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You're not missing anything. 23 posts and it's still a secret.I must be missing something. What is the purpose of the relay? When will it be on/off?
I didn't forget. Since things like your package partialities, or the OP's actual operating voltage and current still are not known, I'm in no position to recommend anything specific. My point is that there are many devices to choose from, they are cheap, and they are easy to find.AK, you forgot to give the number of the fet, I looked and the only 88 cent ones were not TO220, and I say that only because I'm partial to them and through hole devices.
Thank you, very easy and interesting solution.If you use a buck regulator, the inductance of the coil itself will mean you don't need a separate inductor.
If you use something like an LM2576HV-ADJ, with an output load of an 8.2 Ω resistor, and have the relay coil in place of the inductor.
The LM2576HV-ADJ will control the voltage on its feedback pin to 1.23 V, so there will be 150 mA flowing in the load. All the load current comes through the inductor, so its average current will be 150 mA as well.
If you do that, you must not put any capacitors or diodes in parallel with the coil, and it's a good idea to have the coil close to the buck regulator. The resistor will generate a bit under 0.2 W of heat, so a 1/2 W or more resistor would be best.
The circuit diagram is on page 12 of https://www.ti.com/lit/ds/symlink/lm2576.pdf with L1 as the relay coil and "Load" the the 8.2 Ω resistor.
I didn't forget. Since things like your package partialities, or the OP's actual operating voltage and current still are not known, I'm in no position to recommend anything specific. My point is that there are many devices to choose from, they are cheap, and they are easy to find.
ak
Hi AnalogKidA pair of 50 A or 75 A power MOSFETs (for redundancy) and a simple latching circuit will cost waaaaay less than an equivalent mechanical relay. You can use about 10 microamps of current from a battery to maintain the latch state, not nearly enough current to upset the stack charge balance. What is the voltage across the relay contacts now, 42 V?
ak
YUes it will turn off the motor. The voltage will never drop under 2.8v per lifepo4 cell.Is this MOSFET supposed to turn of the motor when the battery gets low? If so, when the tap voltage drops the internal resistance of the MOSFET will increase and it will melt.
Tell us what the circuit is required to do and we might be able to help.
Mike.
Hi MikeIf you look at the datasheet for the IRF3205 you will see that it will start to conduct with a gate voltage between 2 and 4 volts. It will be fully on at 8-10V. At 5V (2nd line from bottom) it will have a high resistance and at 30A will have 10V across SD. Therefore it will dissipate 300W and will melt. The circuit above is putting 2.2V on the gate which may turn it on but it will be in ohmic region and will not work.
Here's the table to consult.
View attachment 100998
Mike.
Thank you for the very specific circuit diagram. The motor has to run free when power is not given to the motor. Will the diode prevent this, and then brake instead? I do believe the motor is brushless, since it had 3 wires in, but I cant guarantee this, but it probably am.Other bikes I've played with free wheel when power is off and rely and mechanical brakes to stop. Is this the same? The transient voltage diode will stop the motor, assuming the motor is permanent magnet with brushes.
What type of speed control does this bike have, can you post a photo?
On my drawing, it would be wise to use a STPS20M60S 60 volt 20 amp diode