New answers tagged

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Any time you are dealing with Point of load converters always start from the load you will be feeding. Once you understand your load and its requirements you may start working your way back. Load requirements may include Maximum Voltage Ripple, Maximum Tolerance Band, etc. There are more to be considered here but for hobby use this is enough to get you ...


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The LM7805 is basically an NPN Emitter Follower with feedback so removing the input makes the regulator high impedance unless you wish to block the ~4mA bias current . You may verify my assumptions.


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Modern single-inductor buck-boost converters seem to fit your requirements with efficiencies exceeding 90 % at 50 mA and much below. The one I looked at was TPS63021 from Texas Instruments. According to the datasheet, with power-save mode activated and input voltage of 3,5V, the efficiency should be above 95% with currents ranging from 1mA to 2A. This is ...


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I'd consider using a low-drop-out buck regulator that works almost down to the wire. Such as this one: - When the input voltage drops to 3.3 volts, the switch inside basically remains on hence, it acts as a regulator down all the way to virtually 3.3 volts. There are probably several examples of this tech from the usual vendors. Note that the efficiency ...


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I suggest you use a 9V linear regulator as a 'pre-regulator' to power the Arduino through the Vin pin or DC input jack. That way any noise on the 12V supply has to get through two stages of regulation before it can affect the Arduino. The Arduino Uno's DC input jack has a diode in series for reverse voltage protection. This has the advantage that the power ...


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The screen spec says "Input Power Adapter: 5Vdc-12Vdc More than 2A" - which I presume means that it may draw up to 2A. The XL6009 boost converter is rated for a maximum switching current of 4A. The 4000mAh Lipo cell has no current spec, but it can probably do 0.5C (2A), and the 2mm JST PH connector on it is rated for 2A. So it should all be good, right? Not ...


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All 24v panels can hit over 37v and as high as 50vdc at Voc MPPT has no bearing here, so the many mentions of that are not particularly helpful. An inexpensive PWM regulator is fine for loads, MPPT just maximizes panel output which is not necessary or useful here since there are no loads mentioned beyond the fans which could make use of the extra energy ...


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It looks like you can manipulate the enable pin of the TPS7A4701 (see chapter 7.4) (On the breakout PCB a pad with in silk screen "EN" sits in the corner of the PCB where the inductor with marking 2R2 sits). You should check the connection with that pad to pin 13 of the IC. You can use a low quiescent current comparator to pull down the enable pin when the ...


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Do the math on your power budget: The Pi needs up to 2.5 A @ 5 V = 12.5 W The motor needs up to 3.0 A @ 12 V = 36 W That's a total of 48.5 W Taking the efficiency of the converters into account, which is going to be 90% or less, that becomes at least \$\frac{48.5 \text{ W}}{0.90} = 53.9 \text{ W}\$ that you need from the battery. But you say that the ...


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You may misunderstand how fans work You need to check the spec sheet on the fan in question, but a great many fans do not regulate speed rheostatically. In fact, in residential wiring a very common mistake is to try to control fan speed with a lamp dimmer, which is itself not rheostatic and is a triac leading-edge or trailing-edge device. You should be ...


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A Solid-State Relay (SSR) is just that: a solid-state (semiconductor) version of a relay. A relay is an electro-mechanical switch that either connects or disconnects contacts together under control of an electro-magnet within it. So your SSR is an on-off switch. Unlike an electro-mechanical relay, it has no moving parts or electro-magnet. Its functions are ...


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I think you've misunderstood how a solid-state relay works. The coil, or input, will work on 3 to 32 VDC. The contacts, or output, are rated for 24 to 380 VAC. When the "coil" receives its required voltage (SW1 closes below) it closes the internal "contact". In a normal relay, this is an electromechanical process. In an SSR it is an electronic process. ...


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You need to lookup the latest version of the datasheet on the ST website, where it shows VO=VADJ(1+R1/R2) Later on it gives you the specs for Vadj, typically 800mV.


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It is standard. The regulator adjusts the output until the voltage at the feedback input pin matches the internal voltage reference. Therefore, you just make a divider that steps down your desired output voltage to match the internal reference voltage and feed that to the feedback pin.


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The datasheet states: This capacitor need not be an expensive low ESR type: aluminum electrolytics are adequate. In fact, extremely low ESR capacitors may contribute to instability. Tantalum capacitors are recommended for systems where fast load transient response is important. Where the regulator is powered from a source with high AC impedance, a 0.1 ...


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Is there a better, more efficient way to handle this? There are brownout protection and power monitoring chips that solve this problem in a single chip. I am not sure whether there is one tuned to your specific requirements (5.4 V nominal voltage, TBD threshold voltage for turn on). Is there a risk that when the OpAmps are at a negative voltage output (...


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Schematic The original schematic is the result of experimentation in the simulator, and it makes the circuit looks more difficult. For the purpose of understanding, let me redraw it. As we see, the circuitry contains three building blocks: A Zener diode as voltage reference. An opamp as a non-inverting amplifier. A BJT as emitter follower for our output ...


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Almost there! :-). This is a MUCH better attempt than most people make. For guaranteed voltage levels you want Vpin ideally inside Vdd-Gnd range and at worst not more than 0.3V above Vdd or below ground. Older data sheets tended to use hard Vdd/gnd limits. More recent ones tend towards a typically 0.3V over/under voltage allowance. Nothing has changed in ...


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\$R_1\$ and \$D_6\$ present a very basic zener voltage concept. You must become very, very familiar with it. It should be immediately obvious to you every time you see it. The basic idea is that the resistor, \$R_1\$ helps to "set the operating current" of the zener diode (and possibly supply a relatively much smaller current to something else attached at ...


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LDOs are not converting voltages, they are dissipating excess voltages into heat. Thus choosing a different LDO won't change anything regarding that. Your selected LDO has a thermal shutdown feature. It could be that you are triggering that due to overheating. Have you checked that your LDO can sustain what you are asking it to provide ? To find out if your ...


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Here is why it doesn't work. The reason it works if you apply the load after the supply has stabilized is because of the 3.3k resistor across the PNP. Without the 3.3k it wouldn't work because your 2.5V reference would never come up. With a load applied before the power, it draws all of the power and reference stays at zero. Remove the 3.3k and get the ...


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I Assume there's no other loads for the 7805. You can take the input to the 7805 from +50V through about 40V zener diode. I guess the 7805 sinks few milliamperes only, so the dissipation in the zener diode is well below one watt. It would be one watt if the current were 25 mA. To be sure how much the zener dissipate, measure the current consumption of the ...


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All the LM317 can possibly do is act like a (controlled) variable resistor between the input and output, it cannot create energy out of thin air. And it can't even do that, being a bipolar part it will have some significant voltage drop at best with even moderate currents (some CMOS-output LDO regulators have a few mV dropout at low currents such as a few ...


3

An LM317 is defined by many parameters, not just the equation you provide. All of the required operating conditions must be satisfied, and one of the those is the "minimum dropout voltage". This is the minimum difference between the input and output voltage, where the input voltage must be greater than the output voltage. The datasheet for your specific ...


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Are you building your circuit on a solderless breadboard? If so, it's very easy to miswire your feedback resistors and see strange problems like this one. I've seen similar problems before. Also, I see you have a decoupling capacitor at Vadj, the junctions between resistors and capacitors can be pretty confusing on a breadboard, and can be easily miswired. ...


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The packages are just containers for various bits of silicon, sized based on how large the silicon is, and how much heat it needs to dissipate, There will be a part number and manufacturers logo printed on 1 face of the device, generally its not too hard to work out what it is by searching for that number, e.g. 7805, or LM741, But there are some harder ...


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The 120 and 200 ohm resistor values calculate an output voltage of 3.33V, not 4.2V. Maybe you are using a defective or fake IC from ebay?.


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As you didn't tell more about the circuit . That must be a noise problem mostly could be a ripple problem you have to stop ripples from being amplified by choosing the right capacitor values . 8.2.2.1.2 Adjust Terminal Bypass Capacitor The adjust terminal can be bypassed to ground with a bypass capacitor (CADJ) to improve ripple rejection. This ...


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You lack much of the detail necessary to provide a viable answer. What is the range of 12V output that can be tolerated? Can the 12V supply go above 12V or is that an absolute upper limit? What is the range of 9V output that is expected? What is the current range for the 12V output? What is the current range for the 9V output? Are you expecting low loss (...


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It was written: "6 * 1.414 = 8.484 VacRMS No load factor 8.484 * 1.64 = 13.91 After rectifier 13.91 - 1.4 = 12.51 Vdc" Because it is a 'short-proof' transformer, it will be inefficient. I interpreted the calculation this way and presumed to fix the apparent transposition of 1.414 and 1.64. I hope it makes sense: 6 * 1.64 No load factor = 9.84 VacRMS (no ...


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As with any regulator, increasing the size of the output capacitor reduces over- and undershoot magnitude but increases duration of the transient response. So the datasheet is saying it's detrimental to widen a transient spike or pulse, why is this? No - the data sheet says that the duration is increased - not that the pulse is widened. The difference ...


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Heat is not the problem. According to other things I've been told on this site, MOSFETs suffer a thermal runaway problem (hot spots on the silicon) when they are used as voltage controlled resistors. Is it okay to use a MOSFET in its resistive region with a heat sink? If you use a linear regulator which is made for the purpose, you will avoid the hotspot ...


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the answer of @DKNguyen is correct. it is normal. so please attention : Pout = 50ma*5.2v = 265 mw Plost= 50ma * (13-5.2)V = 0.400 TO-220 thermal resistance = 100C/W T = 25C + 0.4*100 = 65c it is posible then we have to use heatsinking


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If you want to avoid heat, a buck converter is the correct thing. Without that, you're going to be generating heat somewhere. Sticking a sufficiently chunky wire-wound resistor in series with your circuit would lose some of the heat that would otherwise be generated in the mosfet. Perhaps a 40 ohm resistor, which would drop the 40V to 20V at 500 mA. But ...


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11-13V to 5.25V @ 50mA linearly regulated is 0.2875W to 0.3875W dissipated. If ambient is 25C, then a temperature of 49C is a rise of 24C above ambient. That works out to a junction to ambient thermal resistance of 126C/W. The BD139 has a listed thermal resistance of 100C/W without heatsinking. So that seems pretty close.


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The 12V supply does not need to be very accurate as it is being used for two things The above 12 V, If it can be say 11 V (10% tolerance) I believe the 4 - 20 mA sensor can still work. Consider, LMR36006 4.2-V to 60-V, 0.6-A Ultra-small synchronous step-down converter for example. The drop out voltage is about a couple 100 mV. This is measured ...


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I don't know about the TPS55165 specifically, but buck-boost controllers are specifically designed to smoothly transition from the input being lower (or about equal to) the output to the input being higher. One would hope they'd do the job with ease, but then, it's not the easiest job. Scrutinize the data sheet, while reminding yourself that it was written ...


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So the datasheet is saying it's detrimental to widen a transient spike or pulse, why is this? Because a transient will no longer be regulated, it will spread out and affect the regulated voltage. This can create a problem for loads that need a stable regulation. It also increases the time spent away from the set regulated voltage.


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It can be detrimental. Or it may be better to have a shallower pulse, but not as good as it would be to have a shallow short pulse.


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There are ICs dedicated for this purpose, called voltage supervisors. Texas Instruments make a few, with different voltage and reset options. I've not checked other manufacturers, but you can just search for voltage supervisors, and a few should come up. TI has also made a comparison between the discrete and IC based supervisors (with some obvious interest ...


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The part actually supports 40V, it says 0 to 36V because that is the voltage range that the specifications of the part will work. The part has been tested for 0 to 36V. The absolute maximum ratings are ratings for which the part will fail you can't go more than 40V relative to Vout So you could do what you suggest if the voltage drop across a resistor was ...


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Even though this has a lot of steps, it's still an outline. You need to go through the circuit very carefully and make sure you're not accidentally powering a gate or the processor through an input pin -- so you need to insure that all pins are undriven, or driven to ground, when the processor is off. Find a really low-power CMOS single-gate logic family ...


1

You have to use a logic level shifter IC. You can use a dedicated IC which can translate the signal from one level to the other. you can also use the below circuit if there is only one signal to convert from one voltage level to the other. The MOSFET can be logic N type MOSFET. Image and many other solutions from this link: https://next-hack.com/index.php/...


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Chances are high that in addition to feeding 6V into a pin that's not even 5V tolerant, you're trying to run a 74HC part with woefully inadequate pin voltages. The easiest solution is the proper voltage divider, run your sensor at 5V, and use a 74HCT165 (note the 'T' in the part designation). The 74HCT line is made to run from 5V and accept TTL input ...


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While the datasheet is not clear on what the absolute maximum ratings for the IO's are, I am certain that it's 3.6V. Usually "+0.3V" means that the inputs are protected with a diode that turns on at Vdd+0.3V. So this means don't tie Qh directly to the port of the ESP32, it could burn out the diode. Source: https://www.espressif.com/sites/default/files/...


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This Is Battery Operated by a 12v Lithium Ion battery pack Using a linear regulator from 12V to 5V would be inefficient (about 60% power lost as heat) so I presume you used a switching regulator. If you don't know how to design one, you can get a canned DC-DC module for a few bucks from the usual suppliers. If the arduino is the only load on your battery, ...


0

It depends on current draw. For a linear regulator you must be very careful of the power dissipation in the regulator (Vin - Vout)*Iavg, and then look at the thermal dissipation calculations. You really want to minimise this heat, so if your 5V is a switcher and has enough current capacity, it's almost always going to be better to run the 3v3 linear from 5V....


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+1 for the Clear question, associated data and response for queries in the comment. Can I power a Raspberry Pi with \$24V\$ through a TLE4271? Let us look at some tables in the Datasheet. The TLE chip comes with several features. Do you need all of them? Do you have many TLE in stock? anyways, let us see whether... The TLE can supply sufficient ...


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Here is a useful number for you: standard PCB foil (default is 1 ounce of copper per square foot of area) has thermal resistance of 70 degree Centigrade per watt, per square of foil. how much heat so you need to remove from the IC? How much heat do you need to remove from schottky diodes? From the MOSFET switch? Per the datasheet, the main heat-exit ...


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On top of what @TimWescott already answered, you can also solder along the traces to allow for more current to pass through that trace. This is common practice especially with UPS board that usually have high current traces running through it. Here's a good example of a tinned PCB that can handle more current than untinned traces. Img src: https://www....


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