uber newb

[Clonus]

hi there

been sometime since i read about ohms law.. looking for a memory boost.

im not an electronic expert but i have a pretty good understanding of the concepts


however, either im tired or just forgot something here

im trying to figure this out.

i have a pic and a sensor

i want the pic to use 3 volts and the sensor needs 5 volts

so, using resistors, what would i do?

i thought i seen a resistor drop voltage calculator or something, can't find it

i need to limit the current on the pic and leave the sensor as is with the 5 volts.

also..

if i left the pic to use 5 volts, would it be less efficient than if it ran at 3?

thanks
clonus

 

[Clonus]

Ok, tell me if im hot or cold.

looking at the 12f675 datasheet i see that the pic uses a max of 700microamps at 3 volts, so with my power supply of 5 volts should i be using a 2.7k resistor?

 

[Hero999]

You can't use a simple dropping resistor because the current consumption of the PIC continiously changes with each clock cycle. What would happen if the current consumption at one point dropped to 350:mu:A? The voltage will just increase, this spike could cause undesired operation.

Your best option is to use a small linear regulator to power the PIC, one with a very low quiescent current.

 

[Nigel Goodwin]

Why do you want to run the PIC at 3V, it runs quite happily at 5V (which is it's normal voltage).

The lower the voltage, the lower the current consumption - but it's still low at 5V - and what current does the sensor take?.

 

[Clonus]

I figured if i ran the pic at 3 volts i could conserve energy from the battery.

The sensor needs 5 volts but the pic doesnt need that much of coarse.

 

[Nigel Goodwin]

I figured if i ran the pic at 3 volts i could conserve energy from the battery.

The sensor needs 5 volts but the pic doesnt need that much of coarse.

Except you're wasting energy dropping the 5V down to 3V, best result 60% for the PIC and 40% wasted as heat, but probably more than 40% wasted.

As I asked before, WHAT current does the sensor take?, for that matter what does the rest of the circuit take? - try posting a circuit.

Conserving energy is only worthwhile if it's a reasonable percentage, no point struggling saving uA's if the rest already takes mA's.

 

[Clonus]

ahh, ok, thanks.

I'm not very good at electronics as you can probably tell.

I'm trying to learn, and yes I've heard this "wasting as heat" before.

The sensor I am trying to use is the HVW PIR Motion Sensor

here's the PDF

**broken link removed**

It seems to only tell me the "standby current" at 350 microamps.



----

I am probably confused here.

Lets say that I have a DC Voltage 5 volts, if i use a pic that consumes 700 microamps, what happens to the rest of the current?

I would think that if i just shorted the terminals on a battery the battery would burn out pretty fast.

If i stick a pic between them, which is what we seem to do with all circuits, why does't the battery burn out as fast?

I am thinking that resistors are limiting current but i am not grasping the fact that although they limit they also burn off what they are limiting? is that correct?

Is there something here I am not understanding?

 

[Hero999]

Lets say that I have a DC Voltage 5 volts, if i use a pic that consumes 700 microamps, what happens to the rest of the current?

It goes to power the rest of the circuit, i.e. your sensor, and any output decives connected to the PIC.

I am thinking that resistors are limiting current but i am not grasping the fact that although they limit they also burn off what they are limiting?

Nigel was talking about how a linear regulator wastes power. A 3V linear regulator running from a 5V source needs to drop the voltage by 2V. This voltage voltage drop wastes power. Suppose your circuit uses 1mA, the input power to the regulator will be 0.001*5 = 5mW and the output will be 0.001*3 = 3mW, this is only 3/5 = 60% efficient.

 

[hjames]

At 3 volts, the active current of the device drops by about half, so there would be a net savings in power - even with a linear regulator - With a switching regulator there would be even more power savings.

Easy way is to 2 series diodes to drop it to ~3.6V, 3 series diodes if you want to play it risky and drop it down to ~2.9V. If this is a battery powered device, you'll have to do a lot more than this though.

Do you have an initial schematic or block diagram of what you want to build?

 

[philba]

It seems to me that the best thing for conserving power is to arrange for all events to be external interrupts and then you can put the 675 to sleep. The ints wake it up. Sleeping allows you to conserve alot more energy.

by the way, using the numbers from above and from the PIR datasheet, the whole thing will pull about 1 mA. That's going to last a while with decent batteries. which, of course, begs the question "where is the 5V battery coming from?"... (hint, see if the PIR can run on 4.5V)

 

[Clonus]

thanks for all the input here!

The 5 volt battery is actually a 9 volt with a regulator. I am now thinking this is also a bad idea considering it's just wasting the rest as heat.

---

Let me get back to the rest of the responses after I ask this question.

take any battery Xvolts and put a resistor on it Yohms

The resistor itself will expend so much as heat but what about the rest?

what happens to the rest of the energy?

lets also assume it's just a closed circuit, simple as you can get.

what i am trying to understand here is this.

lets say

1) take a simple circuit, a single component of that circuit uses 500microamps
2) take that same circuit and change the component , now it used 700microamps

What i dont understand is if this simple configuration was able to supply both requirements, is there an amount that is being wasted?

there is a 200 unit difference between 500 and 700, does this mean that the circuit was supplying more than both but the component was burning of , say, 500 but wasting the rest somehow?

i really hope that makes sense.....

 

[kchriste]

take any battery Xvolts and put a resistor on it Yohms
The resistor itself will expend so much as heat but what about the rest?

There is no rest (with ideal components anyway). The resistor will dissipate all the energy (watts) as heat according to ohms law until the battery is dead:
Watts = Volts * Volts / Ohms

what happens to the rest of the energy?

It stays in the battery until it is turned into heat by the resistor. How long this takes depends on the capacity of the battery, battery voltage and the resistance of the resistor.

What i dont understand is if this simple configuration was able to supply both requirements, is there an amount that is being wasted?

Nothing is wasted in this example. The energy difference just stays in the battery. ie: The circuit will just operate for a longer time period.

 

[Hero999]

The 5 volt battery is actually a 9 volt with a regulator. I am now thinking this is also a bad idea considering it's just wasting the rest as heat.

What regulator are you using?

If its something like the LM7805 then the quiescent current can be as high as 8mA, which is a lot more than your PIC.

 

[Clonus]

ahh, ok, thanks again for all the feedback.

So...

from what I've been reading, the different charges, pos and neg have an attraction.

when i stick a resistor between them, is this really lessening the attraction or is it truely "limiting" it?

These "lessen" and "limit" do sound the same however, my idea of lessening would mean the actual pull of the attraction whereas the limit would be the same attraction but the current would have a bottleneck...

at times i try to relate this to the plumbing idea, but i think it's confusing me.

i think im at a point where i need that "click" of understanding, you know what i mean.

im the type that needs to understand what is happening at an electron level i can't learn from theory books.

 

[kchriste]

at times i try to relate this to the plumbing idea, but i think it's confusing me.

Think of a resistor as a restriction in a water pipe such as a partially open valve. The valve doesn't limit pressure (voltage) but does limit the rate of flow(current). There will only be a pressure (voltage) drop across a valve (resistor) if there is water (current) flowing though it.

 

[audioguru]

A 9V battery drops to 6V. A 7805 regulator needs a minimum of 7.5V.
A low-dropout 5V regulator works fine when the battery drops down to 5.5V.

 

[Clonus]

Think of a resistor as a restriction in a water pipe such as a partially open valve. The valve doesn't limit pressure (voltage) but does limit the rate of flow(current). There will only be a pressure (voltage) drop across a valve (resistor) if there is water (current) flowing though it.

Ok, but what about this "heat" issue. Does this pipe have a leak?

I was under the impression that the resistor releases the non flowing current/voltage as heat? is that correct?

 

[kchriste]

Ok, but what about this "heat" issue. Does this pipe have a leak?

You can only carry the pipe analogy so far. ie: an electrical circuit always needs to be a complete loop where as a pipe can just spill the water out onto the ground. But I would guess that a restriction in a pipe with water flowing would create a certain amount of heat also.

I was under the impression that the resistor releases the non flowing current/voltage as heat? is that correct?

Correct, if you take out the word non in non flowing .
The heat that the resistor releases is equal to:
HeatInWatts = VoltageAcrossResistor * VoltageAcrossResistor / ResistanceInOhms

 

[Clonus]

re:non flowing .. I meant, the energy that doesn't make it past the resistor is released as heat.

This is quite interesting.

It seems to me that manipulating energy seems a bit crude. I would assume it's simply because we can't really change the laws of physics.

We can supply voltage but if we need less then we have to consume it somehow, even if this somehow doesn't benefit the circuit. We simply have to waste it.

I know it's more than this, every system has it's ingenious designs.

=====

So, I have another question.

If this doesn't sound correct, please correct me!

If a power source can "source" so much current and an electrical component like a pic uses only , say, 10% of that current. What does this mean for the life of the power source (being a battery) verses the same circuit consuming 20% of the current?

What happens to the 90% or 80% of that sourced current? Is this also wasted?

or does this mean that the 10% or 20% is really all that is being used?

 

[Hero999]

A 7805 regulator needs a minimum of 7.5V.

That depends on the output current you want to draw.