New answers tagged

1

The chip tries to charge the batteries, but it can't know if there is only battery or load too, so it is unlikely that the charger chip can determine when the batteries are full. The charging chip does not output anything, it just charges batteries. The output is the battery output. You can't use a resistor to drop voltage, you need a buck converter. You ...


0

First off, this is a not a well designed product so the datasheet is incomplete. If you can't find the answers it is probably because of the datasheet and product information and no fault of your own. In the diagram above, you'd want a step down regulator if using 2C (7.4V) or step up regulator if using 1C (3.4V) where the resistor is. A resistor will not ...


1

I'd consider using one of these: -


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This was done with the trick LED series cct. Each series switch sends some HF current and each series LED receives some parallel LC HF voltage to drive LED , so a single wire loop can select the matching LED and the C conducts the other freq. current and while the L blocks the voltage. So the switch connects the L//C while conducts then each LED has L//C ...


2

So, sounds like a bandpass filter, followed by a power detector. Detection can be as easy as a diode rectifier followed by a simple RC lowpass. Feed that into a transistor as switch. Amplification might be necessary at some step. Sorry that it's not more specific than that, but you probably know that in RF engineering, things very much depend on the actual ...


1

You can use a voltage reference (e.g. MCP1501) for a precise reference; for instance: simulate this circuit – Schematic created using CircuitLab In this case the LM will output a 1 when the voltage on the battery drops below 1.2V * (27k+13k) / 13k = 3.69V If you need a less precise way of measuring, you can use a zener or even a diode (even less ...


2

Here's a circuit I did for a similar question you raised yesterday. It uses a very low-power voltage reference such as the MAX6006 (it consumes less than 1 μA) or the LT1389 which is also sub μA consumption. Because they are shunt references they will work down to quite low battery voltages. For instance, with 0.5 μA flowing to feed the reference, R3 drops 1....


0

A zener diode is probably the easiest, at the expense of what could be a significant amount of current running through it. If your circuit can be modified to use a different reference, a bandgap voltage reference uses less power but 3.7V options aren’t available.


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Potential divider will do exactly that divide voltage supplied to it based on a ratio. You need a zener diode with an appropriate resistor. This way Zener voltage stays constant even when the battery voltage drops. Of course, within the limit at some point, your battery might drop below Zener voltage and your reference will not be valid anymore. ...


0

You need a voltage reference or a LDO regulator. In the simplest (and not very precise) case, this can be a zener diode with a resistor.


1

BMS is a balancing that depends greatly on how well matched the batteries are. If they are mismatched there will be problems to overcome due to heat dissipation limits when the weakest cell has reach full charge while the others must be in CV mode where the current is declining. If it is still in CC mode, you have a big heat problem to dump 4.2V*CC amps. ...


1

Yes They protect against over charge, over discharge, over current and some of them balance the cells which is needed every once in a while. Never charge them unattended, don't charge in cold weather (below 5C), don't charge in hot weather (above 40C), ensure the connections are well secured, ensure the BMS is correctly set for the voltage and current ...


0

A gain of 500 is pretty aggressive for a single stage. You're essentially multiplying your input offset error and bias current/impedance errors by the gain. At a minimum, you should be adding a resistor to the input to ensure that both inputs are driven from similar impedances; if you don't do this, your inputs will see different voltage offsets, since they'...


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C29 should be removing the bias voltage from your input and since you’re running the op-amp on a single supply you should be seeing clipping of the negative half (with respect to the bias). In my mind, the easiest way to get a consistent bias at the output would be to use a summing stage as the last stage; you might be able to lump that in with your ...


4

Sometimes the people just use a 1M resistor in series with the power; personally I'd also add a small zener to reduce risks simulate this circuit – Schematic created using CircuitLab However, personally I don't like this. Remember that in this case the ground is tied to the neutral wire, and so the rPI is no more insulated. This is usually a thing I ...


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It's possible with an 555 CMOS timer IC, but only for light loads. The lower frequencies may require a higher C2 value.


0

I doubt that your setup is going to work. I see several issues: The USB-C-to-HDMI adapter most likely expects a power supply of 5V on VDD. If your YZ12PDN board negotiates 12V, it will likely kill the adapter. Video data is transferred using USB-C Alternate Mode. In order to activate it, it needs to be negotiated via CC1/CC2. This is in conflict with ...


1

There are a surprisingly large amount of things to tackle for input protection: over voltage (constant), over voltage (ESD) and reverse voltage protection. One of the easier solutions is to use a fuse or polyfuse, and a TVS or zener shunting to ground in the case of overvoltage. The tricky part is sizing the fuse to allow proper operation, but prevent damage ...


2

High Voltage contactors need to switch as fast as mechanically possible to reducing the harmful effects of arc currents on the contact surfaces at extreme temperatures ~ 6000’K. When a simple relay or inductor is clamped by a reverse diode, the same coil current travels thru the diode with the reverse voltage. When such clamping occurs the force on the coil ...


3

The fuse and zener is a good and commonly used solution ... Others include, marking clearly the DC voltage intended (or AC otherwise) to be used, close to the power supply input. This would normally be on the product case, next to the supply input. Also making it difficult to otherwise plug anything else but your intended voltage, such as a mini-USB or USB ...


2

An economical solution is Toshiba 5.5V (5.1 Vth) TVS ($0.036 @ 1k) with desired 1A Polyfuse


3

I would consider something like this ciruit: simulate this circuit – Schematic created using CircuitLab The fuse will blow when the input voltage exceeds 5.1 V If replacing a blown fuse is an issue, consider a polyfuse which self-resets Instead of the zener diode many other solutions can work, a zener + Thysristor solution like this: Borrowed from ...


1

Use a micro USB socket for the power connection. All USB shaped power connectors provide 5V. So the 9V plug doesn't fit.


2

The most common solution is to have a voltage regulator in your device. Your circuit itself runs on 5V, but the regulator can accept a much higher voltage (say, up to 40VDC) and make 5V from whatever input it gets. That's what most consumer devices do. If you want to protect against more extreme mistakes, then a circuit such Andy aka describes can handle ...


9

Probably the most common method is to use a fuse with any (and sometimes all) of the following: - A regular power zener diode A power shunt regulator (like a zener but programmable) A crow-bar circuit (on over-voltage, it clamps to close to 0 volts) So, if the voltage exceeds a certain level, too much current will flow into the aforementioned devices/...


0

You would want 3. For the reason you give for 1, and because floating the ground would not be the best way to do it. But also because you don't really want the ground to be connected on the audio like that due to ground loop issues common with audio and devices that have both internal batteries and are powered via mains adapters. And 2 isn't good because the ...


0

Option 2 is the way to go. We'd draw it like this. simulate this circuit – Schematic created using CircuitLab Figure 1. Double-pole, double-throw switching. Connect the mic lines together as you suggest. Both the phone and the laptop will provide a small bias voltage to the mic but it shouldn't cause any trouble.


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simulate this circuit – Schematic created using CircuitLab The circuit essentially works. Each resistor can be set to modify the differential voltage between the signal VCC's and they can also be used to set the offsets. It is essentially a balancing bridge. In my case all the variable resistors are 10k which match the pullup's on VCC and the pot. I ...


0

I recently reverse-engineered a powered speaker that used this circuit to mix between "dry" (no effects) and "wet" (effects, reverb, etc.) versions of a signal. Keeping a constant volume throughout the range of this control is fairly important. simulate this circuit – Schematic created using CircuitLab They used values for R1 and ...


0

You are going to have to experiment to find the best panning curve for your application (and preference). Start by changing R3 to the 50k pot that you want to use. Increase the gain of the circuit by increasing the value of the feedback resistor (the extreme right-hand resistor) from 10k to 47k. Adjust that value to suit your prreferences. If you leave ...


1

If your intention is TO DESIGN a control system for a given existing plant using only passive components, then it would be difficult to achieve your goal as you would lose at least two fundamental building blocks in control systems: (1) Subtractor (\$e=r-y\$), requiring an inversion in \$y\$) signal; (2) Amplifier with gain > 1. Also, that inversion in (1)...


0

replace the pot with a 3 throw rotary switch.everything then becomes much simpler. or you could try rthis. simulate this circuit – Schematic created using CircuitLab you'll need to adjust your pot's mounting so that the resistance can go down to zero and up to at-least 40K. it might need two diodes in series.


4

The valve works by attracting an iron armature into a solenoid. This means that when un-energised, there's a large airgap in the magnetic circuit. When it's energised, the airgap is much smaller. The large initial airgap means that a large current is required to develop enough force to move the armature. When the airgap is smaller, less current is needed to ...


1

Here's the initial k-map: $$ \begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} &\overline{A}\:\overline{B}&\overline{A}\: B&A\: B&A\: \overline{B}\\ \hline \overline{C}\:\overline{D}&1&0&1&1\\ \overline{C}\:D&0&0&1&1\\ C\: D&0&0&0&1\\ C\:\overline{D}&0&0&1&1 \end{array}\...


0

You could use two pieces 20x3 2.54mm pitch headers (120 pins total but only two parts to stuff) and one shunt for each power pin. Samtek ones are in stock at Mouser. Just run the GND and Vdd along the outside connections and link each center pin to a pin on the ZIF. But I think I like @DkNguyen's daughterboard approach better. Less likely to fry a chip, for ...


0

Diode tests vary with instruments but 0.5mA current source is about what I expect so 0.100 implies 200 Ohms. A diode would be closer to 0.6V. Reversing the polarity tells you if the difference is nonlinear from diode current as opposed to linear , (equal value) for resistance.


0

The diode drop test works by pushing a small current through the diode and measuring the voltage. The diode you pointed out most likely has a resistor (possibly 120 ohms?) in parallel, which is passing the test current with a 0.1V drop.


0

If you are measuring the forward voltage of the diodes with the diodes in the circuit, then you are probably seeing effects from the rest of the circuit. Take the diode out of the circuit and try it again. You'll probably measure the expected 0.7V


0

Use pull down resistors on the Arduino input pins. Assuming the output of the encoder is logic high.


0

Breadboard is likely to have higher parasitic capacitance between tracks than either of your filter capacitors (10 pF and 5 pF) - this means trouble The 100 uH inductor is quite possibly self-resonating below the frequency you are aiming for and this means you won't get sustained oscillations. For both of the above try lowering the inductor value (L1) by ...


2

Oscillators work because the output gets fed back to the input with just the right phase shift to be the same phase coming out again. In the case of the that circuit, that means that the output is -- roughly, because the chip has delay -- in phase with the input. It's close to being in your "acts like a resistor" mode. The weird thing about that ...


0

I'm wondering why a 20mA current draw would drain 3800mA battery in 2 days. Simple mathematics indicate a good week of standby time with the module always on. It's very likely that the battery itself is cheap/beated, being not able to provide the capacity it's rated for. Or, maybe the pump draws too much power. Calculate the consumption of the motor by it's ...


0

assume I would need to use two functions generators You would not need to go through the trouble of buying signal generators, rather construct your own oscillators as they are easy to make even on a breadboard, depending on what is your frequency of interest. If you are looking for fixed frequencies for your signals, a phase shift oscillator would be good ...


1

You need to add these two voltage signals. There's many ways of doing that, for example using an opamp in summing amplifier configuration. I assume I would need to use two functions generators, how would they need to be set up? Well, you don't need a function generator to generate an oscillation. Any oscillator does that, and there's hundreds of different ...


0

OpenGL fix fixed the problem for me. Just download the openGl fix file, then paste the dll file on your program files if its 64 bit and prog*86 if its 32bit.It should work


0

The chip uses internal regulators so all you need is 3.5V Min at the end. If you get this it works for 5m but not 10m. Thus you can distribute AWG16 or suitable wire to either 2 starts or 1 start 1 end or all 3 Nodes. Testing the end supply voltage with 1 Start will tell you when all RGB are on what your drop is on both V+ and rise on Gnd which may be equal. ...


0

you may consider this [on-line calculator][1] to get you started (am sure there are many more other good articles explaining this). Yes the differential impedance should be of the order of 100 Ohms. Other considerations, well, try to keep the lengths equal as suggested, and as short as possible, without obstacles or vias changing layers. If its all on the ...


1

The description of loop compensation in the UC3842 datasheet is probably the best description you will find of how the closed loop control should work. The loop bandwidth is limited by the frequency. Gain is controlled by R69, in conjunction with the CTR of the opto-isolator, gain of the TL431 error amplifier etc. See equation 58: Based on the datasheet, I ...


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