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42v to 12v

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simple and inexpensive, doesn't take up a lot of space, be sure to use a heat sink**broken link removed**
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THANKS
 
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Another search at Digi-Key: An 80 V, 84 A N-channel power MOSFET is 88 cents - in ones! It is not clear if the OP needs galvanic isolation or high speed switching, but even with all of that I think you can do this task for way less than $110.

ak
 
I must be missing something. What is the purpose of the relay? When will it be on/off?

Mike.
 
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.
Jeff
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Well said ak. Too many times we are left fumbling round in the dark trying to help because the OP will not tell us the full story.

Les.
 
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.
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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
 
ok, but I wasn't asking for OP, I was asking for myself, OP's got himself lost in the back ground as you say
 
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.
Thank you, very easy and interesting solution.
 
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
A 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
Hi AnalogKid

The voltage is 42 yes. It has to acces the load from the batterypack (42v charged), it does that through the relay, that switch on as soon as all the cell´s are above minimum voltage.

The idea you come up with, if understood correct. I can use an N channel mosfet with 42v connectet to drain, minus on batterypack to source, and then 1-2 cell´s (3,3-6,6v) to turn on the mosfet on the gate? Ive never ever used a mosfet before. Is it this simple? And it cost so little power so using a single cell, wont "damage" the balancing?

I dont need no latch circuit. I allready have a latching relay on the balancer, that switches state for Open/closed to a potential load. This relay can handle 2 amp´s tus why I need it to switch on something that can handle atleast 50 amps.

I tried to paint the circuit :)

Thank you
 

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That's pretty much it. I would increase the battery tap for the gate from 6.6 V to something between 12 V and 18 V. This will increase the efficiency of the MOSFET as a switch and reduce the heat it has to dissipate. I recommend a MOSFET with an 80 V rating for Vds, a current rating at least greater than the peak current it will have to handle, and use two FETs in a simple parallel arrangement: gate to gate, source to source, drain to drain. The FETs will figure out the load sharing.

ak
 
A couple of points come to mind from parts of this thread.

1) If you use a mechanical relay, make sure that the contacts are rated to handle the voltage that you're switching. And make sure to look at the DC rating, not the AC rating . Don't try to use a regular (12 Volt) automotive relay to switch 42 volts. Especially with an inductive load like a motor. It might work for a while, but the inductive energy at switch off from the higher voltage will soon trash the contacts. The ultimate failure will probably end with the contacts stuck together. Be aware that the coil and contact voltage rating are not related.

2) If you use MOSFETs (and that would be my choice) be aware that their current rating is right on the edge of failure. It is what current will produce enough heat which, due to the junction to case thermal resistance, will have the junction operating at it's max temperature, while the case is at or below 25C. That requires active cooling, or and infinite heatsink. Not practical for most applications.

So, the answer is to generously derate the current rating. Or, what I usually do, is to use thermal management as the basis for the design.

One of my products switches 50 Amps at 28 Volts DC. It uses four 100 Amp, 100 Volt mosfets in parallel. The four have a combined capacity of 400 amps, so I'm running at less than 15% of capacity. One nice thing about having such a large margin, is that no additional heatsink is required. At least, other than that the tabs (Drain pins) of the TO-220 cases are firmly bolted to the 2mm thick copper plate that is also the 60 Amp current conductor. The short Source pins are soldered to a 1mm thick copper plate.

Also, make sure that you have some sort of transient voltage clamp to protect the mosfets from the turnoff spike from wire and load inductance.
 
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.
 
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.
YUes it will turn off the motor. The voltage will never drop under 2.8v per lifepo4 cell.
 
added to your schematic, one IRF3205 110 amp 55 volt FET should handle your load, voltage divider will spread the load over the whole battery stack. If your equipment has an adequate piece of metal to be a heat sink, you can mount the FET to that and insulate it if needed or mount to FET to a store bought heat sink or use piece of aluminum plate, I use alum channel fairly often.IMG_20160102_135310.jpg here's an example of using a piece of alum channel as heat sink and conductor that is isolated from the body of my vehicle.
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the diode across the motor is for transient voltages
you will also need this or similar
**broken link removed**
ebay to220 insolator
 
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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
 
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If 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.
IRF3205.png

Mike.
 
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If 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.
Hi Mike

It seems that the circuit above was using some from my schematic. The out1 6.6v was something I have written, at Kinarfi's circuit, this voltage is 42v, not 6.6v.
I think this is where we have a misunderstanding?

Do you believe his circuit would work, if it had the correct voltage input on gate?

Thank you
 
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
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.
 
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