New answers tagged

0

The LHS 2k resistor start the turn on the LHS PN2222, the 10k res gives turn on hysteresis. The LHS PN2222 is on, this turns on the main xtors. The output rises until the RHS PN2222 turns on which lifts the emitter R, turning off the LHS PN2222 as its base is fixed to 2v5. Sometime later current flowing L to R thru the 39k and 1uF cap turns off the RHS ...


0

The module you are using has an LM7805 5V regulator on board. It will provide 5V as long as the input stays above about 7.5V. Your 12V battery should never drop that low. You can use the module as planned with your 12V battery without worry. The 5V output doesn't "switch" depending on the input voltage. If you supply 12V to the module, it makes ...


-1

If you use 5v charger for charging the powebank, it will give 5v as output. if you use 9v/12v charger for charging the powebank, it will give 9v/12v as output.


0

1 - You are applying about 15 volts and drawing 1.8 amps. 2 - Your regulator output is 5 volts. That means that the regulator is dropping 10 volts across it. 3 - Since the device is passing 1.8 amps, it is dissipating 1.8 x 10, or 18 watts. 4 - You are using an SOT23. From the data sheet, P 2, the maximum thermal resistance from chip to ambient is 203 deg C/...


0

The current sense and output circuit should be like this. This is why the inductor is causing a reverse spike in the circuit when the input voltage sags below the MOSFET output, causing the fault. : simulate this circuit – Schematic created using CircuitLab


1

The destruction of the regulator may be caused by a transient overvoltage you get when you connect your circuit to either the battery or the PSU. I think the story is the following: when you connect the circuit, there is a large inrush current in capacitors C1 and C2 because of the stray inductance between the voltage source (battery or PSU) and the ...


3

You have an oscilloscope as mentioned in the comments. Check the battery voltage - most likely higher than 14V by a good bit. Hook up your AL8871Q with the LED to the battery, but with the AP7370 and the processor removed. Check the battery voltage with the oscilloscope while the LED is lit. I think you will find that there are peaks on the battery voltage ...


1

Can you do this? Yes. Is it done often? Yes. Should you? Probably not unless you understand how it behaves. As the power draw drains the lower half of the battery pack, there will be voltage issues for the microcontroller, possibly reverse voltage into the drained batteries, and battery acid leaking. Not horrible failure with AA batteries but messy and ...


3

There is only 100nF capacitor at the output of the LDO, so perhaps it is oscillating. This should not result in its destruction, but who knows. The datasheet isn't very helpful as it contains no advice on capacitor selection, which is a red flag, but all the examples use a 1µF cap on the output, so perhaps replacing the 100nF cap with 1µF could help.


6

If the LED is drawing its power through the 7805, the regulator will require a large heatsink. If the bench supply has an adjustable current limit restricting the output to a max of 1 A, this might be why the regulator survives bench operation. A lithium battery has no such limiting. BTW, the 7805 is not a low-dropout regulator. It is a second-generation ...


1

From an AC perspective there's no difference as the capacitors' distance is minimal (ignoring parasitic inductance of the trace). However, from a DC perspective it's much better to use the 2nd option as you have a wider trace which means less DC resistance.


1

The problem with most any switching regulator is that they don't give good efficiency at light load. They also have quiescent current that further reduces efficency. A more reasonable approach is to specify your powered logic to support a voltage range compatible with your battery, say 3.0 to 3.6V. Then, use an ultra-low dropout LDO to take care of the range ...


1

As Tony mentioned, you might consider running at a lower voltage, so that you can take advantage of an ultra efficient step down regulator, designed for energy harvesting applications. For example, the ADP5304. If you can run at 2.5v instead, then your efficiency could be 90% - 95%, depending on your battery voltage: If you really need 3.3v out, then you'd ...


1

If the component is LM317TG from OnSemi, it says typically at least of 3.5mA current draw is needed to stay within regulation, but it can be as much as 10mA. Your resistor network does not guarantee this. Basically, it makes very little sense to try using so large resistances between Vout and Vadj to try conserve idle power consumption, and then add an extra ...


0

The datasheet for the LM317 (the TI one, which acquired National) states a recommended 10mA minimum load; probably it won't work with the stated tolerances if you pull less than that from it. The typical value states is 3.5mA. Considering that this regulator is floating its whole supply need to come from the output and the feedback terminal, consider these ...


4

Works okay for me except output is not very accurate (-5.32V, which is out of spec). Maybe your symbol pins do not match the model or you are on the wrong node- it is better to name the nets you are plotting (F4 and, say, Vout) so there is no question of what node it is. Also note that if you edit the schematic the node numbers may change but the plot will ...


4

This might not be a 100% correct component model. However, you're not loading the output, and without any resistive load, the LM7{8/9}xx series isn't known for accuracy. So, add a 1 kΩ resistor from OUT to GND, and observe again. If it still doesn't do anything, either the SPICE model is wrong, or you're not actually observing the right node.


0

I see other people are not liking the lack of power switch too. I just got into raspberry pi and noticed it. I wrote this schematic to solve my own issue. Too bad the guys on that site don't have a good forum to post this schematic. I would suggest a variance of this schematic below, but add a 5V relay on the USB 2.0/3.0 port 5V pin in place where I have a ...


0

Limits for absolute maximum Voltage Current and Temperature are “absolutes” and all these and others must be avoided. However for reliability, you will learn how to derate these limits or stick with suggested operation levels. If you think about it, it’s a little ironic to have a 34V drop for a “low dropout regulator” so if that really was a requirement, a ...


1

But why do some devices have a current limit and some a power limit? The current limit will be determined by the maximum allowable current density in the bonding wires and internal components of the chip. As current density increases so will heating and there has to be a physical limit to that. The power limit will also be due to thermal considerations but ...


2

In general, you can run this at 37V input and 380mA. This would still comply with the absolute maximum ratings where neither a maximum current nor maximum power dissipation are specified directly. But: as you might have seen, this device has some protection features, like short circuit and overtemperature protection. Those implicitly limit both the current (&...


2

How much current it can pass is one property, and how much it can heat up by the power dissipated is another property. You can input 5V and make 380mA 3.3V, and the regulator would heat up only at 0.65W, which is certainly doable. If you input 37V, and output output 3.3V at 380mA, the poor chip has to dissipate 12.8W, which is completely absurd amount of ...


0

I’m using a TLV755 LDO in a circuit I’ve been working on to boost the output of a supercapacitor to 3.3V as it is discharged. The IC that it’s powering does some internal voltage regulation, and without the LDO, the circuit runs for about 45 minutes before the voltage drops too low to keep things running smoothly. When I first introduced the LDO, I used the ...


2

Put a high-power fixed-voltage shunt in parallel with the TEG. I'd set it to let the voltage rise to 50 or 55V -- that should protect your LM2576HV, and still give you headroom above the TEG's maximum power point of 48V. Then let the LM2576HV do its thing. If you use the right value for your stack of Zeners, the shunt should stop conducting before the ...


0

You could use a linear supply, but wasting 36W means a sizeable heatsink. You don’t mention environmental factors like size and operational temperature range. You could use a buck to get 12V from 24V and use a relay to bypass the buck if the input voltage is less than, say 13V. Another option is to use a dc/dc converter with a suitable input range and 12V ...


1

Always refer to datasheets for the information you are looking for. From the datasheet: On the MIC29xx1 and MIC29xx2, the ENABLE pin may be tied to VIN if it is not required for ON/OFF control. CMOS compatible control input. Logic-high = enable, logic-low = shutdown. MIC29xx1 and MIC29xx2 versions feature an enable (EN) input that allows ON/OFF control of ...


1

Yes. The two specifications referred to in the documentation are ratings for aircraft power systems, often nominally 28 volts. Your converter will operate normally up to 60 volts. Then, per the spec, the one-second transient period begins when the input voltage exceeds 60 volts and can be as high as 80 volts throughout the one-second period, but should ...


0

Appart from the discussion about 12V from 11.5 V, you can limit to 12 V with minimum drop with a simple modification of your first diagram. Also a P-channel mosfet for Q2 will improve a little.


0

If you know exact value of current and wire's resistance you may choose proper voltage at source above 12V to create nessessary current to drive solenoids. Solenoids are driven by current, not voltage. Rated voltage is for choosing current source. But for fire safety wire's gauge should be chosen properly, keep it from heating. Wire's insulation max ...


5

You can do this. The pot will be easy to set (range is only about 10mV). Output impedance is < 275 ohms. Adjust values if that's too high. simulate this circuit – Schematic created using CircuitLab Of course if you happen to have a reference that is < 2.5V or an offset that is in the wrong direction this won't help.


0

You can use a 12V MPPT (Maximum Power Point Tracking) charge controller, such as: https://www.amazon.es/Victron-Energy-MPPT-75-Controladores/dp/B075NQQRPD/ref=asc_df_B075NQQRPD/?tag=googshopes-21&linkCode=df0&hvadid=199604873408&hvpos=&hvnetw=g&hvrand=16566384393254120720&hvpone=&hvptwo=&hvqmt=&hvdev=c&hvdvcmdl=&...


1

Get rid of R1, it wastes power. Get rid of the 7812, it wastes power. Get rid of ZN1, it may waste power. Replace the regulator with a switching 12 V output regulator.


1

Here is a buck converter circuit I quite often use. If you have the parts you can try building it. Edit: I have a typo in the design. L1 is 47uH, naturally not 47uF.


-1

Let me help you, this is quite simple, put the clock and LED in series along with a 10ohm resistor. And that's it!


2

Rather than use words, here's a design you can play with. That should please your inner ten-year-old, or your actual ten-year-old. Try this (simulate it here): What's going on? I've set up the sim with a 3V LED. The 510 ohm provides about 4.2mA to the LED. This is a good current to start with for most LEDs. If you want less or more brightness, adjust the ...


3

Having it shut on and off with the printer may be a bad idea- corrupted file systems and so on will likely haunt you unless you manually do an orderly shutdown every time. But you could use a DC-DC converter to get 5V from 24V for the Raspberry Pi. There may well not be enough excess capacity in the 3D printer power supply to provide an extra 15W or so. ...


1

If it is not a high power LED, all it needs is a resistor. The clock current draw is most likely so small it can be handled with resistor voltage divider down to 1.5V, perhaps with a capacitor to provide low impedance. One way to regulate voltage for the clock is to have a red 1.6V LED powered up via a resistor, and provide the clock from the 1.6V over the ...


0

Use 3 diodes in series to get roughly +3V from +5V. Use 5 diodes in series to get roughly +1.5V from +5V. R2 is a small load in order for diodes D4, D5, D6, D7, D8 to turn on. R1 is a current limiter for your led. You may use diodes 1N4148: https://www.digikey.com/en/products/filter/diodes-rectifiers-single/280?s=...


1

The simplest is to get a 1.5V linear voltage regulator to supply the clock. I assume that with "small led" you mean a standard 20mA led. In this case, use a current limiting resistor of some hundred Ohm and drive the led directly from 5V. Alternatively you can use a zener clamp circuit to get a (more or less) constant voltage. But this only works ...


-2

Use Nexperia PESD1CAN for transient over voltage event: https://www.nexperia.com/products/esd-protection-tvs-filtering-and-signal-conditioning/automotive-esd-protection-and-tvs/lin-can-fd-flexray/PESD1CAN.html They are excellent and I successfully tested them with an EFT tester. They are automotive qualified and cost nothing. For reverse polarity use diode ...


1

The Iadj does not matter for two reasons. First, if you use the recommended 240 ohm resistor between Vout and Adj pins, it will result into about 5mA of voltage divider current, so the 50uA typical current is so small that it can be just ignored in the calculations, as the difference is about 100 times. Another reason is that you should not use LM317 to ...


1

Iadj should be given in the datasheet. It's the current that flows out of the adjustment pin, and is related to the quiescent current of the device. (from page 8 of the datasheet you linked in the question) However, with 5 volt input, you won't get a good 3.8 volts out of a 317; even at just 20 mA you'll need a voltage differential of at least 1.5 volts ...


2

It's in the datasheet-- use the typical number (50uA). Usually it will make little difference compared to resistor tolerances, especially if you're counting on the resistors to draw the minimum load current. As others have mentioned, there isn't enough voltage for a LM317. Try a LM1117/AMS1117 (quasi-LDO) or better, and be sure to carefully follow the ...


1

For a high-side switch, when your input signal level is smaller than the load voltage, you generally need two transistor stages. The resistor values and transistor types are approximate, you need to analyze for your requirements. simulate this circuit – Schematic created using CircuitLab


1

I have successfully done it but I had a batch of 5v regulators and tested each one to 1/100th of the same voltage output before parallelizing them to ensure pretty even distribution. I needed to do this with heat sinks too and it works well but the trick was perfectly matching the regulators. My requirements were super small space and just heat dissipation, ...


0

am trying to build a circuit which provides 3.3V power to a micro-controller (Nordic Semi nRF52840). Absolute maximum range of voltage should be 1.8V to 3.6V. Maximum power consumption would be 140mA. It should use battery power when USB is not connected and use USB power when connected (ideally without resetting). Efficiency is not super critical because I ...


1

Your LED drivers are controlled by the same micro, so everything must have a common ground. I don't understand the "series" confusion, there is nothing to be wired in series here. What you need is a 5V power supply for 4x 3A loads, so a 5V 12A load. You can replace your 24V power supply with a 5V 15A mains power supply, or use a sufficient number ...


0

Why and where do I choose linear vs step regulation? low heat low noise? A linear regulator is typically used when a low noise output is desired and/or when the PSU is required to not emit a lot of EMI. A linear regulator is a very simple type of regulator which works by dropping excess input voltage to maintain a constant output voltage. Linear regulators ...


1

The picture you took from the service manual clearly says the power supply input is 15 VAC, right next to the connector. It can't work with a DC supply.


2

You can chain regulators as you suggest BUT the lower voltages will incur large inefficiencies & also large power dissipation in the regulators. Efficiency can be improved with switching regulators. This (10 year old!) SE answer of mine to "My linear voltage regulator is overheating very fast" should be useful. Linear regulator dissipation can ...


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