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8

A simple resistor that burns before more expensive parts do is called a fuse. They are available for pennies at every major and minor electronics seller.


5

A constant current driver will raise the voltage until it is able to force the mentioned current through the load. Voltage can't rise to infinity because of circuit limitations/configurations. Your driver will want to push 600 mA irrespective of load connected to it. In your case, a 120 ohm resistor, it needs to set the voltage to 120x0.6 V = 72 V. 72 V is ...


4

That resistor sets the frequency of the DC to DC converter, the frequency will be set at 100KHz. Shown below is a table from the datasheet of other values for the Rt resistor that can change the switching frequency. Source: https://www.analog.com/media/en/technical-documentation/data-sheets/3757Afe.pdf


4

R = V^2 / P = 36^2 / 500 = ~2.5 Ohm It looks like you've assumed it's a 36v motor. It will generate about that when spun at its rated speed, so neglecting losses, this calculation is correct. If you use very much less than this resistance, then two things will happen. 1) At its rated speed, at 36v output, much more current will flow than you expected, so ...


4

I would say no. The effects of temperature coefficient can be expressed as $$ R= R_0 \left[1 + \alpha\left(T_0-T_{\text{ref}}\right) \right]$$ So, only the change in resistance (not the resistance) will be proportional to the temperature difference, and that holds best for small temp changes about a working point. To be even more specific, $$ R=\frac{\...


4

First of all, PT100 is not thermostat, its a type of resistance thermometer. And that by the way looks like an PTC thermistor. Technically yes you can do so, It's very wrong and dangerous but you can do that. You are asking why it's dangerous? well, your heating device is using the resistance thermometer to get feedback from real temperature. When you are ...


4

Assuming you start with 0V across the capacitor. When the switch closes, the capacitor will charge through the 1k resistor, and will have an RC time constant using the C and the 1k. Eventually the capacitor will charge to 5V. When the switch opens, the capacitor will discharge through the 1k and 10k resistors with an RC time constant using the C and the ...


3

That will work, provided you meet a few criteria. The current set must be less than the maximum allowed current in any LED. The current set must be less than the fanout of your TTL minus the fan-in of all the gates you are driving. You must be willing to accept that when all of the LEDs are on, they will be much dimmer than when only one is on. The current ...


3

If you'll notice in the features section of the linked datasheet: It states that one resistor can be used to set its operating frequency. Evidently, a 140k resistor sets it to its minimum of 100kHz. On page 11 of the datasheet, Table 1 (shown below) provides resistances to be used for several example frequencies.


3

Overview Let me start with some thoughts about using headphones. Typically (I think, but I'm open to correction), headphones are arranged as \$32\:\Omega\$ devices. (They will range upwards towards \$600\:\Omega\$, today, though I commonly used very sensitive \$2\:\text{k}\Omega\$ headphones "back in the day." (Hard to find, now.) The power output offered ...


2

Additional to previous answers that in forward bias, the diode has about 0.7 voltage drop and very low impedance which ignores the parallel resistor, then according to capacitor charge, time=R*C, it will charge quickly because of low R (impedance of diode). If you remove the diode the charge time will be long because here R is the resistor which is high.


2

A direct way to analyze this is to define \$Z_1\$ as the equivalent impedance of \$R_1\$ // \$C_1\$; and \$Z_2\$ as the equivalent impedance of \$R_2\$ // \$C_2\$. This leaves us with a simple impedance divider, where: $$ \frac{V_o(s)}{V_i(s)}=\frac{Z_2(s)}{Z_1(s)+Z_2(s)} $$ Converting to admittance to simplify the calculation, we get the equivalent ...


2

Yellow and green input waveforms are changing at the same instant in time and this presents an EXOR gate with a resolution problem - if green slightly changes before yellow then the EXOR output will change but this will be rapidly extinguished as yellow changes. So either live with the anomaly or find another way of achieving your goal. You might be able to ...


2

As with all perfect current sources, their effective impedance is infinite so it, and all the components to its left are redundant in the calculation of the time constant. The effective resistance seen across C1 is R5 in parallel with (R3 plus R4).


2

No, it is not correct in general. There are many ways that "quanity \$y\$ increases with quantity \$x\$", e.g.: \$y=ax+b\$ (with \$a > 0\$) or \$y=x^2\$ or \$y=\sqrt x\$, etc. If you say, however, "\$y\$ is proportional to \$x\$" you especially mean that the relation between the quantities is \$y= cx\$, where \$c\$ (slope) is a constant. As you can ...


2

Just realized I measured the current. Just answering the question for anyone suffering from too little sleep and missing the obvious... I measured the current and can calculate voltage drop across the 330 Ohm resistor with current. e.g. 330Ohm x 2mA = 0.66V Then I can subtract that drop with Vdd and that's the drop across the MOSFET.


2

They may well be special high precision, high resistance and/or high voltage resistors, but there is no standard coding for such features. High precision (0.01%) resistors from Vishay are often robin's egg blue. I've seen non-magnetic resistors that are green. It would help with inspection since such resistors may otherwise look the same as ordinary ...


2

We have a number of Android mobile phones permanently on charge. Note that the screens are on permanently at full brightness. Phones are not intended for such high duty cycle operation. They will overheat. This will cause the battery to be damaged. Would this strategy of trying to balance the charging current of the USB charger with the power use of ...


2

Note new REFDES (..reference designators) allowing for slight dimmer with low battery @ 8V and < =20mA max at 6.2V nominal for two Blue LEDs with some wider tolerance possible depending on parts and current. Collector R's determine the current limit say at 20mA or maybe you prefer 10mA. Let R4 = R2= (9-6.2V nom)/ 0.02A = 140 Ohms (assuming Vce=0) ...


1

Yes you could do that, assuming it’s actually a resistive sensor. The part looks like a PTC silicon resistance sensor to me, similar to an RTD. If it’s a diode or other non-resistive sensor it may not work. The thermostat would then act like a switch if you crank it one side or the other of the fixed temperature the resistor is pretending to represent. ...


1

It's a tapped inductor (autotransformer), used as part of a switchmode power converter. Inside, it has a ferrite core shaped like a bobbin with some wire wrapped around it, and then the whole thing was covered with heat-shrink tubing.


1

If the machine has a brushed or universal motor, you can use a triac dimmer to control its speed. However, triac dimmers are annoying to control. They fire on each mains cycle, ie every 10ms on 50Hz mains which inserts a large phase lag in the control loop. Also the input-output curve is very nonlinear. If you want to control needle position you will need ...


1

Green and yellow outputs are clearly being generated by a divide by 4 counter, or the two LSBs of a divide by 2^N counter. In a synchronous counter (like HC16x for instance), these outputs change at nominally the same time. However, there will be slight differences due to slight differences in propagation delay, current drive, loading etc between outputs. ...


1

When data is being sent from one GPIO pin from one microcontroller to another on the same circuit board with short traces, How long exactly would it take for the other microcontroller to receive the data? The time taken for the signal to travel down the wire will be minuscule compared to the time taken for the other MCU to respond to it. The AT89C4051 ...


1

Your differentiator circuit can also be thought of as a High Pass Filter. This might be a little easier to understand. There is a Corner Frequency for the High Pass Filter which is; F = 1 / (2 Pi R C) At the Corner Freq R = Xc where Xc = 1 / (2 Pi F C) At frequencies below the Corner Freq. the response is increasing at 6 db/octave. As long as there ...


1

Well, we have the following general circuit: simulate this circuit – Schematic created using CircuitLab Now, using KCL we can write: $$ \begin{cases} \text{I}_\text{in}=\text{I}_{\text{R}_1}+\text{I}_{\text{R}_3}\\ \\ \text{I}_{\text{R}_1}=\text{I}_1+\text{I}_{\text{R}_2}\\ \\ \text{I}_{\text{R}_3}+\text{I}_1=\text{I}_{\text{R}_4}\\ \\ \text{I}_{\...


1

Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab The input resistance can be found: $$\text{R}_\text{in}=\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\tag1$$ Now, the input current is given by: $$\text{I}_\text{in}=\frac{\text{...


1

You can get the value of R by using the Wheatstone bridge equation. Which is applicable when the voltage between the two midpoints is zero. So in your case R = 12/4 * 5 = 15 See more at the wiki page for the Wheatstone bridge: https://en.wikipedia.org/wiki/Wheatstone_bridge


1

Actually, I'm not sure that Tony's idea will help, because the surge that is tripping the over-current fold-back of the power supply is occurring when the string is first powered. At that time, the PTC resistor will be cold, and have low resistance; by the time it heats up, the supply may have already shut down. Actually, a simple small-valued resistor may ...


1

I figured out that double bouncing the headphone plug as you plug them in can trigger the higher output on my V30+ Slowly insert the plug until it's almost ready to snap in, as soon as the phone detects the headphones, let off a little and then push it all the way in... Works better than 50% of the time.


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