# Tag Info

0

I added two transistor same as q3 and q4 it worked. but it can improved with change some parameter!

2

There is no way (at least on my knowledge) to do that using only DC voltages all the way. At some point you have to convert them to AC in order to use (for example) a voltage doubler. In fact, that would be my solution if I have to do this: dc source -> convert to AC --> a series of voltage doublers or multipliers -> rectify/filter the output --> enjoy ...

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Usually an opamp or ADC doesn't need 1A current for negative rail, therefore a charge pump is used. In your application, also two LDOs from +/-15V would suffice. However if you need dual supply, I would recommend you to use already made DC DC module for that. They come from various manufacturers: Murata, Traco Power, TDK Lambda, Recom, ...many others. You ...

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Yes but redrawn looks like this. simulate this circuit – Schematic created using CircuitLab If done with 1 IC on a custom PCB it would look like 2 independent circuits with Nch for postive switches and Pch for negative. Some are independent adjustable, others tracking single control adjustable and others fixed Buck regulators. You can create a ...

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I am 7 years late, but I have to add my answer for others who come across this issue: The very high ripple of 680mV (if you didn't mistype it) at the output seems to me like your Co (output capacitor) is either faulty or is not the low-ESR (equivalent series resistance) type. ESR is basically a "resistance" of capacitor seen at high frequencies. If your ...

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The manufacturer doesn't tell you the current consumption, but they do give you a story problem to find it: Minimum supply voltage (VDC) = 17 + 0.02 x (Resistance of receiver plus line). From this, I infer that the current is 20mA. You could always measure, or ask the manufacturer. You could use the filter shown. It looks like it's there to keep the ...

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One thing to consider is how long the cable length is between each transducer and avoiding ground loops. If the system doesn't share a common ground (and the transducers are far away (meters) from each other, it might be good to buy isolating DC to DC converters (in this case you could use a single supply). Is there a way to determine whether this ...

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only one curly brace should be used in a statement: v={if(v(a)>{{vthresh}},{{vdd}},{{vss}})} should be v={if(v(a)>vthresh,vdd,vss)} also, another thing that you need to remember is you can only reference nodes. So if vthresh is a node, and you want the voltage, then do this v(vthresh) (I don't know what your voltage nodes are since you have not ...

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Here is a schematic that will achieve the stated goal: simulate this circuit – Schematic created using CircuitLab When the GPIO is floating (i.e. micro-controller is off / booting) the transistor is turned on by R1 and EN is driven low. When the GPIO is asserted low by the microcontroller, it turns off the transistor, and EN is floated. Not necessary,...

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An open drain buffer (like the NC7WZ07) would work. When you pull the buffer high, the buffer goes to high impedance and enables the TPS54302. To keep the buffer from operating during startup the pullup can be used before the buffer. simulate this circuit – Schematic created using CircuitLab Source: https://www.onsemi.com/pub/Collateral/NC7WZ07-D....

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Yes, you're fine with your approach. The EN pull up current is less than 2uA, so a 10K will allow you to remain below the threshold, and the microcontroller output will be able to pull it up to 5V (or at least greater than the threshold. Just stay below 7V on the enable pin and you'll be fine.) As you pointed out an open drain output on your micro doesn't ...

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The light load efficiency mode (PFM) greatly improves the efficiency in the range where it's operational, at the expense of higher ripple and variable switching frequency. It's a hysteretic mode, so frequency depends on the load current. There are other parts from TI and other manufacturers that are PWM only, or selectable between forced PWM and auto-PWM/...

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Such converters exist. That link goes to the datasheet of the Texas Instruments LM2621. It takes in from 1.2 V to 14V, and puts out a regulated voltage at up to 14V. It can deliver up to 1A. This is the datasheet of the LT1073. The datasheet includes an example of boosting 1.5V to 9V. It is only intended for low current, though. Like 16mA when ...

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Usually the gain crossover frequency is defined as the frequency at which the open-loop gain crosses zero dB. The phase crossover is usually the frequency where the phase crosses -180 degrees. The difference in frequency between the two can give hints about stability (E.g. if the phase crossover is much higher than the gain crossover it's likely that the ...

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It's hard to guess what is going on inside the black box, but I think you are right in the first sentence; the concern is to keep out of overcurrent protection. I believe this is a buck converter. We don't know what the high-side FET is but the concern is that the voltage across this FET is not significant so that it does not dissipate any power. Because ...

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Knowing all this, how do I choose my fuse from all the existing models? Typically fuses are for short circuit currents, and good for removing the circuit from operation if the current goes well above it's intended operation. If I were sizing a fuse for a circuit, I usually use at least 50% of the rated current as a starting point, and go to the next ...

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ESR (equivalent series resistance) isn't the same as impedance, although it is frequency dependent in ceramic caps. ESR represents the real part of the overall impedance, which may also contain capacitance or inductance. The overall impedance includes capacitive and inductive effects, dominated by capacitance below the part's resonance. You can reasonably ...

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Multiplying $I_{RMS}$ by $V_{RMS}$ gives apparent power. This is the hypotenuse of the power triangle, where the other two sides are the real (average) power (in watts) and the complex power (in VARs). If, and only if, the load is purely resistive then the complex power is zero and the apparent power will equal the real power. In general, multiplying \\$...

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Multiplication of Vrms and Irms gives average power into a resistive load only. Into other loads which result in out of phase waveforms, capacitive, inductive, rectifier for instance, the result is not average power. If you multiply Iavg by Vavg, the caveats are similar. For well behaved waveforms like DC, the result is average power. It's easy however to ...

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Two silicon diodes in series will knock 1.2V off the supply, getting you 15.3V perhaps use a 15A rectifier bridge connected in reverse, add a heat-sink. simulate this circuit – Schematic created using CircuitLab

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What about a voltage divider? Could that work? It would waste a lot of power, and would not regulate the voltage For noise? use an adjustable linear regulator, which you'll need to parallel a few of them for the current output. This will also waste power, which probably isn't the best thing for a battery powered application. A DC to DC in combination ...

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Step down converter like LM46002 is good and effective to get 12V or 5V with >80% eff. But note that with small load like your case (28mA), eff will lower.

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You are concerned about bulky capacitors but can trade a little loss for the benefit of lower ripple, so have you considered using a MOSFET gate follower after the RC filter? A capacitance multiplier? The capacitance at output will be amplifier gain x C4. You'll need to have a steady load current and good heat dissipation, but you already said you wanted a ...

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Huisman was spot on. The issue was the improperly sized capacitor. Just posting this to close out the question. Thanks!

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