I started to build the relay but I am having problems.

Here is what I have.

Signal coming into a invertor and to MOSFET 1. From the invertor to a MOSFET 2.

RLoad is connect from Power source 1(ps1) to Drain of MOSFET 1. MOSFET 1 Source is connect to negative of ps1.

RLoad is connected from ps1 to drain of MOSFET 2. ps1 is in series with ps2. MOSFET 2 source is connected to negative of ps2

The problem i am encountering is when i send the signal to turn one of the MOSFETs on, there seens to be a short thats causing MOSFET 2 to overheat. I'm getting this when i start grounding the ps1 and 2 with the ground from the invertor chip, etc...

any ideas?

ref to previous topic: http://www.electro-tech-online.com/v...ic.php?t=10238

 

If you are trying to extend battery life by series connecting dead batteries, it is a waste of time. The voltage of the battery drops like a stone at the end of its useful life, so the increase in life is not worth the effort.

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change the power supply's from batteries to a dc supply.

whether its batteries or dc supply, its shorting out somewhere causing MOSFET 2 to overheat and burn up.

Thanks,
bob

 

Are you using the circuit I posted? If not, to which supply terminal do you have your control signal referenced, i.e., which node in the circuit is your local ground? What is the voltage swing of your control signal? What two voltages does your inverter switch between, relative to this ground? What are the voltages of your power supplies?

 

I built it like yours only I didn't include the BJT portion or diodes. I used an invertor and has the signal sent to the second MOSFET and the inverted signal sent to the first mosfet.

I'll build it again and make a schematic of what I have and describe the problem in more detail when I get home.

But is the BJT and diodes necessary for this to work?

Thanks,
Bob

 

Ok...I rebuilt it and this is what I came up with.

I put 2 loads. The first load with the first mos. and the second load in series iwth the first load connected to the sencond mos. hooked the invertors up and it works fine with onlhy one supply or the other.

If I take away the sencond load, and have the system work under 2 power supply in series, the second mosfet starts overheating. Burned myself a few time.

Now my question is why is it over heating? on my power supply it says its pulling .83mA on the second MOSFET.

Thanks,
bob

 

Trying to visualize a circuit from a verbal description is only slightly easier than trying to imagine how a banana tastes by having you describe it to me. I need a schematic.

 

schematics are worth a thousand words. i agree with the previous post. its easier to help with a visual display of what you are working with.

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Here is a windows paint schematic

Both n channel
Make the power source what ever voltage you want

The problem I'm getting is when the 2nd mosfet is turned on, it heats up but if I put another load in series with the first load, then it works fine....I only want one load.

Attached Images

untitled_112.jpg (12.5 KB, 651 views)

 

unless I have really missed something, it seem to me that when both FET's are on, the lower battery is basically shorted by the 2 FET's.

also the gate for the FET on the right needs to be higher by whatever the lower battery voltage is. IE, if the FET needs 5 volt to be completely on, non-linear, and the lower battery is 12 volts, then you will need to have 12+5=17 volts at the gate to get it out of its linear region. If its only partially on, then its dissipating part of the power, and if the load is a fairly small resistance, it could be basically dissipating the majority of the load. Yes it will get hot then, and quick.

 

no...the switch is the input to the invertor.....the invertor output will be the opposite of the input signal so they're not both on at the same time unless I'm missing something...

It works fine if you have 2 load hooked in series...nothing heats up...but if you take away of the second load that is connected to the source of the 2nd mosfet, then the 2nd mosfet heats up....I have both power supplys set at 3V. The load I'm using is a LED.

Thanks,
Bob

 

What mosfet are you talking about?

The one on the left is the one that gets hot when it is turned on.

When the top is turned on, nothing gets hot, it works like it should.

Thanks,
Bob

 

What is the inverter? Is it an IC, or is it made up of discrete parts? The inverter in the circuit I posted in the previous thread is also a level shifter. This is important!
BTW, why did you start a new thread? It makes the discussion more difficult to follow.

 

the Invertor is a IC.

why is the level shifter important in this case?

Thanks,
Bob

P.S. I started a new thread cause its habit for me....sorry if its hard to follow...

 

See below. The level shifter gets the gate-to-source voltage on M2 to switch between zero volts and 9volts, and it inverts the logic levels so the MOSFETs turn on in complementary fashion. It won't work for low battery voltages (below about 6v).

Attached Images

battery_switch1_531.gif (7.5 KB, 569 views)