New answers tagged

1

I need to regulate it to 1to 4ma maximum output without having much voltage drop... Can I just use a resistor in series? No. A resistor works linearly according to Ohm's Law. Resistance = Voltage / Current, so for a maximum (short circuit) current of 4 mA the resistance needs to be 12 / 0.004 = 3000Ω. However at 1 mA the resistor would drop 3000 * 0....


1

I don't understand why the total voltage is equal to the voltage drop over the 3 Ω and 2 Ω resistors. Normally on these exercises we are dealing with ideal voltage sources. This means that they will hold their output voltage no matter what the load is. That means that any pattern or combination of resistors can be connected across the voltage source and it ...


1

If you are familiar with Kirchoff's Voltage Law, then you know that any path that connects your two (ideal) battery nodes (assumed to be at the far left and the far right of your diagram) has the same total voltage across it as the battery. Thus the 2R and the 3R path has the same voltage as the battery. The same is also true for the path containing the 5R ...


0

Here you have a big tip. Use this diagram to solve for all the unknown resistors values:


3

The voltage across an LED is about 2V and stays at just about 2V. With a fixed power supply voltage it means that you have a fixed voltage across (and therefore a fixed current through) the series dropper resistor. The resistor is acting as a constant current source and its current is shared between however many LEDs you have in parallel. Lets say you ...


0

With the classic resistor with LED in series circuit, the current is mainly determined by the resistor. So in parallel they should draw more current and stay the same brightness. doesn't happen, the total current stays more or less the same. So when going from one LED to two LEDs, the current halves. That current is divided between the two (or more) LEDs. ...


1

From the specs, I am assuming 5V is the voltage drop so V = 9-5 = 4. Plugging this into Ohms Law 4 = R*.2 So that would work out ot a resistor of 20 ohms. Is this right ? 200mA is the maximum current. The minimum is zero (device off), so your voltage will vary between 9-(20*.2) = 5v and 9-(20*0) = 9V. Since average current is usually a lot less than ...


1

On the topic of powering LEDs with a constant voltage: LEDs have a negative temperature coefficient. This means that their resistance decreases as temperature increases. Providing a constant voltage will cause an ever increasing current until an equilibrium is reached. As a result, LEDs are recommended not to be used without some form of current control. ...


5

None of the above. If you wire LEDs in series, all of them will receive the same amount of current. They must, since current is number of electrons flowing through a wire per second, and the electrons must pass through all the LEDs in a series circuit. If they all get the same amount of current, identical LEDs will all light with the same brightness. (If ...


0

I'm not entirely sure, but i think i've come across a similar situation once: The status byte of the parallel port always reading the same even after changing electrical inputs at the port. What i did then was to select, in the PC's BIOS setup, an option for the parallel port OTHER than 'standard'; i believe i chose ECP. After that (and no other change), it ...


0

Your first assumption („current flows from left to right“) is not true and the actual direction in the diagram is meaningless (depending on the convention the current in question just has an inverted sign). Therefore, I2-I1will actually be 15A-(-30A)=45A while in another convention it would be 15A+30A=45A. Yes, the resistors can be replaced with a parallel ...


1

Each resistor has same voltage over them, so you know how much current flows from power supply to each resistor. And if you get the same result by calculating the combined parallel resistance, then you have proof whether they are parallel or not.


3

Any other improvement I could make? There are resistors specifically designed to handle surges. In fact all will but, few have the information in their data sheet that allows you to make the right choice. For instance, the Bourns CRS0805 range of surface mount resistors have a nominal power dissipation value of 0.25 watts but, the data sheet tells you ...


0

Does this mean only a resistor rated over 11 watts is suitable? Resistor power rating typically indicates maximum steady-state power at which resistor reaches maximum temperature at some ambient temperature. If you want it to be cooler power must be lower and if power is intermittent, it can be higher. This document describes how to calculate power and ...


0

Its simply a 2-in-1 potentiometer of the value indicated at the back. As in 3-pin potentiometer, the outer two pins are for fixed max resistance, then the middle pin is the wiper. The pins at the back, is just a replica of whats at the front


0

I found an answer by watching one of Dave's videos. He needs similar 10k resistors and his solution is to check a couple of the resistors which he has with the multi meter to find resistors which match as good as possible. Sounds like a great low tech solution to me. That's what I will do with the resistors which I need. https://youtu.be/8-qar5vgnbc?t=173


0

Replace the batteries with fresh alkaline ones. What you describe is normal behavior for batteries that are either capable of supplying the current draw or are close to the end of their life. The "recovery" is due to polarization.


0

simulate this circuit – Schematic created using CircuitLab Figure 1. How to measure battery and LED voltage. When taking the measurement across an LED measure at the points shown on the left most voltmeter. From the comments: LEDs are flickering yellow 2.1 V - 2.6 V around 20 mA and operates as low as 1.8 V and as high as 3 V. ... VUPN527. WOAH! ...


1

I know this is an older question but I found it when I searched for a similar answer. I found in the TL081datasheet that the maximum output current is 60mA, and minimum is 10mA. These values are for output short-circuit current so I take them as the maximum that I can drain from that opamp.


0

Specify a realistic accuracy, then work to that. Hint, to find out what's realistic, do an error analysis including drift over time and temperature, as well as initial tolerance of the components. Initial tolerance can be calibrated out, drift between calibration and measurement cannot be.


0

My understanding is the resistors should drop voltage from 4.2 to 2v but my volt meter shows no change. Allow me to clear up your misunderstanding. The battery under a light load drops from the charge voltage of 4.2 to 3.8 fairly quickly. (minutes) The current in each LED is always defined by the voltage drop across each R. or I=V/R I think you would ...


0

If the resistor leads are SHORT, and the 2 leads are soldered to large areas of copper so the heat is removed very near the resistor body, then why not go for 225mW? Given each square of standard-thickness copper foil has 70 degree Centigrade thermal rise per watt, then a resistor dumping heat into 0.5" by 0.010 (50 squares) in each lead, has 25 squares ...


0

Don't exceed 50% of the rating, if you're concerned. I have no issues with an upper threshold of 75%. If you only have a large stock of 250mW resistors, parallel them ( i.e. 4 1K resistors in parallel for 250ohm, for 1W). Using resistors rated for higher power dissipation shouldn't really be necessary. In what usage scenario is this a problem?


1

My safety margin for resisters is 50% ,unless the client specifies otherwise.If the circuit boards are going into relatively small numbers of high value equipment being sold far away then it is better to be safe than sorry .Complex thermal analysis would be needed to justify pushing things harder .Product numbers would need to be very high to make this ...


1

Considering 60'C can burn your finger and 85'C case temp is a prudent reliable max temp, consider your maximum ambient and max temp rise you can allow to meet 85'C . Usually I derate resistor power 50% to avoid hot spots near other temperature sensitive parts. That allows it to rise 45'C from a max ambient of 40'C to reach 85'C rated at 75% so I have 25% ...


3

the power spec is not really a power alone specification per se, you gotta think of it as heat as well, if the resistor has proper cooling it should be fine. However, depending on the application you spec everything with a certain safety margin, I think for high grade circuits that would not be acceptable since they are in the range of 50% or so Now, you ...


1

The key to getting a multidrop interface to work, like this one in the OP's post, is to manage the lengths of the unterminated stubs. They are the "vertical" connections that come off the main signal lines and go the line receivers on the bottom. How long they can be without compromising signal integrity depends on the edge rate of the signals. An RS-422 ...


1

Those diagrams are like schematics, they are correct but missing actual wiring details. The bus could be 100 meters long, and the branches where to put a receiver might be only few millimeter stubs, so regarding the electrical signal they really are not branches at all when the receivers are just in-lined on the bus. And the termination resistor must be at ...


4

When you send a high speed signal down a cable, the current that initially flows is dictated by the voltage applied AND the characteristic impedance of the cable. For the types of cable recommended for RS485 and RS422, the cable characteristic impedance is circa 100 ohms. So if 1 volt is applied at the sending end an initial current of 10 mA flows and all ...


1

To avoid reflections which mess the signal unreadable matched load is needed at the receiving end of the line.Bidirectional (=half duplex only, no more than one talker at a time) traffic needs matched load at the both ends of the line. The receiver in the RS422 or RS485 IC doesn't be the needed load, the voltage sensing circuit takes very little current ...


2

Simplifying series/parallel resistive circuits you use two formulas. For N resistors in series, \$R_T=R_1+R_2+...+R_N\$ and for N resistors in parallel, \$1/R_T=1/R_1+1/R_2+...+1/R_N\$ or transpose to get: \$R_T=1/(1/R_1+1/R_2+...+1/R_N)\$ You can solve these circuits by identifying series connected resistors and combining them, then parallel ...


0

simulate this circuit – Schematic created using CircuitLab You will need to use a 12V LED strip, just using a series resistor with a 4.5V strip will probably not work. If you use a 12V strip, the resistor can be omitted. Here is the problem with putting the resistor in series with the LEDS: simulate this circuit The problem is that you need a ...


0

if you ask me the best thing is to look at each project with oscilloscope if you think it is critical. You can do the math, but... it is better to measure it :) because impedance (length of lines and so on) have a big factor. There is a gif animation of effect of different values on my website: https://small-roar.com/i2c-inter-integrated-circuit/ As you ...


1

Calculate voltage at t=0 and t=1, then use P=V^2/R? Nope, power (or voltage) at t=0 and t=1 gives no information about what happens between t=0 and t=1. Use P=V^2/R and integrate between the limits to calculate energy dissipation Yes, you have to calculate \$ Energy = \int_{t=t_0}^{t=t_1} power dt = \frac{1}{R} \int_{t=t_0}^{t=t_1} v^2 dt\$ Also \$ P_{...


1

First of all, if your circuit itself does not have a Ground (0 V) reference, just probing a wire anywhere and finding voltages just isn't gonna work because there is no reference voltage to compare it to. That is why voltage across something is called Potential Difference. Now assuming the negative terminal of the battery is connected to Ground and the Vout ...


4

What is the second/lower 15 k-ohm resistor for? I understand why there is a 12 k-Ohm resistor in the top, that's for the V-out. The arrangement shown gives \$ V_{OUT} = \frac {15}{12+15}9 = 5 \ \text V\$. But why is there a 15 k-Ohm resistor in the lower right? You wouldn't need it if the load required 5 V and was exactly 15 kΩ. This could ...


2

That is a voltage divider, one of the most basic building blocks in electronics. The Vout node is 5V in respect to battery negative terminal. If there was no lower resistor, Vout would be 9V because without current there is no voltage drop in a resistor.


2

There is no universal answer on your question. Normally in low power electronics you try to use available space as efficient as possible by taking in consideration possible interaction of closely located components. For example in radio circuits you would try to shield oscillators from amplifiers, make traces as short as possible to avoid antenna effect ...


0

I wonder why this has not been suggested, but I think you could simply use a diode bridge (section "rectifier") as you would do to rectify AC voltage. I would use schottky diodes, because they have a very low forward drop and therefore also low power dissipation in a continuous current situation.


6

What they should have said, was that in practice you work with what you got and it often isn't necessary to use networks just to get a non-standard value. For that, you really want to get used to the E12/24/48/96 series. For example, if you calculated you need 18800 Ohm, a 18k7 would usually be good enough. Often enough 18k will do fine. Not even talking ...


4

Generally speaking combining resistors is either irellevant or self defeating, but when prototyping one does what one needs to so of course it is done extensively. For example it is easy to solder an smt resistor on top of another and saves a step vs desoldering if I want to quickly try a value smaller than what I chose . Consider that resistors values ...


1

A common rule of thumb for improved long-term reliability is to overrate electronic components by 100%. Examples: For a 12 V system, use a FET rated for 30 V or more. For a peak load current of 3 A, use a FET rated for 6 A or more of continuous drain current. Because your operating voltage range is relatively low (in FET terms), select a "logic-level" ...


0

These are LED power supplies. I don't know what application you have , but these power supplies do not need polarity protection. It is the loads that need it. Also 12V LEDstrings do not need reverse protection because they can handle it. I suggest you use polarized connectors rated for 10A. If you only need 1 , these are pretty common and can be ...


0

You want a MOSFET that turns on with less than 5V on the gate, so "logic-level", for the zener it should be less than the max Vgs of the MOSFET your MOSFET needs to be able to handle the max current draw of the device - perhaps 10A, perhaps less. the resistor is less critical 1K or 10K or 100K makes little difference. simulate this circuit – Schematic ...


7

If you are purchasing parts for an amplifier design that you will be tweaking resistor sizes to optimize performance I strongly suggest that you invest in a resistor assortment kit. Such a kit will prove useful for you for many future projects as well. I still regularly use resistors from an 0805 SMT kit that I purchased over 16 years ago. Kits such as ...


21

The pros about using a single resistor over either parallel or series: Less cost Less wiring Less space needed Less error prone (one component less to fail) The pros about using a combination of resistors: You can select resistors from the ones you have if you have a limited number of values When putting them parallel, you increase the power (Watts), see ...


1

It seems that the type is SLN5TTED50L0F from KOA Speer (http://www.koaspeer.com/slw07-slw1-sln3-sln5/). The F stands for 1% tolerance.


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