Trying to figure out the current used

[lilimike]

Hi,

I have a circuit composed of a PIC operating at 5V, a 2 stages OP-AMP(LM833) and a comparator (LM393) both operating at 9V. both are driven by 7805 and 7809 respectively.

When I measure the current at the main input of my circuit (12V DC adapter) I get 23.4mA.
When I measure the current at the output of 7809 I get 4.8mA
When I measure the current at the output of 7805 I get 5.27mA
I conclude that my circuit excluding the regulators consumes 10.07mA (4.8 + 5.27)
Since I get 23.4mA at the main input and my circuit consumes 10.07mA is it right to say that my two regulators are consuming 13.33mA (23.4 - 10.07) ?

Mike

 

[Neal]

yes, that is correct

 

[ljcox]

Did you supply the 7805 from the 9 Volt line or the 12 Volt line?

I have not studied the 7805 data sheet, but I would expect the current to be a bit lower if it is supplied from the 9 Volt line.

 

[lilimike]

If I use only the 9 volts regulator with resistors as voltage divider to supply 5 volts to the PIC, will the 5 volts be regulated or safe enough for the PIC ?
The PIC is driving a Piezo, a transducer at 40KHz and 2 LEDs.

Note that the LEDs and piezo were not connected when I took my measurements above.

 

[lilimike]

As it is now, both are supplied by the 12 Volts line.
But I find it is a waste to loose so much current by just regulators!

 

[ljcox]

If I use only the 9 volts regulator with resistors as voltage divider to supply 5 volts to the PIC, will the 5 volts be regulated or safe enough for the PIC ?

You misunderstood what I wrote.

What I'm suggesting is this:-

Connect the input of the 7805 reg to the output of the 7809 reg.

In other words, the 5V reg is supplied from the 9 Volt line.

 

[Pommie]

As it is now, both are supplied by the 12 Volts line.
But I find it is a waste to loose so much current by just regulators!

You do realise you are talking about wasting 0.12W or to put it in monetary terms 0.04c per day or 14c per year.

Mike.

 

[mneary]

Thirty years ago when the 7805 and 7809 were created, they used the best low-cost technology of the times.

If you want better performance, they are available. The parameter to search for is quiescent current.

 

[ljcox]

I looked at the data sheets for the 7805, 7809, 78L05 & 78L09.

The first 2 have quescent currents of 5 mA typical and a max of 13 mA. Thus your calculation of 13.3 mA is reasonable.

However, the 78L05 & 78L09 have quescent currents of 2.8 mA typical and 6 mA max.

Thus you can reduce the reg currents by using these instead.

 

[lilimike]

You misunderstood what I wrote.

Len, I got your response after I wrote my message, sorry for the confusion.
I tried driving the input of the 7805 from the output of 7809 and the current went down by 0.1mA.

Mike, I am not really worried about the cost, I am trying to improve to make my circuit battery operated.

This being said if I use resistors as divider for the 5V would that be safe for the PIC?
If so, based on my browsing my circuit would last about 50 hours on a 9V battery?

Mike

 

[lilimike]

Sorry,
I take out what I said, I see no way to make my circuit battery operated because although the PIC can sleep, the op-amp and comparator can't and so it is a device that will have to be on at least 50% of it full time.
I will just try to find a better way to supply 5V and 9V, maybe a regulator that can supply dual outputs to make the circuit it cleaner (in term of smaller)

 

[crutschow]

For higher efficiency you could use a switching regulator such as the Black Regulator. The disadvantage is that the switching noise can affect the analog circuits so you may need extra filtering for the 9V.

 

[ljcox]

Mike,
Your measurement is in line with the reg data sheets. Sorry, I should have looked at them before opening my mouth.

However, there may be an alternative solution. What awakes the PIC from sleep?

I have just designed a garage door opener using a PIC and some hardware.

The power is turned on by either the up or down button. The PIC starts and turns on a transistor that holds the power on independent of the button.

The PIC operates arelay to drive the motor and another relay to turn the light on.

When the door is up, the PIC stops the motor & then waits for 3 minutes (to keep the light on).

It then switches the light off & turns the transistor off and so the power consumption drops virtually to zero until the next button press.

Perhaps you could do something similar?