Hi there,
Another question:
Suppose a user by mistake put a battery in reverse in the battery holder of your circuit. Can your circuit be damaged? If yes, how to protect your circuit against such mishaps.
Regards
Use a diode.
Leon
leon_heller:
Use a diode.Leon
Leon,
Thanks. I understand what you mean. But do you know of an image of a circuit depicting diode’s use for the same purpose? Thanks
Regards
+ve ---->|-----
-ve ------------
Leon
Greetings smdFan,
smdFan:
Suppose a user by mistake put a battery in reverse in the battery holder of your circuit. Can your circuit be damaged? If yes, how to protect your circuit against such mishaps.
A power diode is series with the positive lead will block
accidental reverse connections. However, a silicon diode
drops a half volt or so, and wastes power.
A power diode in parallel with the supply (cathode of
diode to positive supply) will “short out” the supply
and protect the circuit from reverse polarity. No power
is wasted or voltage dropped for correct connections.
A zener diode in parallel with the supply will protect
the circuit from both reverse polarity connection and
over-voltage. The zener voltage should be slightly higher
than the worst case normal supply voltage.
If the circuit draws a high current, or the suppply is
capable of high currents, a fuse should be included
before the diode.
If the circuit is isolated from the ground of the supply
a common bridge rectifier in series with the supply will
work correctly for either polarity connection, but there
will be two diode voltage drops (allow 1.5V).
Comments Welcome!
using a FET is another way you can do it. I used a FET in combination with a zener here.
A zener diode in parallel with the supply will protect
the circuit from both reverse polarity connection and
over-voltage. The zener voltage should be slightly higher
than the worst case normal supply voltage.
If the circuit draws a high current, or the suppply is
capable of high currents, a fuse should be included
before the diode.
IMHO that setup is a great balance of simplicity and effectiveness.
A resettable fuse could be used.
Leon
[This app note from Maxim has a few other variations.](Mixed-signal and digital signal processing ICs | Analog Devices)
Andrew02E:
A zener diode in parallel with the supply will protect
the circuit from both reverse polarity connection and
over-voltage. The zener voltage should be slightly higher
than the worst case normal supply voltage.
If the circuit draws a high current, or the suppply is
capable of high currents, a fuse should be included
before the diode.
IMHO that setup is a great balance of simplicity and effectiveness.
I don’t disagree, but know that on an over voltage condition, you will almost certainly need to replace the zener since it will be toasted. Leon’s suggestion of a resettable fuse is a good one as well, especially given the current requirements for the examples (i.e., okay for lower current, not quite so good for high currents).
Thanks a lot folks. Your responses were insightful and led me in the proper direction.
I think for the simple design I am doing, I will settle for a Schottky diode in series (see below). The forward voltage drop of Schottky diodes are small and they are cheap. Thanks for your help.
My battery is 3V
The load voltage requirement range is: 1.8V- 3.6V
Bat ---->|-----
GND ------------
Regards
smdFan
smdFan:
Thanks a lot folks. Your responses were insightful and led me in the proper direction.I think for the simple design I am doing, I will settle for a Schottky diode in series (see below). The forward voltage drop of Schottky diodes are small and they are cheap. Thanks for your help.
My battery is 3V
The load voltage requirement range is: 1.8V- 3.6V
Bat ---->|-----
GND ------------
Regards
smdFan
Also another issue about it.
The battery I have is 3V. But the operating range of the device, as you can see above, is in the range of 1.8v to 3.6v. The supply voltage falls in between 1.8v and 3.6v . Would this be a problem? Thanks
Regards
You’ll be dropping about 0.2v across the Schottky so your battery limit is around 2v. Many batteries’ output drop off rapidly near the end of the useful life, so I’m not sure that the extra 0.2v will give you much more time.
riden:
You’ll be dropping about 0.2v across the Schottky so your battery limit is around 2v. Many batteries’ output drop off rapidly near the end of the useful life, so I’m not sure that the extra 0.2v will give you much more time.
I thought it would be like this:
3.0-0.2=2.8v
some Schottky diodes have even lower Vf like 0.1v
How long of a battery life I might get?
.
It should last quite a long time if you put the batteries in backwards :-)smdFan:
How long of a battery life I might get?
emf:
It should last quite a long time if you put the batteries in backwards :-)smdFan:
How long of a battery life I might get?
Never dealt with battries so I do not know if it is somehow calculatable. I think it should be.
Let’s say you have the following:
Bat = 5V
AVR MCU operable in the range of 2.7-5.5v running an LED (blinking).
There got to be a way to calculate precisely how long the bat will effectively operate the MCU to turn on the LED. But I do not know where to begin. Any thought, books, tutorial et ct. ?
Regards,
smdfan.
Greetings smdFan,
smdFan:
Never dealt with battries so I do not know if it is somehow calculatable.…
There got to be a way to calculate precisely how long the bat will effectively operate the MCU to turn on the LED. But I do not know where to begin. Any thought, books, tutorial et ct. ?
The useful life can be calculated easily, but it’s
a situation with many variables so the answer
isn’t very accurate. A better method is to measure
the useful life over several different batteries and
under worse case conditions, and calculate the
average.
Before starting you’lll need to know about the
battery. In particular, the energy capacity measured
in Ampere-hours at a given load, and the behaviour
of the battery at the end of it’s life.
Use Google to find the datasheet for the battery you
will use. Ideally, the battery will give constant voltage
regardless of current demand or temperature, and
when exhausted the voltage will drop suddenly.
A real world battery will gradually decline and do so
faster at elevated temperatures. Also, batteries have
a self-discharge behaviour that will kill the battery if
it is left unused (storage) for a long time. Fresh
batteries are always preferred!
The ampere-hour capacity is known as “C” and the
rate of discharge is often C/5 or C/10, because
a higher current load will shorten the battery run time.
An end-of-life voltage should be selected, and even
if the battery has more energy to give up the load
should be removed when the EOL voltage is reached.
Some rechargeable batteries will be damaged if the
battery is discharged too much.
You can easily rig an experiment with a DMM and
stopwatch (or wall clock) to measure the battery life,
either running the target project or using load resistors
to draw the desired current.
Comments Welcome!
Which battery, AVR, LED and resistor? Clock speed?
Leon
Now we’re starting to get somewhere. The main thing you should check is how many mAh (milliamp-hours) your battery will provide. They’ll usually say on the package. Let’s say you’re using four rechargable AA batteries in series. NiMH are 1.2V each, so that gives you 4.8V, pretty close to your 5V. Capacities vary, but mine are fairly old so they’re 1800mAh each, so all together, you have a 4.8V 1800mAh battery. Let’s also say you’re just lighting a basic red LED with a 330 ohm current-limiting resistor. That will draw about 10mA (less, but 10 makes for easy math). Your battery should power the LED for 1800mAh / 10mA = 180 hours = 7.5 days. It’s never really that simple, but that’ll give you a ballpark estimate.
The microcontroller varies quite a bit. The current it draws depends a lot on how fast you run it and what type of tasks it is doing. If you’re just blinking an LED a few times per second, you can leave the microcontroller in a low power mode and it could consume less than 1uA, so little it’s not worth trying to calculate. If you are running the processor as fast as it can go with all the peripherals enabled, it could suck up 20-40mA. The datasheet should list the current draw at various speeds.
If you’re blinking the LED, calculate the average current draw. If it’s on for one second, off for one second, then call it 5mA and enjoy your 15 days of runtime.
I think the answer to “how long does it last?” is uninteresting if you haven’t considered the question: “How do I make it last longer?”.
The simple answer to the former question is measure the current draw of your device and divide the battery capacity by that. This is an upper bound, probably 10-20% optimistic.
There are a myriad of tricks and techniques that you want to use for a battery powered device to minimize current draw. Designing your software to be interrupt driven and keep the processor in sleep mode except when processing interrupts will help a lot. Running at the lowest effective clock rate will also help. Some processors also allow changing the clock rate so that when you need the horsepower, you run fast and when you don’t you run slow. The TI MSP430 is particularly well designed for that but the AVR line isn’t too bad.