Hi,
While building a kit using PIC16F876A, I realized that the power consumption is very high. A new 9V battery will last about 24 hours. Knowing that the power consumption is dependent on the circuit components and design, I have attached the schematic diagram in PDF.

Could anyone please tell me how I can calculate the power consumption?

Thanks,
kfchoong

 

I think you will find that most of the power is taken by LCD backlight.
If you run the cct off a battery turn off the backlight.
Also having the power LED on all the time does not help matters. To get maximum life out of your battery pull that out as well.

 

Thanks for the reply. During one of my test, I removed no only the LCD's backlight, but the entire LCD. Even after doing so, the power consumption is still very high. It probably only added a few more hours to the battery life.

I'm new to circuit, not sure if the variable resistor used to control the LCD brightness is causing the problem. Depending on the resistance set, doesn't it cause some sort of short circuit?

 

ok.
here's my rule number one of batteries.

9V VOLTS BATTERIES ARE CRAP.

do not ever use them. period.
using a step up converter with a set of 2xaa cells would be waay better.

a single led will use more power than a lcd display.

as far as having any longevity out of a 9v battery don't expect nothing.
They are landfill demons, they have no capacity and are expensive.
I cannot stress this enough.

 

Thanks for the advice, I will give it a try. But will changing the battery from 9v to AA makes such a big difference? Or is there something wrong fundamentally that causes the battery to only last a day?

Is there a way in which I can measure the mA consume by the circuit?

Thanks again

 

Thanks for the advice, I will give it a try. But will changing the battery from 9v to AA makes such a big difference? Or is there something wrong fundamentally that causes the battery to only last a day?

Is there a way in which I can measure the mA consume by the circuit?

Thanks again

If you can obtain a Digital Multimeter you can measure the current with that.

hgmjr

 

Your basic application seems to be a LCD clock or a programmable timer.
I have downloaded your diag. let me study it & will post the bugs in the same.
I have been working with low consumption systems for years. The simplest method to start with is to know the min. curr consumpton of each device which will be powered up by the battery. Take the total consumption , & divide the battery capacity ( mAH ) by cir. consumption to get the life of battery. If the life is unacceptable you have no choice but use a higher capac. battery or change the devices used where avail. in a lo curr. type.
e.g if you are using a pwer hungry LS or similar gate , try using Cmos or
HC mos devices , for normal 555 use C555 .

 

How do I measure the current with a Digital Multimeter? Must I measure each component individually?

Also, if powered by a 9v battery, how long should I expect the device to work? Articles from different website seem to indicate that the PIC16F876A consumes very little power and could last for months on battery power. Is this true normally or are they stating the best case scenario?

BTW, I have a link to the document, showing what the kit does and how it works. Maybe it will provide a clearer picture of the device.

http://www.cytron.com.my/attachment/Details Description/PR12 v4.pdf

 

The best indicator for current draw would be on the wall adapter. The visible plate gives the output voltage and the current. It's too small to see in the photograph in the PDF file, though.

 

You are wasting battery capacity in 9 v to 5 v dc regulator. So i advise you to use 3 x 1.5 v pencil cells or 4 x 1.2 nicads to get around 4.5 v without the need for regulator. The higher mAH cap will also work in your favour.
Also keep the back light & leds off after a set time interval till the display function is activated by user pressing a button. This will have to be added to your cir. Use the pic with as low osc freq as possible to reduce active power consumption. Otherwise , as your basic RTC function is in a micropwer chip
the pic can also be powered into standby mode till user presses the wakeup button as noted above. Also the lcd bias resistor should be disconneced in low per mode to maximise battery life. For leds use hi efficiency types with a suitably higher val resis to reduce active cur.
These modifications will keep your battery alive for several days or more depending on your implementation.

 

The best indicator for current draw would be on the wall adapter. The visible plate gives the output voltage and the current. It's too small to see in the photograph in the PDF file, though.

Is it true to say that if the adapter is not designed specifically for the kit, but a normal switchable voltage adapter, then we will not be able to get this information?

 

You are wasting battery capacity in 9 v to 5 v dc regulator. So i advise you to use 3 x 1.5 v pencil cells or 4 x 1.2 nicads to get around 4.5 v without the need for regulator. The higher mAH cap will also work in your favour.
Also keep the back light & leds off after a set time interval till the display function is activated by user pressing a button. This will have to be added to your cir. Use the pic with as low osc freq as possible to reduce active power consumption. Otherwise , as your basic RTC function is in a micropwer chip
the pic can also be powered into standby mode till user presses the wakeup button as noted above. Also the lcd bias resistor should be disconneced in low per mode to maximise battery life. For leds use hi efficiency types with a suitably higher val resis to reduce active cur.
These modifications will keep your battery alive for several days or more depending on your implementation.

I was told by the company selling the kit that if I were to remove the regulator, I would most likely kill the PIC after some use. They advise against removing it. How true is their statement?

I don't quite understand the portion about "lcd bias resistor". If I were to remove it, will my lcd still show? Is it true that by having the resistor, the battery drains faster because depending on the resistance set, it is similar to a "short circuit"?

Also, does led consume so much power? I always thought that they used very little power... maybe because of their size

Finally, do you foresee that at the best case scenario, the battery life will only last a couple of days? How likely would it be to make the battery life last for 3 months?

Thanks again.

 

You could remove the regulator and run your circuit of 3X1.5V batteries BUT if you ever were ever to connect it to a plugpack or a 9V battery you would blow the cr!p out of your chips. Running it from 6X1.5V batteries and keeping the regulator may be more expensive in the short term but its much safer.

As for power consumption, the PIC if its running all the time and not going into sleep mode will pull about 10mA.
Each lit LED with a 220 series resistor will take about 20mA, you could up the series resistors to 470 and reduce the current to 10mA without effecting visibility too much .
The LCD display without backlight takes about 1 mA, the bias may add another 1mA so its really not that critical to disconnect it. LCD backlighting usually takes 70mA-90mA so unless you really must have backlighting I would keep it off.

A 9 volt battery driving your circuit with all LEDs and backlighting disconnected should last for about 48 hours. With LEDs and backligthting on you would expect about 6 hours life.

If you use 6X1.5V D cells and leave the backlighting and LEDs, off you could expect the batteries to last for about 90 days. With backlighting and LEDs on you would be looking at about 7 days.

Of course the other alternative is to use a plugpack and run it all off the mains.

 

Thank you so much for the detail explanation!

Using 6 x 1.5V D cells will occupy too much space for the device. And even doing so with all the rest of the modifications will get me 90 days... Maybe I should consider an adapter. But it is real sad that I couldn't get it to work with normal batteries and still get decent battery life.

If I were to use an adapter, is there a way that I can still hook up a 9v battery to act as a standby power source? I don't want the device to go blank after a short power failure or something. BTW, I have modified the code, it is not running as a simple clock, otherwise the button battery would be sufficient.

 

Running from a plugpack with battery backup is dead easy. See the attached file to get the general idea.
As long as the plugpack voltage is greater than the battery voltage the plugpack will supply all the current. If the mains fails and the plugpack voltage drops below the battery voltage, the battery will automatically take over.

 

Running from a plugpack with battery backup is dead easy. See the attached file to get the general idea.
As long as the plugpack voltage is greater than the battery voltage the plugpack will supply all the current. If the mains fails and the plugpack voltage drops below the battery voltage, the battery will automatically take over.

Thank you so much. Your help is very much appreciated.

 

The main component that consumes the most power seems to be the PIC. Does it mean that PIC are generally not suitable for battery powered devices? I have read that there are some low-power PIC available, are only these designed for batter power?

 

The main component that consumes the most power seems to be the PIC. Does it mean that PIC are generally not suitable for battery powered devices?

Please read up on sleep modes. The PIC in active mode will consume 7 to 20mA just idling. If you do not need it to do anything then put it into a sleep mode and have it wake-up when needed. In sleep mode the current draw will be in the 10 to 30 uA range. so if it is only in active mode 1% of the time the average current will be very low. This is just one of the many techniques for lowering power.

Yes there are "nanowatt" PIC devices that are designed for better/lower power usage. The have more options in the low power modes and such.

 

Please read up on sleep modes. The PIC in active mode will consume 7 to 20mA just idling. If you do not need it to do anything then put it into a sleep mode and have it wake-up when needed. In sleep mode the current draw will be in the 10 to 30 uA range. so if it is only in active mode 1% of the time the average current will be very low. This is just one of the many techniques for lowering power.

Yes there are "nanowatt" PIC devices that are designed for better/lower power usage. The have more options in the low power modes and such.

Thanks. May I know where can I find more information on programming sleep mode using C? Since the device will be on at least 12 hours a day, I doubt it will help much. But I will still give it a try.

 

Running from a plugpack with battery backup is dead easy. See the attached file to get the general idea.
As long as the plugpack voltage is greater than the battery voltage the plugpack will supply all the current. If the mains fails and the plugpack voltage drops below the battery voltage, the battery will automatically take over.

Hi Alex,

Thanks for showing the example. Just a little question from me. How would it be different if I were to expect a current supply of larger than 1A from the source? Would it work if I just change the diode with a much higher forward current rating? And what about the voltage rating? 1N4004 can sustain up to 280V, is it chosen because of the mains supply voltage level? Can we choose a lower voltage rating one?

Thanks.

eric