AC adapter to battery switch transient

Tonyr1084

Joined Sep 24, 2015
6,270
Admittedly I'm no expert on OpAmps. But wouldn't a diode effectively drop the battery reference by 0.7 volts? I'm confident the PS will maintain at least that much above the battery as long as the load (which I haven't read in any posts yet) doesn't draw the PS down below that threshold. And if we can't assume 0.7 v will be enough - then why not two diodes? (1.4 volts)

When I joined this website it was to learn. I've found many here to be vastly more knowledgable than me. So if I offer a suggestion it's in part from what I do know (or think I know) that I offer it. I'm willing to be told I'm wrong if it means I get the opportunity to learn something more. And trust me, visiting this website has taught me MUCH.
 

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electrophile

Joined Aug 30, 2013
166
@Tonyr1084 Thanks. You may be on to something. I could place a buck-boost regulator on the load side that provides the right output to the load and the adapter necessarily has to be the higher value one. But if I were to do all this, I might as well spend and get the LTC4412 :)
 

Tonyr1084

Joined Sep 24, 2015
6,270
As I said, I'm here to learn. So I drew this schematic to visualize whether it would work or not. I'm also rather new to using MOSFETs and I'm not even sure if I've drawn them right or if I've even used the right MOSFET symbols correctly.

I don't know why you would have to use an over-rated wall wart (WW). If the battery voltage is dropped 0.7 volts (or 1.4 volts) then the comparator won't switch until the WW has dropped - um - lets assume the WW is rated at 12VDC @ 500mA. Assuming the load does not drag the WW down, with the battery fully charged (12.6v) and the diode dropping 0.7 volts, the comparator sees the WW at 12 volts (or greater). As long as the WW output doesn't drop below 11.9 volts then the comparator holds the P-Chan MFET on, supplying power via the WW. But if power fails and the WW drops below 11.9 volts then the P-Chan MFET drops out and the N-Chan MFET kicks in, turning on power from the battery. The only thing I haven't worked out in my diagram is how the comparator is powered. If via the battery - and with no change in state it should present nearly no load at all to the battery. If powered from the WW then as soon as power drops out - so would the comparator.
 

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electrophile

Joined Aug 30, 2013
166
I'll try simulating this. Btw, in my earlier schematic I replaced the P-MOSFET with an PNP transistor and added a capacitor on the outout as suggested by @hp1729. Now the transient drop is abot 1V on switch over. Something like this:
 

AlbertHall

Joined Jun 4, 2014
11,506
I'll try simulating this. Btw, in my earlier schematic I replaced the P-MOSFET with an PNP transistor and added a capacitor on the outout as suggested by @hp1729. Now the transient drop is abot 1V on switch over. Something like this:
You need a resistor in Q1 collector else vast currents will flow from the battery through Q2 emitter-base, and Q1 to ground. Perhaps 4k7 but it depends on the load current. That won't affect the transient but it will protect real world transistors.
 

grahamed

Joined Jul 23, 2012
100
At first glance -

adapter goes off
current flows battery Q2E, Q2B, Q1E, R1, R2, GND

Q1 and Q2 are both on, but what limits Q2B current?

damn - beaten to the post.....
 

crutschow

Joined Mar 14, 2008
27,770
Note that the base current of Q2 should be about 10% of the load current to fully turn it on (saturated) so that will reduce you battery life by that factor.
For example, with the 10Ω load you show, R3 should be no more than 100Ω.
To minimize that current you can make Q2 into a Darlington stage (using another 2N3906 for example).
The disadvantage of a Darlington is the minimum transistor ON voltage goes from a tenth of a volt or so to about 0.7V.

An alternate would be to go back to the MOSFET and use a comparator to more precisely determine when to do the switch-over.
 

johnmariow

Joined May 4, 2016
20
This will be eventually implemented in the real world. I see what you mean. I placed a 1k resistor there and now I see the 2.6V transient again.

@johnmariow I simulated it with the square wave instead of the adapter and you are right, the width of the time delay is the transient width as well.

So how do I build this without the LTC chip?
The difficulty lies in the time it takes for a change in the each input signal to reach the output. The time for a new voltage level to get through a component is called the propagation time. The MOSFET increases the propagation time from the battery to the load. Hence the propagation time from the battery to the load is greater than the propagation time from the adaptor to the load. This difference in propagation time is causing the transient you are seeing. I think HP1729's suggestion of adding a capacitor from the cathode of D1 to ground would increase the propagation time from the adaptor to the load. I don't know if the discharge time through Rload will cause a problem.
 

EM Fields

Joined Jun 8, 2016
583
Unfortunately the battery and AC adapter are more or less the same voltage since when the power goes out, the battery takes over.

If I were to eliminate the P-MOSFET, both the AC adapter and the battery would share the load current and this is not something that is feasible since when the AC adapter is present the battery is being charged.
I'm confused...
if the AC adapter is charging the battery and, at the same time, running the application in parallel, why doesn't the battery just take over - a la UPS - if the adapter output dies?
 

grahamed

Joined Jul 23, 2012
100
I'm confused...
if the AC adapter is charging the battery and, at the same time, running the application in parallel, why doesn't the battery just take over - a la UPS - if the adapter output dies?
That is the idea I guess. However two-diode switching requires that the mains-derived supply (when present) be higher than the battery and it seems this may not be the case.

As you say such a scheme usually goes along with charging the battery. though how this can happen when the line supply might be lower than the battery is a mystery.

In the end there is a basic problem somewhere - the circuit is misconceived or inappropriate given the relative voltages.
 

EM Fields

Joined Jun 8, 2016
583
That is the idea I guess. However two-diode switching requires that the mains-derived supply (when present) be higher than the battery and it seems this may not be the case.

As you say such a scheme usually goes along with charging the battery. though how this can happen when the line supply might be lower than the battery is a mystery.

In the end there is a basic problem somewhere - the circuit is misconceived or inappropriate given the relative voltages.
Indeed.
If the adaptor is used to charge the battery while powering the application, then the connection / circuitry between the adapter and the battery isn't shown.
Unfortunately the battery and AC adapter are more or less the same voltage since when the power goes out, the battery takes over.

If I were to eliminate the P-MOSFET, both the AC adapter and the battery would share the load current and this is not something that is feasible since when the AC adapter is present the battery is being charged.
Unfortunately the battery and AC adapter are more or less the same voltage since when the power goes out, the battery takes over.

If I were to eliminate the P-MOSFET, both the AC adapter and the battery would share the load current and this is not something that is feasible since when the AC adapter is present the battery is being charged.
Can you explain how the [separate?] charger is connected or, if you're using the adapter - instead of a dedicated charger - to charge the battery how you're doing it?
 

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electrophile

Joined Aug 30, 2013
166
I was using a TI design based on their BQ24650 chip (attached). I never built cause other things took priority and I thought I'd explore this again now. This is essentially a MPPT solar charger. This particular design was for charging multiple 18650 cells and the charge voltage was 14.4V (I think). I had put in the LTC4353 which is a supply OR'ing chip. However, in my parts of the world, this chip is not easily available and is quite expensive. Though not shown in the schematic, P4 and Vin would be connected to supply adapter power (something that TI calls PowerPath). Now that I'm looking at this circuit, the adapter voltage is slightly higher than the battery. I also found this solution on stackexchange.
 

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electrophile

Joined Aug 30, 2013
166
Ah I just read your post again and realized that you might be referring to the 4353 chip. I think that chip essentially is just a switch but cannot actually pass higher currents through it, hence a mosfet.
 
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