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Automotive Generator Universal Digital Voltage Regulator

CK3

Member
...

You should use a much larger Gate-Source Voltage than the threshold voltage. The threshold voltage is the voltage at which the MOSFET starts to turn on. To get the best resistance you need more voltage. The NTMFS5C612NL has a threshold voltage of 1.2 - 2 V, but that is at a drain current of 250 μA. To get the typical 1.2 mOhm you need 10 V on the gate.

If it's a negative-earth vehicle, when the dynamo is off, the MOSFET source will need to be a 0 V, so the gate will have to be at 0V as well. When the dynamo is generating, the drain and source will be about 7 V, and the gate will have to be at 17 V to get the best resistance. You don't really need any current at 17 V and you will have the big advantage of little heat being generated, but you certainly won't be able to use the 3.3 V output of the GPU to drive the gate.
Thank you for that lucid explanation! When you put it that way, it all makes perfect sense.
There is an IC designed to drive an N-MOSFET in this way. It's an LM5050 (https://www.ti.com/lit/ds/symlink/lm5050-1-q1.pdf). One of those and an N-MOSFET would probably work as a cut-off with no further connection to the regulator part.
That part looks perfect for this application. Unfortunately, they seem to be scarce, like everything these days. I have ordered the single one I could find to experiment with. The price is reasonable: $2.75 in qty. 1 from Newark.
 

CK3

Member
TI also has the LM74700-Q1 3.2-V to 65-V, 80-uA IQ automotive ideal diode controller, Low IQ Reverse Battery Protection Ideal Diode Controller, which seems pretty comparable to the LM5050, but not any easier to find.

Update: The LTC4359 seems to available, and looks like it would do the job. It costs a little more: $6.52 from Mouser in quantity 1. Analog Devices has a few other, similar chips, with features that I don't need.
 
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CK3

Member
The LM5050 data sheet says it's a high side driver, but could it also be used as a low side driver? I'm thinking of the field current switch. In my 6v automotive voltage regulator application, a "B" circuit regulator energizes the generator field by grounding its end of the field coil. If I want to control that with a digital signal, I again have the situation of the signal not being high enough voltage to fully enhance the power MOSFET. If I use a simple gate driver, I'll get 6 or 7.5 volts at most from the car battery circuit; maybe less depending on state of charge. Not really enough, right? Is there a way to use the charge-pumped LM5050 to solve that? Or a more appropriate charge pumped low side driver? I've looked at a bunch of low side gate drivers and they all seem to be limited to (almost) rail-to-rail, which would be an improvement over a 3.3v logic signal, but probably still marginal in a 6v system.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
The simplest things would be either photovoltaic FET drivers (though they are slow switching), or integrate a small boost converter such as a "simple switcher" type IC that could be used to power the FET gate drive circuits & allow fast drive for PWM on the field control.

Though the LM5050 or 74700 do look to be perfect for the "cutout" control!
 

CK3

Member
I've played around with the IRL40B209 HEXFET® Power MOSFET and I can use it to turn a 6v headlight on and off with a 3.3v logic signal from the STM32 MCU directly on the gate. So, a rail-to-rail driver might be fine for that MOSFET (see Fig. 1 in the data sheet). But, I can't find an equivalent surface mount part.

[EDIT: After doing some more looking, I think there are plenty of MOSFETs that will meet the requirements for the field current switch, and can be driven directly by the 3.3v logic signal on the gate. For example, for US$0.2043 from LCSC, YJD90N02A. Or, if you're made of money, IRL60SC216 for $4.20 from Mouser. Or, maybe the $2.07 SiR404DP.]
 
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CK3

Member
I've come up with a design for a power board: https://oshwlab.com/carlk3/volt-reg-1-1_copy_copy_copy_copy_copy_copy. First, I want to send off my controller board and test that out. But I wanted to take a stab at the power board to see if I missed any dependencies on the controller.

I went with the idea that I mentioned above, of making everything surface mount and attempting the thermally couple through the board to the voltage regulator frame. The bottom of the board is as smooth as a baby's behind:
Screenshot from 2021-11-19 22-05-26.png


I'm thinking this could be placed on a thermal pad (e.g., Thermal Pad, 100x100x0.5mm Highly Efficient Thermal Conductivity 12 W/mK, Insulated Heat Resistant High Temperature Resistant Silicone Thermal Pad) and bolted down to the smooth steel floor of the voltage regulator frame.

PXL_20211120_051047832 (1).jpg

I think with a driven gate MOSFET Cut-out, it might just be cool enough to work.

Another concern I have is the ability of PCB traces to carry 50 amps. I looked at some online calculators and they suggested that the traces need to be inches wide. About 200 mils is the best I've been able to squeeze in. JLCPCB can do 2 oz. copper*; that should help. I guess I could go to 4 layer board and run more redundant traces in the middle layers. But those can only be 1 oz. And the extra layers might add to the thermal difficulties.

It's all pretty speculative. I might just have to try it to find out.

* Ouch! The price for the PCB jumps from $4 to $21.4 if I select Copper Weight: 2 oz. You can also select board thickness ranging from 0.4 to 2.0 mm (default 1.6). (Of course, the prices shoots up again.) Maybe a thinner board would transmit heat better?
 
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Diver300

Well-Known Member
Most Helpful Member
For high current traces, you can just have no solder resist on the bits that take high current, and add solder when the components are fitted.

2 oz copper is 70 µm thick, or 0.07 mm. Tin has a lot more resistance than copper, about 5 times, but you can easily get 0.5 mm of solder onto a wide track, which will reduce the resistance by a factor of more than 2 compared to 2 oz copper.

You could even solder down bare copper wire along the track. A bit of 2.5 mm2 solid wire* is about 1.8 mm diameter, so can be soldered along a 3 mm wide track quite easily. A 3mm wide track on 2 oz copper is 0.21 mm2 so the copper wire on top of it reduces the resistance by over 10 times.

A thinner board will transmit heat from front to back better, but it won't make any real difference in conducting heat across the board, as the copper does that. If the heat is conducted from front to back, unless there is something on the back of the board to take heat away, it's not going to make any difference.

Tracks inside the PCB don't add to the thermal problems. They can make it a bit better by conducting heat sideways better. Redundant tracks don't help much if the copper is only 1 oz and the heat dissipation from an internal track is worse than from a surface track.

*2.5 T+E cable (Translation to American:- 14/2 Romex) is a good place to get solid wire like that.
 
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CK3

Member
For high current traces, you can just have no solder resist on the bits that take high current, and add solder when the components are fitted.

2 oz copper is 70 µm thick, or 0.07 mm. Tin has a lot more resistance than copper, about 5 times, but you can easily get 0.5 mm of solder onto a wide track, which will reduce the resistance by a factor of more than 2 compared to 2 oz copper.

You could even solder down bare copper wire along the track. A bit of 2.5 mm2 solid wire* is about 1.8 mm diameter, so can be soldered along a 3 mm wide track quite easily. A 3mm wide track on 2 oz copper is 0.21 mm2 so the copper wire on top of it reduces the resistance by over 10 times.
Some great ideas there!
A thinner board will transmit heat from front to back better, but it won't make any real difference in conducting heat across the board, as the copper does that. If the heat is conducted from front to back, unless there is something on the back of the board to take heat away, it's not going to make any difference.
I'm planning to bolt the board down to a steel plate with a 0.5mm piece of something like "SIL PAD" in between. (I didn't know it was called that until I watched this video yesterday:
.) Hopefully, that will take some heat away.
Tracks inside the PCB don't add to the thermal problems. They can make it a bit better by conducting heat sideways better. Redundant tracks don't help much if the copper is only 1 oz and the heat dissipation from an internal track is worse than from a surface track.

*2.5 T+E cable (Translation to American:- 14/2 Romex) is a good place to get solid wire like that.
The guts of the old regulator are a mine of good solid wire:
OZk7ZVgD1EhX1FqBkfMiPUOl4LzLv0vXpJ3cTlvFxRvUZHivM8IQGpeUx3pRuSsFrZH1rIGwJux_3ktUBRU2pm926DVBW5ph5X8wr-UEXczwgS4EF8isDpvpLhD2TcisDuYGQKmbXuw9SrRVJ5gEafmqCJ8FIlSOhilFZ7icMLdm9A8fw8cd54yJBCb27wJovpXZFlyzw7uaImE5Bsf1ZeEzRLFu3jKAa7e3G374tb890FS-XtkIOdKAB-Ua2UMbJn6UkZZow4gpeTlDQmYOOIfoYHlgTE2uLL2wkEuv6ZNo9PTMb87r1-8lZc_zr75zLFVb8isYZoyK_vYgk9RC54L2ZoeLZ8aW4KfvHl7A9ixd9ppmXqSvuAOi2AnKMOmk7osam25y_vwOaBOJjwRlyg4sXoG6M9g22cu7tQkhpqyuCjFUQ4uTQZB08Wtk-352SZ2ztpu8nenqNhsytG54F938NQtByirGotsVSVbQQkPq5ws1oLecH36NFwUtWdfJb7W7N3ubSQOLN6bSkaHqMWSXBNsHJPZ57HSxOwIV5DQ-_Ps-3AbOUZZqAWHXrOFeFvBZMzEmkbRJ5mbtagQ5JW4xT1VJwlg8pd0nZltOjfRDW9_7hnCi3imU2RFHc6k6QAmXQsjnfiRfyQ-KuIfQqiznYl9kTvjWl_VKcsf1vmIcEitePD7XOEH3Cn0mGp5rVffBPD1toIxIqcDcwotN4qYdNQ=s949-no

That current coil is really thick stuff. Probably too thick to lay on a PCB track. But I have lots of Romex scraps around, too. I will have to reconsider my board layout (again). I put the 100+ mil tracks on the back side, but I want to keep that side smooth. So, I need to get some big, straight tracks on the top side. And then, maybe I won't have to go so crazy with vias, which I have been liberally sprinkling around, hoping they can take the current.
 

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