Switching voltage sources ...Help

[The Real MicroMan]

I have an application that needs to switch from one battery to another when one source gets too low. I tried Pch mosfet switching to switch the high side but with no luck. Does anyone know how to do voltage source selecting to a load?

 

[Nigel Goodwin]

Simple diode OR gates are as easy as it gets - but your question doesn't give any details to make more explicit suggestions possible!.

 

[dknguyen]

Can't you just stick diodes on the batteries with relays? And have a slight overlap between the switching of the relays (so there is a time when both batteries are connected)?

You said you tried something...what exactly was it that you tried and how did it fail? I am assuming that the entire system is powered by just those two batteries and you require a "switchover" while still keeping everything powered? NO external 3rd mini battery to hold the system over during the switchover right? Tell us what you tried.

 

[Papabravo]

What exactly was wrong with your P-channel switches? I've used this technique many times with great success.

 

[The Real MicroMan]

What exactly was wrong with your P-channel switches? I've used this technique many times with great success.

When I used the p-channels I had the Drains hooked together to a common load. My simulation showed large current draw. How did you do it and could you provide a simple schematic? just how the common load hooks to the p-channel.

 

[The Real MicroMan]

Can't you just stick diodes on the batteries with relays? And have a slight overlap between the switching of the relays (so there is a time when both batteries are connected)?

This has to be a low cost, and low power. Relay's are just too current hungry and not very cost effective. uA's of current is what I'm dealing with.

 

[dknguyen]

How were you using to control the gates of the PMOS? Is there any reason you can't just parallel the batteries? I assume one is meant to recharge while the other is running? You said large current draw? How large? Where? How small should it be?

 

[The Real MicroMan]

Simple diode OR gates are as easy as it gets - but your question doesn't give any details to make more explicit suggestions possible!.

More detail: Well I have 2 sources one starts at 7 volts and the other at 6.4. When the votage on the first gets to 4.0 volts I want to switch the source voltage to the second. Then when the 1st source is charge back up switch back to the first source. The forst source could be a super cap or a rechargeable battery. What I did was set up 2 p-ch mosfets to a common load. On the source of each mosfet was the voltage source in question, on the drains were connected together and the gates were controlled by another circuit that basicly turned on fet on and shut off the other. I know the model I was using was an enhanced mosfet(ie diode intergrated in the fet) so I figure I have to use a non-enhanced fet. I also know the turn on delay is differnt from the shut-off delay..I can correct this with a resistor and diode on the gate side, but what I don't know is Can i connect both drain's together to a common load?

Thanks to all of you who gave feed back to this question. You comments are greatly appreciated

 

[dknguyen]

Hmmm...yeah it could be that the body diode is turning on in the MOSFET that is meant to be off...that's probably what is happening The potential difference between 7V and 4V is definately high enough to cause the body diode to conduct. It's causing the full battery to discharge into the empty battery through the body diode on the "off" MOSFET.

Have you tried sticking a diode at the drain (or source) of each MOSFET to stop this from happening? THis is what was meant earlier by using diodes for each battery. It prevents one battery from discharging into the other.

Maybe this is a silly question because I don't know your application...but are you able charge up one source before the other empties, can't you jsut power it straight off the source? The only (obvious situation) that I can think of is for battery to power the device in sleep mode where very little current is used so that the second battery can be charged up- presumably the sleep mode uses less current than is being charged. Because if you use energy faster than you can charge the auxillary power, the device ends up dying. If you can charge up the auxillary power as quickly as the device uses power, then you might as well just power the device directly.

 

[The Real MicroMan]

Hmmm...yeah it could be that the body diode is turning on in the MOSFET that is meant to be off...that could be what is happening. I assume you know how to apply the proper gate voltage to switch a PMOS on and off right? The potential difference between 7V and 4V is definately high enough to cause the body diode to conduct.

Have you tried sticking a diode at the drain (or source) of each MOSFET to stop this from happening? THis is what was meant earlier by using diodes for each battery. It prevents one battery from discharging into the other.

Maybe this is a silly question because I don't know your application...but are you able charge up one source before the other empties, can't you jsut power it straight off the source? The only (obvious situation) that I can think of is for battery to power the device in sleep mode where very little current is used so that the second battery can be charged up- presumably the sleep mode uses less current than is being charged. Because if you use energy faster than you can charge the auxillary power, the device ends up dying. If you can charge up the auxillary power as quickly as the device uses power, then you might as well just power the device directly.

In my simulation I set one source to 6 and the other to 7, then uses a third source set to 7 volts/ 0 volts to turn on and off the fets. I used a pull-up on both fets from gate to source to get a full tuen off on the fet. Will the body diode cause the fet to turn on is the drain voltage is 6 and the source voltage is 7?

 

[dknguyen]

If the voltage from the voltage difference from drain to source is ever higher than the body diode forward voltage, this will forward bias the diode and cause it to conduct current from the drain to source of the MOSFET. If your body diodes have a forward voltage drop of less than 1V this will happen.

Just stick some diodes in series with each MOSFET's source-drain (doesn't matter if it's at the source or drain). This will block current from flowing backwards through the body diode. I think the forward drop across these diodes will cause you to have to change your voltage thresholds when switching between batteries.

Use your simulation to measure the current in the source-drain of the transistor that should be off.

 

[eblc1388]

Common MOSFET can block current flowing in one direction only.

Your special case calls for blockage of current in both directions and the only simple answer without incurring diode drops is to use two MOSFETs connected up in series Drain-to-Drain for each voltage source. This is commonly done in circuits I've seen.

 

[Hero999]

I've done a quick google and found some interesting stuff.

https://www.eng-tips.com/viewthread.cfm?qid=149960&page=7
https://www.electro-tech-online.com/custompdfs/2006/09/dt94-8.pdf
https://www.google.com/search?clien...rity+circuit&sourceid=opera&ie=utf-8&oe=utf-8