# Tag Info

2

Yes, you need the diode. It should be mounted reverse-bias across the solenoid coil. This minimizes the current loop of the coil discharge path.

0

Kirchoffs current law - make the assumption of whether currents going into VA is positive or negative. Remember the voltage source VR is flipped. Assume the diode is forward biased, subtract the 0.7 voltage drop, calculate the total input resistance to the op amp, calculate the gain of the opamp (-R2/Rin)

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You've made a bridge rectifier. If you put a sinusoidal 7 V RMS into the bridge rectifier and add a smoothing capacitor to the DC then you'll get about 12 V DC. The RMS voltage of a sinewave is $\frac {1}{\sqrt 2} V_{pk}$ so the peak voltage in your case will be $7 \sqrt 2 = 9.9 \ \text V$. From this we must take away two diode voltage drops (0.7 V ...

0

ideally rail to rail Rail to rail won't happen because of the diode drop. Use a PNP bipolar transistor with base on op-amp output, collector to ground and emitter connected to VF3 via the diode. If your supply rail is genuinely only 2.5 volts then you probably won't require the series diode. Any supply rail greater than 5 or 6 volts requires a diode to ...

0

Since a non-intrusive solution hasn't been proposed yet, I'll recommend one. Basically: don't build a charger from scratch unless you actually know what you're doing and understand how lithium cells/packs are charged, balanced and protected like they are in commercial products As far as intrusive solutions go, adding a diode is probably the worst possible ...

0

Your idea is good, but it won't work with a diode. Lipo batteries have specific charging requirements that are critical for safety, and the charger won't work properly if it doesn't see the expected battery voltage. A diode will add at least 0.3V to the battery voltage when charging, which will prevent the the battery from getting a full charge. Also the ...

1

If you power the comparator from the 3S battery and divide the two voltages to be compared with a resistive divider of say 2:1 ratio it will do what you want. Note that the LM339 has an open collector output and the output must be "pulled up to operate correctly. HOWEVER As Chris says, you need to charge Lithium Ion batteries "correctly" or you will ...

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2

Perhaps you are being misled by the fact that circuit simulators model a diode as a voltage source in series with a resistor. That is a fiction that is required to make the circuit simulate correctly. A Real diode does not produce a voltage across itself, but will produce a voltage drop when you pass current through it in the forward direction.

11

The guys in forensics had a difficult job enhancing the dodgy photos but the problem is clear. The meter is set to measure DC. Volts. (Barely visible in this rendering) the range selected is 'V'. The correct switch setting. Since you selected DC V and a diode doesn't generate any voltage the reading is zero. This is correct.

11

In voltage mode, a multimeter just measures what voltage is present between its leads. What you want is diode test mode, which is usually indicated on the dial with a diode symbol. On your meter, it's the option one to the left of your voltage mode--set the dial to that and press the mode button a few times to put it in diode mode; it'll say on the LCD. In ...

6

Nope. You need to set your multimeter to diode test mode. One click anticlockwise, and then press the "mode" button until the LCD shows a diode symbol.

3

The problem here is that you've stumbled into two different classifications of diode functioning. Sometimes authors get messy and confuse them. There is a first classification that takes into account bias polarity: the diode can either be forward biased or reverse biased. The second classification is the working mode of the diode: a diode may be either ON (...

0

Consider that you most likely already have flyback diodes in the relay module, and you don't seem to notice that your relays are "slow". Your valves won't appear any slower with flyback diodes unless the EM field decay is the limiting factor, as opposed to mechanical movement, and I doubt that will be the case.

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You're on the right track. Inductive loads, like solenoids, usually need a flyback diode to prevent voltage spikes when they are switched from the on state to off state. Tenths to hundredths of a second isn't all that fast for electronic devices. I suspect you'd be fine with nearly any diode type. Even relatively slow diodes will still have a reverse ...

0

Using a zener diode should solve your issue while keeping the speed fast enough. Since solenoids have inductivity fast switching causes the solenoid do demagnetize into the circuitry. You should definetly use a flyback diode when working with induction load. Zener should do the trick well.

2

On the left is a full-bridge fully-controlled thyristor rectifier. With that circuit it's possible to turn DC into AC if the outer DC voltage is ever higher than the rectified AC voltage. The current flow is reversed then. D1 stops this reversed current from the DC connector. In addition, D2 and D1 function as a crowbar for reversed polarity on the DC ...

0

You won’t get 50Vdc out of this design as provided since the Vac peak voltage varies from 130 to 340Vp. These all use active tabs that must be insulated with mica and silver oxide grease for electrical insulation, then you can earth ground the heatsink. But worse, you have huge surge currents into a low ESR cap and huge dynamic range of DC for input ...

1

Do I need to use insulation to isolate the diode's thermal tab and the heatsink from each other? If yes, should I connect the heatsink to Ground? If insulation is not used, should the heatsink be connected to the rectifying diode's thermal tab terminal (cathode) through the PCB, through the heatsink's mounting screws? Summary: Heat sink isolation of ...

0

You would need three of them to get a full bridge. There is no electrical necessity to isolate the heatsink from all three dual diodes but you do need to isolate at least two since the common cathodes/heatsinks must be connected to three different places. There may be safety reasons to isolate all three and ground the heatsink or leave it floating within ...

0

The TIP121 is not a single transistor, instead it is a Darlington with two transistors. Then the bridge rectifier has two diodes in series and the Darlington transistor has two transistors in series. The 100uF capacitor has such a high value that it takes a long time for a continuous loud signal from the amplifier to charge it to about 2.4V when it will ...

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The bottom diode provides positive feedback to the first opamp so it oscillates.

1

Steps to get your answers: Look at the diagram. In the simple diode model (fixed 0.6V drop) the diodes act like pressure (voltage) operated valves: any voltage over 0.6V (difference between the two diode pins in forward direction) will be shorted (bleeded). When there is less voltage over a diode, it will not conduct, it might as well not be there at all. ...

0

The board should have reverse polarity protection for both battery and external source... My Question: The FET (FDN340P) and diode (1N4001) work as reverse polarity protection for board? No. In this circuit the MAX6326 is powered directly from the external source. Absolute maximum input voltage rating is -0.3V to +6V, so reverse polarity will kill ...

0

That circuit does not provide full reverse polarity protection from the batteries but it's hard to connect AA batteries in reverse also at low supply voltages it may not prevent charging

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In this schematic D1 together with R1 ensures that voltage on pin U1.3 is 0V so that U1 would keep Q1 open until there is a voltage source on External Power. Q1 is there to disconnect batteries from load.

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There are only two options here the zener is ether conducting or it isn't If it is: $$\dfrac{Vo-Vz}{R1}+\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$ If it isn't: $$\dfrac{Vo-Vd}{R2}+\dfrac{Vo-Vd}{R3} = I1$$ Where Vo is the voltage a junction of the diodes, Vz is the zener voltage and Vd is forward diode drop (0.6V). What do these say? Does only one ...

1

1) This is the power schottky rectifier STPS5L60S from ST 2) mattman944 gave an excellent comparison With regards to repairing: Be sure you not only fix the result of the problem (replacing the broken components) but also the cause of the problem (why was the component broken in the first place? Are there more broken components that are not visually ...

1

2) Yes, that is almost certainly a good substitution, the parts are nearly identical. https://www.infineon.com/dgdl/irfr5505pbf.pdf?fileId=5546d462533600a4015356358535210f https://www.infineon.com/dgdl/irf9z34nspbf.pdf?fileId=5546d462533600a40153561228de1de0 If you suspect that the gate isn't driven hard, also look at the Vds I-V curves. The replacement is ...

1

I would replace D1 with a 3.3V Zener diode, and flip it around so that the anode connects to the battery and the cathode connects to R1. This will prevent the battery from pulling the microcontroller inputs below about 0.2V, and it won't draw any significant current when the digital pin is floating (or set to 0). I was trying to draw the schematic for you ...

2

An idea... If the diode anodes weren’t tied to GND, what happens? You get a large, negative voltage through each diode when each transistor turns off and the coil flux collapses. Add a capacitor to capture the charge and you just made an inverting boost converter. Rearrange things a bit: put the npn’s on the low side and connect the coils to the supply, ...

0

In theory, yes, but you want to design the thing so that the highest reverse voltage on the diode is about 80% of the rated value. You may have some trouble finding Schottkeys that'll do that for a 100V supply.

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I'm using two power source and because of "even with 0.0001 V difference, they are going to fry each other when you connected them in parallel" rule. This is not a rule. There are many cases when power supplies can be paralleled. These include battery, mains DC PSUs and AC transformers. Many power supplies including batteries, simple mains rectifiers and ...

0

My question is that if free electrons are not attached to any particular atom, then how come they leave a hole behind when they go to the P side ? They do not. A pure (undoped) semiconductor has an equal amount of mobile charge carriers (electrons and holes). A very useful property of semiconductors is that you can control the electron and hole ...

2

$10sin(\omega t)=0.8$ $sin(\omega t)=0.08$ $arcsin(0.08)=\omega t=0.080086\:rad$ $\therefore \:t=0.080086/200=0.40043\: ms\:\approx 0.4 \:ms$ Note: $sin(\theta)\approx\theta\:$ for small values of $\theta$, where $\theta$ is in radians. So we could get to the approximate answer without doing the $arcsin$.

3

At the instant in which the diode starts conducting, its current will be vanishingly small. So vL will be zero. Then v2 will be 0.8 volts. Assume that Vs = 2 v2, since the transformer is overall 1:1, and solve for v2 = 0.8.

0

The precursor to SPICE at Berkeley was an operating-point software program. That was key in, as you already understand, building a small_signal simulator.

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You are reinventing the wheel. Open source versions of SPICE* already exist. Instead of rolling your own, why not study, and possibly improve, the existing simulator? * If that link rots, just search on "open source SPICE".

1

Not too much for me to say since it's so straightforward other than make sure to read through the derivation, especially the parts related to the first two terms of the exponential series. Source: Microelectronic Circuits 7th Edition, Sedra/Smith It's a good book, though it can be difficult to tell if you're in class getting so much work shoveled onto you ...

1

The greatest possible amplitude of an undistorted output sine is: min(|L+|, |L−|) = 8 V And the amplitude of the corresponding input sine is therefore: min(|L+|, |L−|) ∕ 100 V/V = 8 V ∕ 100 V/V = 0.08 V = 80 mV The minimum (min) value of the absolute values of the supply rail voltages needs to be taken in order not to exceed the supply range and so to ...

0

Your knows are: VS = 15 V. R2 = 1 kΩ. Your unknowns are: I2. VZD. U2 (which should be the same as VZD. If that's all the information you have been given then you can only solve for U2 and I2 in terms of VZD. You won't be able to give a numerical answer as in the case of U1 and I1. Are you sure that you have supplied all the information given?

4

Those two diodes perform a useful function when the external load is connected between the power supply outputs (this configuration doubles the voltage potential of the positive rail). In this case, if the load becomes short circuit then the two outputs will fight each other, one will become dominant and in this situation the diodes prevent one of the ...

1

If you assume that the opamp is a rail-to-rail device you can then say from inspection: The configuration has unity gain. At zero load current the output limits at approximately +/- 4.3V (one V(be) drop from each supply). The configuration will clip at approximately +/-70mA and the output voltage will reduce linearly (the 10 Ohm resistor) above that current....

0

All op-amps "try" to set 0V between its + and - terminals, so connecting the output directly for feedback should make it a voltage follower or buffer (amplifier with gain 1, Vo=Vi). The push-pull part made by the two BJTs is just for power output.

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The reverse breakdown voltage must be greater than the maximum reverse voltage that will be applied to the diode. The 1N4007 is one of a series of 1 amp diodes numbered 1N4001 - 1N4007, with increasing reverse voltage ratings as the numbers increase. It is common to specify the 1N4007, even if that voltage rating is not required, as there is little price ...

0

TA7805F can handle up to 35V input voltage. I would remove the diode or use a continuity tester to see if the diode is parallel to the 12V power supply. Take care, the diode might be shorted so better remove-it from the board. If it was a Zener it should be with cathode to 12V and anode to ground and a breakdown voltage higher than than maximum voltage of ...

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