# Tag Info

54

Power dissipation will be the driver. Using six in parallel allows use of standard resistors which may be a stock item. Using standard parts allows use of automatic assembly equipment. Lower profile. Heat spread out over larger area resulting in lower peak temperatures. Ability to combine to make a non standard value. The 20 Ω in your question is not an E12 ...

44

I'm struggling to understand the exact circumstance where this would be needed. What most folk want with a common emitter amplifier is for the DC collector voltage to be about half the supply voltage. So, when you have an AC signal present at the input, the amplifier's output (collector) doesn't clip the signal asymmetrically. Another way of phrasing it is ...

34

No. Because the notion of averaging a bunch of resistor values only works if you can be sure that the error in their values is random, and has a zero-mean distribution. Typically, neither of these is the case. First, because the resistors have already been selected for their value at the factory. Second, because there's no guarantee that what's coming out ...

31

LED forward voltages can vary some - enough that it would be difficult to set a precise current with them merely tied in series across a fixed voltage. Total Vf for the string can vary by several volts. You could very easily blow up the string just tying it to the supply as you’ve shown. So, yes, you should still have a dropping resistor. Even better, ...

31

BOM optimization is a likely candidate. The value 2k is probably used elsewhere and adding another value would mean one more line on the BOM and one more slot in the pick-and-place machine. Another BOM optimization could be availability. For example two 0603 resistors are easier to source than one 0805 (for power handling). You mention an eight resistor in ...

28

If you connect to the top and wiper, you can use the potentiometer as a simple variable resistor. If you connect to the top and wiper, AND short wiper to bottom, you are still using it as a variable resistor, but with one important difference : if the wiper goes open circuit, the resistance value is 10K rather than infinity. That can: reduce crackling ...

26

You can buy resistor networks that have superior matching characteristic relative to tolerance. However 75 ohms is rather low for that. Vishay has their excellent bulk metal foil resistors in 75 ohms with 0.01% tolerance and (just as important) +/-2ppm/°C drift. If you are looking for a half bridge you can easily trim the midpoint in with a few additional ...

24

Let's have a look at the pcb... It looks like they used several resistors for higher power dissipation. This is quite common, as several low power resistors may be cheaper than one high power resistor, especially if you already use the value somewhere else in the design, which means you already have them loaded on the pick and place machine so you don't ...

23

If you want to convert the voltage to 5 volts, you should not use a resistance-voltage divider. That way you will indeed create 5 volts, but as soon as you apply a load the voltage will drop. Instead, you have two options in general to regulate voltage. The first option using a linear regulator, and the second option is using a switch-mode power supply. ...

23

It's a current shunt - 68 milliohms. That's what the "M" stands for. It's used to measure the current drawn by the circuit which also explains its large size (and therefore high power rating). The resistor is perfectly fine if you measure a short-circuit across it.

21

A long time ago, the technology did not exist or was too expensive to print such small text on parts. Also, you can read coloured rings no matter what angle you are looking at the resistor from or how it is mounted, but as someone who has trouble telling similar colours apart this is useless and I don't really buy it.

21

Each red LED drops about 1.5V. So three of them takes 4.5V, leaving only 0.5V for the resistors. 0.5 / (47+47+47) = 0.0035A or 3.5mA. That's rather low. If you were to remove two of the 47 ohm resistors, you would get 0.5 / 47 = 0.011A or 11mA. That's brighter, but still less than most LEDs can handle.

19

Resistors are all the same size no matter the resistance. It's convenient for the manufacturer to test how much power a certain-sized resistor can take before it burns up. And it's the same for all resistors of that size. Also the power rating is the same regardless of the resistor's exact resistance (within the tolerance). Cables with different current ...

16

Some leaded resistors, particularly precision ones, are marked with the value. Putting color bands on the resistor allows the value to be read regardless of the orientation, and very quickly for a skilled person.

16

Even nominally "0 ohm" resistors have some resistance. The datasheet for the part you linked says that zero-ohm resistors have "< 20 mOhm" of resistance. I don't know how you're using this part, but in most cases for a resistor that small, the exact value of the resistor won't affect the current flowing through it very much, so you can ...

14

Millman's theorem is a kind of multi source voltage divider theorem based on this topology: - So, re-arrange your circuit to use a 3rd source that is actually 0 volts and you're good to go: - Hence $V_{AB} = \dfrac{\frac{10}{3.33333} + \frac{0}{10} +\frac{20}{4}}{\frac{1}{3.3333}+ \frac{1}{10} + \frac{1}{4}} = \dfrac{8}{0.65} = 12.3077$ But equally, ...

14

R1, R2 and R3 form a voltage divider. simulate this circuit – Schematic created using CircuitLab When a 5V signal is applied to the top of the divider (the echo signal), ohm's law states that: $$\frac{5V}{3000\Omega} = 1.67 mA$$ Will flow trough the resistors to ground. This means there will be a voltage drop of: $$1.67mA * 1000\Omega = 1.67V$$ ...

14

I think you're over-thinking this. Here's how the voltage across two resistors in parallel is measured: simulate this circuit – Schematic created using CircuitLab Since R1 and R2 are in parallel they are each connected between the same nets, net A and net B. That means both R1 and R2 experience the same voltage, that is the voltage between those nets. ...

13

Cables are rated for power too, but somewhat indirectly. They have a stated thermal limit as well as a maximum working voltage (insulation strength). Nevertheless it's usually more convenient to think of their current handling capability since their primary purpose is to do exactly that: move current. You would arrive at the cable's current rating from the ...

13

It is known that current in a series circuit remains the same. The current is the same at any part of a simple series circuit. But that doesn't mean the current stays the same if you change the circuit. An LED drops around 1.5 to 3V when it's working normally. If you connect it straight across a 5V supply, an excessively large current flows, and the LED ...

13

A 0.05% resistor is guaranteed to be no more than that tolerance from its nominal value. When you connect for instance four resistors in a series and parallel arrangement to get the same value, there is nothing to stop them all being high, or low, so this will not reduce the worst case deviation. However, if you took many sets of four resistors, then the ...

12

In the first circuit, you have one (single) voltage source and one (single) resistor. This one (single) resistor is connected directly across the voltage source terminals ( terminals $A$ and $B$). Thus, from point $B$ to A the voltage is equal to the battery terminal voltage $V_B$ and because our single resistor is also connected directly ...

12

Cables are rated for power - it's just that it's not a useful way to think about it. The upper rating on cable size will be determined by how hot it gets before its insulation is damaged - and this will depend directly both on conductor size and insulation characteristics. Cables aren't intended to drop voltage, they are intended to carry current with an ...

12

The series Rs resistors aren't for 'protection' pe se, the I2C spec notwithstanding (more about this below.) In any event, for Standard (100 KHz) and Fast (400 KHz) mode they're optional. For the higher-speed I2C modes (1 and 3 Mbit) they are used as series damping to reduce ringing and thus improve signal integrity. For that case, choose an Rs value such ...

12

I am guessing that you are looking at an SMT resistor marked "470" and that it is actually 47 ohms. As to operation- The TL431 draws as much current as required (up to 100mA+) to reduce the voltage at the sense input to 2.495V nominally. As long as it gets > 1mA to use itself it will function properly (400uA typically). If we assume the circuit ...

12

The emitter resistor serves mutiple purposes. It improves the linearity of the amplifier, raises the input impedance, and simplifies biasing. Let us look at each characteristic individually. In the following I will refer to variation of the circuit with $R_E$ omitted as the 'grounded common-emitter amplifier'. Linearity The gain of the grounded common-...

11

If you look at the PCB layout in the 179-140 instruction manual for compressor amplifier 179-130 on page 9 you'll see R56 close by Q1 like this: - And I suspect that Q1 has a heater coil around it that may be called "OVEN". Q1 is a JFET that is used to control the amplitude of the signal (hence form a compressor) and although the heating side of things isn'...

11

Warning upfront: Service manuals of audio equipment often contain slightly manipulated schematics to at least stop the most incompetent counterfeiters. Don't assume everything in the schematic makes sense. but here: the symbol is that of a variable resistor. So, either this is just a variable resistor to set the power of an oven attached to (K) and (O), or,...

11

Another question about LEDs and resistors... Ok, I'll take it. I believe you might benefit from the visual aid associated with the use of load lines. Try to see your circuit as the combination of two one-port elements: the generator and the load. Each of these one-ports will have its own Voltage-Current characteristic and when you combine them together,by ...

11

is the number of LED enough to limit the current to around the 350 mA range? Absolutely not. do I still need the resistor added in the series? Yes, or something that acts like a resistor. A MOSFET as you have would work fine as long as it has feedback to effectively work as a constant-current source, which you don't currently have. There are many examples ...

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