can sign in to vote the answer. Still have questions? Get your answers by asking now.