I have a boombox that has a 12 volt lead acid battery as a backup battery, the thing is that I need a cutoff circuit that will cut off the battery from the circuit at 11.6 volts. What parts would I need to make this happen?
Below is the LTspice simulation of a circuit using a TL431 reference to provide an accurate cutoff voltage.
When the voltage at the Ref pin drops below 2.5V, the TL431 stops conducting and the MOSFET is shut off, removing the load.
For the values of R2 and R3 shown, the cutoff is slightly below 11.6V.
R4 provides about 100mV 0f hysteresis so that the battery wouldn't be turned back on when its voltage slightly rises after the load is removed and oscillate around that point.
The P-MOSFET can be just about any standard (not logic-level) MOSFET with an ON-resistance of 100mΩ or less.
R_Boom-Box is the simulated load.
View attachment 124756
For the R2 and R3 values shown the cutoff is nominally 11.56V.This circuit exactly how it is in the diagram will cutoff at 11.6 volts, correct?
For the R2 and R3 values shown the cutoff is nominally 11.56V.
You can make the cutoff voltage adjustable by changing R2 to a 50kΩ pot (wiper connected to one end of the pot to make it a variable resistor).
You can get away with discharging to 10.8V without damage - but you don't want to leave it discharged that low more than a few days.I have a boombox that has a 12 volt lead acid battery as a backup battery, the thing is that I need a cutoff circuit that will cut off the battery from the circuit at 11.6 volts. What parts would I need to make this happen?
When I make this circuit, the circuit does not cuttoff at the right point, and acts like a dimmer when using a test led if that makes sense,as the voltage goes down the led gets dimmer and dimmer when connected to the NFET. it cuts off at 9 volts and turns back on at 3 volts, what in the world, could the TL431 be blown, or something else?Below is the LTspice simulation of a circuit using a TL431 reference to provide an accurate and stable cutoff voltage.
When the voltage at the Ref pin drops below 2.5V, the TL431 stops conducting and the MOSFET is shut off, removing the load.
For the values of R2 and R3 shown, the cutoff is slightly below 11.6V.
R4 provides about 100mV 0f hysteresis so that the battery wouldn't be turned back on when its voltage slightly rises after the load is removed and oscillate around that point.
The P-MOSFET can be just about any standard (not logic-level) MOSFET with an ON-resistance of 100mΩ or less.
R_Boom-Box is the simulated load.
View attachment 124759
PFET?When I make this circuit, the circuit does not cuttoff at the right point, and acts like a dimmer when using a test led if that makes sense,as the voltage goes down the led gets dimmer and dimmer when connected to the NFET. it cuts off at 9 volts and turns back on at 3 volts, what in the world, could the TL431 be blown, or something else?
I used the NFET as an alternative, thinking it would worked. Could this only work with the PFET with that 1MEG resistor providing hysteresis so that the circuit wont turn on? I am new to PFETS, so I don't know much about these types of MOSFETS.PFET?
Yes, It is for a PFET only. We can help you find one if you tell us how much current the load takes and where you are located. US. Europe, etc..I used the NFET as an alternative, thinking it would worked. Could this only work with the PFET with that 1MEG resistor providing hysteresis so that the circuit wont turn on? I am new to PFETS, so I don't know much about these types of MOSFETS.
I go on a electrical component supplier called Farnell Element 14. I live in the US and the power draw is 6 to 7 amps maximum. Thank you for the help.Yes, It is for a PFET only. We can help you find one if you tell us how much current the load takes and where you are located. US. Europe, etc..
Here is one. Or if you want to PM me your address I'll send you a couple of IRF4905's. I have a couple tubes of them I'll never use.I go on a electrical component supplier called Farnell Element 14. I live in the US and the power draw is 6 to 7 amps maximum. Thank you for the help.
If you don't know much about them, then I'm puzzled why you would think an N-MOSFET would work in place of a P-MOSFET.I used the NFET as an alternative, thinking it would worked. Could this only work with the PFET with that 1MEG resistor providing hysteresis so that the circuit wont turn on? I am new to PFETS, so I don't know much about these types of MOSFETS.
Let me try to give your thinking a boost. You can temporarily imagine an n-fet as an npn transistor and a p-fet as a pnp transistor to get the logic of the circuit correct. Then switch the active component back to a mosfet, correct for the difference in gate voltage and gate current, and then your circuit will work.I am new to PFETS, so I don't know much about these types of MOSFETS.
Thanks, I'll PM You With Details.Here is one. Or if you want to PM me your address I'll send you a couple of IRF4905's. I have a couple tubes of them I'll never use.
http://www.farnell.com/datasheets/6...33.762311853.1497661737-1731933628.1497661737
Likely a wiring errorThe cutoff circuit for some reason will not cutoff at that set voltage, and the circuit is exactly like in the diagram. What could be happening here.
What voltage for the battery do you want me to measure voltages at the nodes?Likely a wiring error
Show us the exact circuit you built with the voltages you measure at each circuit node.
Two sets of measurements at both above and below the desired trip point.What voltage for the battery do you want me to measure voltages at the nodes?