AlbertHall
- Joined Jun 4, 2014
- 12,345
No.Am I wrong?
But we can't guarantee that the adaptor will have a higher voltage than the battery.
No.Am I wrong?
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.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:
Yup.Something like this then:
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.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?
Is D2 required? What does it do? That the (reverse-biased) collector doesn't do?Thanks. Something like this then:
I'm confused...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.
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.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?
Indeed.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.
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?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.
Why two FETs in each switch arm?
by Jake Hertz
by Jake Hertz
by Jake Hertz
by Jake Hertz