# Tag Info

26

What could explain such a short battery life? Could it be the 5v regulator? As mentioned, the 7805 has about 4mA of quiescent current. You need to find a data sheet for the battery (Eveready has nice battery datasheets, if you're using an alkaline cell). It's probably no more than 100mAh -- 100mAh / 4mA = 25 hours, so that should say something to you. ...

11

The idle current of a 7805 regulator is around 4 mA so, armed with the ampere hour capacity of your battery, work out how long it will last with a continuous drain of 4 mA. If you establish that is the problem, you will find that there are plenty of regulators having a significantly lower quiescent current. Once the battery drops to about 7 volts you are ...

11

All those parts can run from 3 to 5V so use a battery that doesn't need a regulator, a 16500 Li-ion cell, or a 3xAAA battery pack are about the same size as the 9V and produce voltages in that range. (or even a Li-po cell) Without the regulator the microcontroller can shut down and the circuit will only need a few microamps.

10

I'm running similar sensor nodes with much better results. My setup has a few differences to yours: I'm running the µc directly (no regulator) from rechargable 1S LiPo batteries (3.7 V nominal) originally sold (very cheap and with a matching USB charger) for mini-drones. The whole voltage range (4.3 V - 3.5 V) is acceptable for the µc. 1 I power the ...

6

Can a 7805 regulator be used without input/output capacitors Short answer: YES ....but there are conditions First you should read a datasheet for the regulator such as this one. The datasheet clearly lays out: If your DC in supply is not too far away (distance/length not specified) you don't need a capacitor on the input. If your input supply is a small ...

4

If you want to know something about a device, consult the datasheet from the manufacturer. The minimum VIN must meet two conditions: $V_{IN} \geq$ 2.3V and $V_{IN} \geq (V_R + 3.0%) + V_{DROPOUT}$. So, Vin must be larger in your case, about 3.3 * 1.03 + 178 mV = 3.577 V

4

Very similar to "how come my solar/battery/inverter system has so little range?" > because the inverter is spun up all the time. Use different loads that work on direct battery and eliminate unnecessary voltage conversion. You've done engineering 101, you've slapped the bits together and they work. Engineering 202 is making them work efficiently enough to ...

4

The action of D3 and D4 is to clamp the input voltage at around negative 13-14V. I see this, and I look at the action of D2, and it makes be believe that the purpose is for some sort of protection against connecting the board up backwards. However -- without a fuse between that part of the circuit and the power input connector, what that circuit implements ...

3

That has not really anything to do with the maximum current drawn. The 7805 just needs the 100nF at input AND at output to run stable. Without these capacitors there is a high probability of your output voltage to oscillate. Don't save on these two cheap components, in the end it will more likely cost you even more because of failed parts you have to replace....

3

Use step up converter instead This is how I do similar projects. I use 3xAA which gives me 2.5V-4.8V this is within the operational range of atmega, I connect this to a step up converter with disable pin, when disabled the converter consumes near to nothing and passes the voltage through. When atmega wakes up and needs to make measurement it will turn the ...

2

You are not back powering the SMPS regulator you are back powering the regulator load. Providing the load current on the switching regulator (Buck or Boost) is much higher than any current from a clamp there should be no problems at all. However if the current drawn from the regulator is less than the current from the clamp (or many clamps) then the VCC ...

2

It is pretty standard practice to clamp a signal to Vcc by using a diode and resistor. However, when this happens, some small amount if current (maybe ~50mA) is flowing back into the Vcc line. When using a switching regulator to supply the Vcc line, what implications does introducing this current source have on the system? Is this safe? Brief: Energy ...

2

If you can find a multiplexer with a high current limit then yes, that is possible. Many chips have a 50mA limit, and multiplexers are typically not used for switching current. It would be better to find a high side switch configuration with a logic level input like this (There should also be integrated high side switches if you need compactness with the ...

2

There may be problems if you have unpowered peripheral devices connected to a powered microcontroller. You need to make sure that all microcontroller pins that are connected to these peripherals are set to a low voltage (logic 0) before removing power to the peripherals. Otherwise, significant current will flow through the input protection circuits of the ...

2

Measure what's the actual current drain in the idle and active states. Use an ammeter between the battery and the 7805 input. A typical new 9V battery has more than 300 mAh, and the 7805 quiescent current alone couldn't really consume it all - something's fishy! I've measured a lot of 9V batteries and they are typically 500-600 mAh. The caveat is that they ...

2

According to your numbers, you're getting expected behavior, between your sensor, your microcontroller, and your regulator (8ma). If you want better, sleep the controller, switch the sensor off, and get a more suitable regulator.

2

We've designed piezo drivers for many years. The piezos are typically driven with a triangular or sine wave for use with interferometer alignment. If this old post is still of interest I can try to have a schematic reduced to a B size.

2

The circuit is easier to understand if it is redrawn so that voltages increase strictly from bottom to top. I've drawn the rectifier as a floating voltage source for simplicity. Since ground is defined as the point between the two output capacitors, then D15 establishes the most negative voltage as -13V. D14 then puts the ADJ pin of the regulator at +12V, ...

2

Power line transient is also a major fuse blower with Z1 OVP protection. The 3 phase bridge produces 6f frequency ripple which is used to drive the FET as a BUCK regulator with Zeners to limit Gate drive, LDO input max and forward current to power tthe load. Given the LDO operates at 1.25 between Vout and Vadj, with a short circuit between these the ...

2

your approaches won't work – the LM7805 is so ancient, that it's voltage drop at reasonable charging currents will be > 1V (and that would be bad already if your alkaline batteries are totally fresh and not half discharged. Also, using a linear regulator in a battery-powered device is a bad idea; you'd be wasting one sixth of your energy. Use a switch-mode ...

2

The simple answer is, the capacitors are not required as long as the input is a clean DC and/or the load does not demand strict transient regulation. According to the datasheet, the two capacitors are required for the following reasons: C(in) is required if regulator is located at an appreciable distance from power supply filter. C(out) "improves" ...

2

As it stand it will clamp any negative voltage to < -12.6V due to the zener+diode in series. This is typically done in environments where secondary induced transients occur. The car industry will typically include lightning protection Reverse protection (due to installation) is provided by D2. Assuming C1 is large enough to absorb the energy from a ...

1

A simple DC motor works on the principle that when a current carrying conductor is placed in a magnetic field, it experiences a mechanical force. Speed of the motor is directly proportional to supply voltage. Speed of the motor is inversely proportional to armature voltage drop. Speed of the motor is inversely proportional to the flux due to the field ...

1

The numbers seem correct, as do the capacitors. Check the regulator pinout that you are using (it is different from similar positive regulators) and double check the resistor values if the pinout is correct. Edit: since it appears you would be better served by a positive regulator, I suggest you do a parametric search at a distributor or series of ...

1

You seem to be mixing up voltage regulators and transistors. There are voltage regulators (like the common 7805) that are packaged in the TO-220 housing. The TO-220 looks like this: That housing is also commonly used for transistors, so it is easy to mix them up when looking at just the housing. A transistor and a voltage regulator a vastly different ...

1

I think sstobbe is on the right track here - something in the supply likely has a “sneak path” to the chassis. I would start troubleshooting this by connecting a high value resistor (100 k perhaps) between the minus output and the chassis ground. That shouldn’t make the supply turn off and you can measure the voltage across the resistance- check for both AC ...

1

From the delay and loop functions looks like you are using Arduino code. The delay function is a active loop, it will not put the microcontroller to sleep! The Arduino API doesn't have support for sleep mode. Read the ATmega328P datasheet and see page 34 for how to put the device to sleep mode.

1

While I don't think for this product we need to technically be concerned about EMI, I'm curious if a little bitty switching regulator like this generates all that much noise? Yes, it does, or it can. And if this commercial product is going to be sold, it will most likely need to conform to IEC 61010 and pass the FCC unintentional radiators testing at an ...

1

If you are powering your modules over a fairly long wire, when the relay actuates there will be an additional voltage drop across the power feed wires. Therefore I suggest you generate the required 5V locally at each module derived from a raw 9V supply derived from the Arduino. A search of Ebay, Banggood or possibly Alibaba should turn up some suitable cheap ...

1

You can get good regulators for this from RC plane sources. It may be called a "battery eliminator circuit" or BEC. Just get one that's rated for more than 12V (or 4 LiPo cells). Connect the grounds so that the Arduino and the BEC input go to the battery connector separately, and try to make the signal wire from the Arduino travel with the servo ground (...

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