Main power/Auxiliary battery Power switching

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yorem_kastor
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Posted by yorem_kastor: Sun. May 7, 2017 - 10:37 AM
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Is it possible to achieve power switching without any current leak?

I have searched on internet but could not come up with a proper design except for recommendations.I wanted to try something and built my own but this one has current leak probably, because battery runs out when i use the DC power supply. Schematic is very basic, there is a motor works with around 5V. Our DC main source is 5V and 7805 Regulator is regulating 9V battery. Diodes are for blocking short circuits, Led is for indication of which power source is being used and 680µF is for keeping voltage steady while switching and BD240A does the switching.

I have tested this circuit with my home made AT89s52 real time clock(connected 89s52 where motor is placed). Microcontroller did not show any faults(reset/stop). Battery in the other hand did not last as i expected and it run out in few days while working with DC main power. If i were to use a logic state mosfet instead of a BJT power transistor would current leak still be the case?

All type of comments, help and reference are welcome.

 

BATTERY IS NOT RECHARGEABLE!!circuit

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Kartman
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Posted by Kartman: Sun. May 7, 2017 - 10:44 AM
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yorem_kastor wrote:
Is it possible to achieve power switching without any current leak?

Technically not - there's always leakage, it depends on how significant the leakage is and how it affects your application.

A switch would usually have the lowest leakage. Anything semiconductor will have more.

According to your circuit the LM7805 is always powered - the LM7805 has significant quiescent current, so it is not a good choice in your circuit.

Transistors are current driven - they require current to turn on, so this is wasting your precious battery power. I'd suggest you use a mosfet and chose a regulator with low quiescent current.

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mojo-chan
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Posted by mojo-chan: Mon. May 8, 2017 - 11:01 AM
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You might be able to get away with a simple diode OR if you can tolerate a few hundred mV drop. Otherwise try googling "ideal diode". I've used various ideal diode power switchers in the past, there are a few designed for this kind of battery/external supply switching.

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awneil
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Posted by awneil: Mon. May 8, 2017 - 11:39 AM
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"power path" is another good google term for this.

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valusoft
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Posted by valusoft: Mon. May 8, 2017 - 11:41 AM
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If you are trying  to conserve your battery, I would not be running the LED from the battery for a starter.

 

This is how I would do it if you have a small relay in your parts drawer.

 

Connect the relay's normally closed contacts in between your positive battery terminal and your 7805 input. Connect the relay's coil across your DC supply input terminals.

 

Get rid of that PNP transistor.

 

Use two Schottky diodes to "mix" the 7805 output and your DC inputs.

 

When DC supply is present, the relay is energised and the contacts open disconnecting your 9 volt supply. The battery will not be supplying any current while your DC is connected.

 

I would draw it if I had a means on this PC, but I don't. Sorry.

 

Ross McKenzie ValuSoft Melbourne Australia

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ki0bk
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Posted by ki0bk: Mon. May 8, 2017 - 01:26 PM
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Do not use a linear voltage regulator (7805) with a battery, as it wastes a lot of power, use a switching regulator that is only powered with the loss of your ac line power.

Jim

 

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mojo-chan
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Posted by mojo-chan: Mon. May 8, 2017 - 01:30 PM
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Linear regs can be okay, but only if you application is low power (<100mA) and if it is an ultra efficient one. Texas and a few others make some with sub 1uA quiescent currents.

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awneil
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Posted by awneil: Mon. May 8, 2017 - 02:27 PM (Reply to #7)
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mojo-chan wrote:
Linear regs can be okay ... if it is an ultra efficient one

Eh??

 

Surely, a linear regulator is by definition inefficient?!

 

Its operating principle is simply to burn off excess energy as heat!

 

Especially in a case like this - where the regulator will be dissipating almost the same amount of energy as passed to the load!

 

Texas and a few others make some with sub 1uA quiescent currents.

That will help in standby - but has nothing to do with the regulation efficiency.

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mojo-chan
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Posted by mojo-chan: Mon. May 8, 2017 - 03:23 PM (Reply to #8)
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awneil wrote:

Eh??

 

Surely, a linear regulator is by definition inefficient?!

 

Well, no... A 7805 is inefficient. Newer designs waste less energy. How else is it possible to linearly regulate 3.6V to 1.8V with 1uA quiescent current from the LDO if inefficiency is some inherent property?

 

And if you have a lot of capacitance after the regulator, they waste even less because they can operate at lower load where the losses are lower too.

 

In a battery powered application a switching regulator is going to be very inefficient if the device spends a lot of the time idle, pulling only microampres. It really depends on how the device is used. In all the battery powered stuff I design, switching regulators are only used for battery charging because while the system is idle their quiescent current is higher than the losses from a good LDO.

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frog_jr
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Posted by frog_jr: Mon. May 8, 2017 - 03:40 PM
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If you are using a linear regulator (LDO or other) then regulating from 3.6V to 1.8V means the power loss in the regulator (not including the loss due to quiescent current) is (approximately) the same as the power used by the 1.8V circuit.

David (aka frog_jr)

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awneil
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Posted by awneil: Mon. May 8, 2017 - 03:49 PM
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mojo-chan wrote:
How else is it possible to linearly regulate 3.6V to 1.8V with 1uA quiescent current from the LDO if inefficiency is some inherent property?

You are missing the point and confusing two different issues here.

 

As frog_jr says, if you use a linear regulator to drop 3.6V to 1.8V, then the the regulator drops the same voltage as the load.

Even if the regulator were ideal - ie, consumed no current itself - that would mean that the current through the regulator would equal the load current.

Therefore the power dissipated by the regulator would equal the power delivered to the load.

Thus, you waste 50% of the power - which means that you waste 50% of the battery capacity!!

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dak664
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Posted by dak664: Tue. May 9, 2017 - 02:16 PM
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If the circuit can handle varying supply voltage then you can simply connect a rechargeable (or not) battery across the external power supply. 3.55 volts is a good match for a single LiFePO4 cell. To protect the cell from low voltage damage the the MCU could sleep below 2.5 volts or so.

 

Make sure the battery can handle the maximum charging current when the load drops to zero, also that there is enough excess current to charge the battery with the average load.

 

4.2 volts with a LiPo cell would be possible but inherently dangerous without a low voltage cutoff and *permanent internal* overdischarge disconnect. Such protected cells are made, e.g http://www.batteryjunction.com/u...

 

Also consider the worst case failure of power supply regulation. One nice thing about 9 volt batteries, you don't have to worry about fire.

Last Edited: Tue. May 9, 2017 - 02:17 PM
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mojo-chan
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Posted by mojo-chan: Tue. May 9, 2017 - 02:39 PM (Reply to #11)
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awneil wrote:

As frog_jr says, if you use a linear regulator to drop 3.6V to 1.8V, then the the regulator drops the same voltage as the load.

Even if the regulator were ideal - ie, consumed no current itself - that would mean that the current through the regulator would equal the load current.

Therefore the power dissipated by the regulator would equal the power delivered to the load.

Thus, you waste 50% of the power - which means that you waste 50% of the battery capacity!!

 

Nope.

 

For example, I built a little device that sleeps most of the time, but does about an 30 minutes of rapid ADC readings every day. Single lithium battery, 10 year life from about 4000mAh.

 

Running directly from the battery at 3.6V, idle was about 3.5uA and active current was about 5mA. I added in a 1.8V regulator, so according to you wasted 50% of the available battery capacity! In reality, idle current remained at 3.5uA and active current DROPPED to 2.5mA.

 

In other words, adding an "inefficient" LDO actually massively increased battery life and allowed me to hit by 10 year lifespan goal.

 

That's because the MCU uses much less current at lower voltage, as do all the other parts of the system like the op-amp and precision voltage reference. Yes, some energy is "wasted" by the LDO, but the overall gain is huge. In fact it's twice as efficient!

 

Okay, what about a switching regulator? They are even more efficient, right? Well, no. At low loads they are usually less efficient, because so much energy is consumed by the regulator itself. Some include an LDO that they switch to under light load to overcome this. Also, you have to deal with the switching noise.

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frog_jr
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Posted by frog_jr: Tue. May 9, 2017 - 08:59 PM
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I do not dispute your results; however, you are measuring two very different environments.

The LDO in this application is providing better overall performance, but the LDO is burning (about) 50% of the total circuit power.

Without LDO: 5mA * 3.6V = 18mW

With LDO: 2.5mA * 3.6V = 9MW (of which roughly 4.5mW is lost in the LDO)

 

It is the lower MCU current at 1.8V that is giving the benefit. (Which is possible because of use of the LDO...)

I am not advocating a switcher for this kind of circuit; just be aware of what losses are occurring in the linear regulator.

 

Edit: I have encountered designs where someone was dropping 12-15V to 3V using an LDO thinking the low quiescent current was giving them an advantage. And they wondered why the LDO was so hot! They just could not understand that 4 to 5 times the target circuit power was being "lost" in the LDO.

David (aka frog_jr)

Last Edited: Tue. May 9, 2017 - 09:04 PM
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mojo-chan
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Posted by mojo-chan: Tue. May 9, 2017 - 10:03 PM
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What is your point? You say energy is "lost" to the LDO, but the alternative is losing 2x as much to the MCU.

Do you have a more efficient solution? Because if you don't, you can't really say that the LDO is "inefficient" because it's literally the most efficient option.

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