New answers tagged voltage-regulator
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You're looking for an OR controller. You'll need an external MOSFET and some other little passives but you can get the controller in something as small as an 8-DFN which is 2x3mm.
Also the NCP1117 has a dropout voltage of 1.4V so your 3.7V Lipo won't be able to power it.
Edit: if you don't care about the losses, you might be able to use a common-cathode ...
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I assume you want to build a proof-of-concept, rather than an optimised turbine that produces a significant amount of power.
From what I understand, a generator behaves more like a current source than a voltage source. Because of this, you can take advantage of the LED's nonlinear voltage-current relationship and dispense with a regulator entirely, as long ...
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Your biggest concern when using a small motor as a generator won't be getting too much voltage or current. Your problem will be getting enough voltage and current.
My son has a small steam engine.
Not that model, but one like it. It produces more power than your home made steam turbine is likely to.
We built a generator from a small motor and used it to ...
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For just one LED, you can use a current limiter circuit built from 2 transistors. Here's an example (simulate it here):
More here: Controlling High Current LEDs with an ATmega328
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The first regulator has an absolute max rating of 1.5 W, yet you're dissipating 4 W through it ((32-5) V times 0.15 A).
In the second case it is dissipating (12-5) x 0.16 = 1.12 W, which assuming a thermal resistance of 100°C / W would give a junction temperature of 112°C + ambient temperature, very close to the maximum allowed 150°C.
With such a high ...
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Yes, the regulators are not suitable for the task you need them to perform. You are exceeding their limits.
Example 1: Drop from 32V to 5V at 120mA. The regulator needs to dissipate 3.24 watts. It is rated for max 1.5W and it will overheat.
Example 2: Drop from 12V to 5V at 160mA. Regulator needs to dissipate 1.12 watts. It should handle it with enough ...
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As soon as I turn on the bulb the voltage drops to 6.43 V
Something is terribly wrong. That is only 0.8 Volts per cell. Much too low. Try to measure the input current. Maybe the boost is not operating properly for some reason and is consuming vast amounts of input current (although it would be getting hot if that were the case).
Also, consider using 10 AA ...
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You measured .85A at 12V. With an input of around 10 volts, the current draw would be right around 1A. You say the cells are 2 Amp-Hour. With those numbers I'd say the most run time you would get is 2 hours.
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Your AA rechargeable cells must be very old, not fully charged or very cheap to have their voltage per cell drop to only 6.43V/8= 0.8V.
A Name-brand AA Ni-MH rechargeable cell can supply 1A for almost 1 hour when its 1.4V has dropped to 1.1V.
A name-brand AA alkaline battery can supply 1A for half an hour when its 1.5V has dropped to 1.25V.
A very cheap no-...
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Increase your resistance a bit
Half a watt at 5 milliohms is 100 Amps, that's rather a lot, so consider going to a smaller wire size. Let's say you use a fine enough wire that you only need 10 A.
Use a current-limited power source
If you ensure that your power source will not deliver more than about 10 A, the voltage is produced is only important if you care ...
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This is a buck regulator, it's physically impossible for it to output a higher voltage than its input voltage. When you request it to output a voltage that it can't produce, it will drop out of regulation and go to 100% duty cycle. It won't get damaged and will continue to operate normally once you request a "sane" voltage again.
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Voltage drop depends on where you measure it and your reference.
Your question lacks two things which makes it ambiguous. The reference and the test points.
Load Regulation % Error is listed in all datasheets for all regulated power supplies. It does not include distribution loss from Ohm's Law with associated cables and connectors.
It is measured as ...
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The difference is context. Voltage drop is measured across loads, for example voltage drop across a length of wire. More current will cause more voltage to be dropped across that length of wire.
Voltage regulation describes the behaviour of a voltage source, where an applied load causes the output voltage of the source to drop. For example in transformers ...
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Adding to other answers: use of a transformer will also provide safety isolation. Especially important if your power source is rectified AC, as found in your average switch-mode power supply. Assume that you will touch the wire at some point.
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If you're serious about wanting a whopping 200W, the M4-ATX-HV from mini-box (the US Pico-PSU distributor) claims to do it. They've also got a variety of smaller options including some lower powered LiFePO4 UPS gadgets.
Realize that 200W means you need to be prepared for 17A sustained current in all of your 12V wiring. Both inside and outside the batteries....
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In practice extra length of wire will give you better mounting options, better electrical options, and allow you to fine-tune the voltage (though PWM is also an option there).
I hacked together a rather larger hot wire cutter - for 6"-thick foam that was delivered at the wrong width and I had no time to send it back. My drive was rather easier than ...
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This is where the old school solution of a transformer is by far the easiest way to get the low voltage and high current that's needed.
Find a transformer you can put a single secondary turn onto, with as thick copper wire as you can manage, or several pieces in parallel. Use a toroidal transformer with an open hole, or indeed any transformer that's unpotted ...
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As others have mentioned in the comments, your wire is thick. Making it thinner will increase the resistance, so setting the current will become somewhat easier, as it would decrease the needed current significantly.
An useful (but somewhat dirty) trick my father used to make such things was to run them on a small transformer. This will limit the power just ...
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Let’s see your wish-list:
Charge a 12V car battery from the “main battery”. <=> Assumed here the main battery is the battery connected to the car starter engine and alternator.
Use of thin cables, to not draw to much power in case “aux” battery is empty.
Here is a problem, as thin cables should not be used to present a high resistance to limit the ...
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As a way to diagnose the part what happens when you set up the 317T as a standard voltage regulator:
Input Voltage = 29V
Connect the six 3.5V leds in series with your 4.7 ohm resitor.
Connect the resistor and LED series string directly to the output pin.
The two resistors to set up the output voltage
VO = VREF (1 + R2 / R1)
(neglecting IADJ current)
VO =...
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It is not possible to draw 5A with a 7824. In addition, it is not possible to regulate with just 1.8 V more. These regulators have a dropout limit of 2 V (typical.) Incidentally, series regulators are the worst choice as the excess voltage has to be dissipated as heat.
Perhaps you can proceed more skillfully and not regulate the voltage, as this should be ...
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If a voltage regulator was the only feasible solution I'd be looking at a low-drop-out buck regulator like the one below using the LT8638S (for example): -
The example circuit above is for a 12 volt regulator but it looks like it should be good for a 24 volt output at up to 5 amps without much trouble.
I'm not recommending that device in particularly (...
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You need a voltage higher than the LED voltage by a few volts for the LM317T to regulate properly. At up to 1A, 2V + 1.25V is typically enough for temperatures above 0°C.
The LED voltage will increase as the current goes up. If you need 21V for the LEDs, then your power supply has to be 24V or a bit more for it to work properly. At that power dissipation ...
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Minimum voltage should be 24.2V. 21V LEDs drop voltage plus 1.2V on resistor, plus minimum 2V IC drop voltage. Also chip requires heatsink. If the chip overheated, internal thermal protection will restrict output current.
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Ignoring the option of low-quality, or internally current-limiting LM317:
Have you verified the resistance of your 4.7Ω resistor? What's the voltage across it during operation?
My interpretation so far: It just gets really hot, and thus changes resistance:
300 mA across 4.7Ω, that's 0.42 W (since V=I·R, the power P=V·I=I²·R), and now the thing gets hotter, ...
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You would use a regulator circuit. Since the current required and the voltage difference is low, a low dropout linear regulator would work. A step down or buck switching regulator would work as well. Worst case you could use a 1n4001 or similar diode to drop it down the nominal 0.7V of its forward voltage drop.
But you don't need a separate circuit. The ...
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You don't need to do this, you can directly power it from your 5V supply using the VBUS pins (JP3 pin 3). The board already includes the necessary power management circuitry to generate the 3.3V module supply from 5V USB (or any other 5V) input.
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5v for the logic shifter: I could do the same as with 3.3v, but what
if the power supply itself is 5v? I was wondering if there is a better way to go about this.
I'd consider using a buck-boost converter. An example: -
If you need more than 1.5 amps, there are others to choose from. Other products made by TI will also do the job. The point I'm making is ...
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If you're using an s2p file that the manufacturer supplied for you to design with the part, I think you should assume that the S-parameters were measured with the device operating, ie biassed, so already have any influence from the other pins of the package taken into account.
Note that the S-parameters have only been measured for the typical Idd of 65 mA, ...
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It was earlier stated that: "So the 240 ohm resistor shown for R1 would not meet the LM317 minimum load requirement worst case."
But I'm pretty sure that "worst case" would be no load, i.e. the 317 supplying no current to a subsequent circuit. Assuming at least a 10 mA load provided by the electronics being supplied, I'm not sure it ...
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I know I can add an external current sense and comparator circuit to
the Feed back pin of LM2576-ADJ version to limit current.
In the fixed voltage versions of the LM2576, the FB pin is still used except there is an internal resistor divider as per the data sheet. I've taken the liberty of highlighting the value of R2 for the 4 different fixed voltage ...
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Of course you can select fixed resistors instead of a potentiometer to make it non-adjustable.
The 79xx regulators are intended to be used as fixed regulators, but can be made adjustable, and the actual voltage of the regulator just sets the minimum voltage, it can be adjusted higher.
The output voltage term always includes the ground or adjust pin current, ...
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What about a 2-transistor constant current sink? Use an NPN as the control transistor. If you put in a 0.2 Ohm resistor, the NPN will choke the pass transistor at about 3-4 A. If current is lower, the pass transistor will be fully on and can be a low resistance N-MOSFET.
This is probably a little bit more efficient than using only a resistor of around 1-2 ...
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Often a device that runs from 4 AA batteries can instead be powered by from 5V from a "USB charger" this is because batteries drop in voltage to a little over 1V before end of life (and rechargables start at 1.25V) so the 5V is within the acceptable 6V-to 4V voltage range of a series 4 cell pack.
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No, generally voltage regulators can't be paralleled, and in this case even the datasheet forbids it.
Some regulators can be paralleled, and they can have a special connection between them to communicate how to share the load.
The problem is the regulators are not identical, they have output tolerance. If one regulator outputs 4.99V and the other one 5.01V, ...
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If you cascade two liner regulators, one to drop the 15V to 12V (which can be used by your 12V circuits) and from 12V to 5V, you will spread the heat and not need a heat sink.
If you don't have problems with size, you can use the TI LM340K liner regulator in a 2-pin TO-3 package for the conversion of 12V to 5V, and the LM7812/LM340 in a 3-pin TO-220 package ...
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You are doing it wrong way. Based on the discussion in comments you expect input power to be 5-12V which you are trying to covert to 15V with DC-DC module, which only accepts 4.5-9V.
Next you want to have isolated 5V & 15V supply. That's ok, but the DC-DC module you've picked (not to mention it doesn't works with +12V supply, as noted above) are extra ...
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A synchronous switcher to drop 15V down to 6V. Shoot for 250kHz+.
A "pi" C-L-C filter consisting of switcher's output capacitor, LDO's input capacitor, and a series inductor, plus a small ferrite bead in series with the inductor. Set cutoff to get rid of some of the switcher's ripple & harmonics.
An LDO like LP2989 - low noise, decent ripple ...
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You can use a hybrid approach: use a DC-DC step-down (buck) regulator to make 5.2V, then use an LDO (low drop out) linear regulator to make the 'clean' 5V for your analog section. Then you get efficiency and low noise.
What's an LDO? It's an improved linear regulator that allows a much smaller overhead voltage than earlier regulators like the LM7805. ...
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Don't reinvent the wheel.
They sell a ready-made solution: 12 V to 12 V battery charger.
https://www.google.com/search?client=firefox-b-1-d&q=12+V+to+12+V+battery+charger
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You have to rely on the data sheet. Some manufacturers provide schematics, but none of them claim that they are good for analyzing performance outside of recommended operating conditions: they're just to aid understanding how the part works. If you find a part that does have detailed performance data you can't depend on another part, from someone else, with ...
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I'm working on ESP32, so I just put a 1R resistor in series with the supply to check power supply current at boot. Yellow trace, 1 ohm, so 1V = 1mA.
Labels on plot:
RST - reset button released.
1 - it enters arduino setup() and the first thing it does is pulse a GPIO (blue trace) so that shows up on the scope.
4 - frequency is lowered to 80MHz, which does ...
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It depends on the internal circuitry of the regulator. If the manufacturer doesn't provide this information then you should not make any assumptions, except that the regulator no longer regulates. One thing you can be sure of is that \$V_{out} \le V_{in}\$
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