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1

Your power source can source (dc converter or the power supply feeding it) can supply a limited power (P=V*I). So in example if a power supply is set on 1W then at 10V it can source 100mA. If a 1k resistor is placed it consumes Pload = V²/R = 0.1W which is less than the power supply can provide and you get all the current you need for it. Now if you have a ...

1

A buck-boost regulator can work with those voltages. These are available as modules. Sometimes these are called "power stabilizers". Here's an expensive one that's a 'marine' version: https://www.boatid.com/newmar/6a-12v-isolated-dc-power-stabilizer-mpn-12-12-6i.html A less expensive one: https://www.ebay.com/c/1683893376

1

how about using a 10amp or 20amp stud-mounted rectifier, in series, to provide the high current. Use an LDO to provide the 0.2 amps. And, if there is spike/trash on the +13, use 1milliHenry and 1,000uF low-pass filter, the inductor in series with VDD, and the cap shunting to ground. Place a 1 ohm resistor in Parallel with the inductor, to dampen ringing. (...

4

If your voltage drop is 15V, and your current is 450 mA, then your linear regulator, no matter whether it's just a transistor, a LM317 or an AMS1117, will have to dissipate 7W. That's an insane amount of power you waste and convert to heat. They will all overheat. So, you're wrong, and all your solutions are inadequate; they are wrong "by the same amount"....

2

All your calculations are correct. The graph is a typical I-V curve for this LED, the drop will be higher for some parts, and less for others. The table lists the maximum voltage drop at 20 mA to be 3.4 V. They did not list a minimum voltage. This is a common practice, they don't want to reject parts because they are "too good" (low voltage drop is normally ...

0

Jack Ganssle, in his newsletters, The Embedded Muse 229 and 230, rants and laments about how Datasheets should be more helpful. In 229, he says Datasheet Rant It's all about the code - unfortunately. As mentioned above it's hard to get concrete information about the Kinetis' L-series power consumption. But that's a common problem today -- ...

0

Buck/boost off of 3x AA's, or buck from 4x AA's. If you read the datasheet off of a common AA battery you will see that more than half of it's life is below 1.1V. If you have an LDO or a buck converter you simply will miss out on most of the battery life. Throw out the 9v idea unless you have extremely low power draw, 9v's are notorious for their internal ...

0

The choice on Vin beeing 4.5 (3x1.5) or 9V mostly depends on how efficient your voltage converter is. If it only "burns" the over voltage one would go with the closest input voltage that is higher then the output voltage. Also have in mind that the "burning" regulator needs a voltage that is Vout + 1V to operate correctly (Might even be higher then 1V). If ...

0

This TI appnote goes into detail about linear regulator characteristics. http://www.ti.com/lit/an/slva079/slva079.pdf tl; dr version: line regulation and load regulation vary with regulator type. Check the datasheet, and consider your line and load transients to estimate ripple. The example you give doesn’t have any load or line transient characteristics, ...

2

Linear regulators (LDO regulators are a sub-class of linear regulators) don't have ripple in the same way that switching power supplies have ripple. You cannot calculate the noise on the output of the regulator just from its output voltage, the current, and the value of the output capacitor. A linear regulator will have some noise generated by the ...

0

For the adjustable version of the LM1084, the input/output differential rating is needed because the regulator's only connection to Ground is through the voltage adjusting resistor network. Yes, the variable output version could be used with a 40 volt input to provide a 24 volt output. Quoting an input/output differential for the fixed voltage versions ...

1

Why not operate a Zener diode at its load current? Then its voltage is spec'd.

4

A Zener diode voltage regulator is highly dependent on the load current. When designing such a circuit, you have to know the current (or range of currents) you will be drawing from this regulated voltage. For proper regulation, you want the Zener diode current to be between a certain minimum value $I_{Z,min}$ (see the datasheet for a working curve, let's ...

2

Your use of the inverting switcher is a reasonable idea. However I would suggest a couple of things to improve the analog front the performance. Have the main converter LMR23625 produce 6V instead. Use an LDO to produce the 5V for your analog. Have the TPS63700 invert the 6V to -6V instead. Use a negative LDO to produce the -5V for your analog. This will ...

1

There are special IC available e. g. from Analog, Maxim or TI for that purpose. Look for "Overvoltage Undervoltage Protection" or similar.

0

Here's a suggested circuit: simulate this circuit – Schematic created using CircuitLab Now for a quick explanation: U1 is a voltage regulator. It outputs a fixed voltage of 5 V, regardless of fluctuation in the battery voltage. This ensures the output voltage stays fixed regardless if the battery voltage is 11 V, 14 V or anything in between. R1 and ...

-1

when designing my led walls (60 amps 5v per section/panel) I always make sure to not go over 80% constant draw. I like my power supplies cool. but it depends on your duty cycle and how much its on though, finding out a cheap power supply that has a peak efficiency rating at 50% load is bad.

3

You need to provide current to the sum of the maximum of all components. Your powersupply must be able to handle it if all of your components try to pull their maximum current at the same time. The only way to guarantee that is to have a powersupply that can deliver the full total of the maximum of all components. You don't need to regulate the current to ...

3

The output capacitor type mentioned is critical. The tantulum capacitor is specifically mentioned in the datasheet. Tantulum has higher ESR compared to ceramic, and will help in loop stability for the regulator. If you are switching to ceramic, there is a possibility of ringing and unstable output. I would choose I higher value of ceramic, indeed the same ...

1

Substitute them for what? I'm sure that you can substitute them for ceramic capacitors if that's what you're asking. They'll have less ESR, and the datasheet for the AMS regulators doesn't say anything about needing a minimum ESR (for what it's worth, sometimes regulators with difficult-to-stabilize control loops may need capacitors with a minimum ESR, but ...

3

Figure 1. The reference schematic shows resevoir capacitors (1) and (2) which your circuit is missing. Figure 2. Without the smoothing capacitors you get the purple dotted waveform. With the capacitor you get the black line. The voltage regulators give out the specified voltage only when the input voltage is greater than the specified voltage plus a couple ...

4

You can get switching converter modules that have a 7805 pinout with a form factor that's about the same as a 7805 heat sink (but if you bet me money, I get to choose the heat sink that it's the same as :). I'd use one of those. They're a few bucks in onsie quantities, but you have to put almost zero effort into designing them in.

1

If you want lower voltage, you can use a MOSFET, for example, AO3400A which has a maximum 32m$\Omega$ Rds(on) with 5V drive. The maximum current draw at room temperature and 5V is 217mA, so less than 10mV drop. However I don't think your series resistor is adequate to guarantee 1.5V +/-100mV during the measure phase, which may be more critical. The ...

1

VCE saturated on the OnSemi PN2222 is less than 0.1V at 150mA or less (datasheet figure 11) With nearly 150mA through the transistor you'll need 15mA or more into the base to meet the requiremnts of that graph your 100 ohm resistor should work fine for this.

1

A simple shunt regulator would be best, since your photodiodes produce a bit more current than you need, and you can control their voltage by absorbing excess current. This is only approximate and won't provide perfect regulation, but it's simple. You will probably need to adjust R2 to hit exactly 1V under typical situations; even then I don't know if it'...

0

I agree with @mkeith's comment to monitor the voltage (relative to ground) at the LM317's power input pin when the fan is attached to the LM317's output pin, and the fan is running. If at all possible use an oscilloscope, not a multimeter, to perform this voltage measurement. (NB: If you don't have an oscilloscope, use a multimeter to measure this voltage.) ...

1

I am 99.99% sure that it's connected to Vout, like every other xxx1117 regulator out there. You should connect the tab to a large area of copper if the regulator is to dissipate much heat, since the die is mounted on the leadframe in such a way as to allow most of the heat to be conducted out through the tab. Image of typical SOT-223 package from this ...

0

This datasheet seems to indicate that the fin isn't connected to anything in particular. The safest thing to do would keep it electrically isolated. You could also use an ohmmeter to see if it's connected to another pin, but that doesn't guarantee it's not connected to an internal node.

2

In general the tab would be connected to the middle pin. So no, there is no reason to think the tab is GND. The tab is Vout, like on many other 1117 type regulators.

0

Take a look at the isolators section on digi key https://www.digikey.com/products/en/isolators/digital-isolators/901 or level shifters. https://www.digikey.com/products/en/integrated-circuits-ics/logic-translators-level-shifters/732

0

RDS3115 needs 1.5A on 7.2V, but look into LM338 datasheet and you will see that Icl current limit is 1A for Vin-Vout=30V. Oscilation is result of current limiting (LM338 limits current by lowering output voltage).

1

Your regulator dissipation is about 14 Watts - you'd need a significant heatsink to handle that. Power = V x I = (12-5)*2A = 14W. Without a heatsink "you haven't got a show" to use technical language :-). The datasheet table 7.5 page 6 shows the thermal resistance in C/W for the T model regulator as 23 °C/W, - say 25 ° C/W, so say 100C rise you'd need a C/...

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