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[SOLVED] Switch positive and negative (Choosing/Building an H-bridge)

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johnwarfin:

--- Quote from: greywanderer012345 on June 16, 2013, 09:57:08 PM ---I always try to do things as inexpensive as possible,
--- End quote ---

the lowest cost solution is to use cheap fets for your h-bridge. in most cases nothing besides these two components is required. one p and one n for each of the 2 legs. one pulls up, the other pulls down. only one is on at a time. since you like digikey heres a couple part numbers:

568-5956-1-nd si3202 n-fet
irf7425-nd 7425 p-fet

ive used these with great success with 5s (18.5v) rc motors. would probably work fine for my n-scale trains too which are dc but havent actually tried. you did get me thinking though.

greywanderer012345:
Great information everyone!

I'm thinking I'll be getting the FETs suggested by johnwarfin.

I'm still confused about how an H-bridge is set up. I'm looking at the basic concept as explained by admin here:
http://www.societyofrobots.com/schematics_h-bridgedes.shtml
and also considering what's been said here.

Forgive my lack of comprehension; I'm only a hobbyist. I still don't understand why each leg needs one P and one N. Are they both turned 'on' by a supplied voltage and 'off' at 0V, or is one turned off by supplying a voltage?

Do I need to get a motor controller or MOSFET drivers too, or will my MCU board and the 4 mosfets be everything? What about the capacitors suggested by admin's sketch? The track is 18VDC for sure, and there's already enough resistance in the circuit that nothing blows even if the track is shorted out.

Also, why does admin suggest adding a resistor before the gate?


--- Quote ---First question - will an H bridge work for this application? Many HO trains run on AC power and just switching the two leads via an H bridge will do nothing.  Have you tried switching the wires manually to see if it will, in fact, run backwards?
--- End quote ---


The control box has a physical switch that just reverses which rail is 18V and ground.

greywanderer012345:
I think that the "Power-Max" Spec was confusing me more than anything on Digikey. The FETs suggested say 20V Vdss and 2.5A continuous drain. However, the Power-Max is listed as 830mW. I was thinking that this spec was the max power that could be going through the FET, but now I'm guessing that this is actually either the maximum power going through the gate, or the maximum power being dissipated as heat by the resistance of the FET. Either way, I probably shouldn't have been so concerned with this spec as I searched.

Also, are there H-bridge ICs that can be bought and have the FETS built into them, or are the ICs just 'drivers' which are separate from the actual transistors?

johnwarfin:
there are h-bridge ics like mentioned earlier in this thread and some do use fets. but they cost a LOT more than plain fets and personally i prefer the basic circuit you just posted. i suggest you read what the other two fellows just posted on h-bridge operation. its accurate and complete. however imo most of those extra components are only needed for highr power applications. for me the fet only solution has worked well.

you simply need to picture the fets as switches. with  one end of motor shorted to neg and the other to pos it turns cw. reversed it turns the other way. some think there should be a small period (deadband) where neither fet is on. ofr me i just tie the gates together and switch from one bit. true theres small spike but in my case no heating or other issues.

btw an alternate p-fet i use is irlm6401 probably also available from digikey. a little below spec but ive had no problems. these and the 2302 only cost 10-20 cents ea in small qty so tend to be the route i prefer.

jwatte:
A few things:

The maximum power 830 mW means maximum power dissipated. Look at the Rdson (internal resistance when on) value. That's the resistance (at a particular gate voltage value, which may be specified as high as 10V!.) Now, multiply I-squared by resistance, and you get power dissipated. This puts a maximum limit on the current.
You actually have to also look at the thermal resistance of the device. It may be that, at the rated power, it will actually overheat in plain air, and you will need a heatsink to actually be able to dissipate that much.

I think the device seems way under-specified for your case, though. MOSFETs may simply die if they see a higher gate-source or drain-source voltage than they are designed for. 20 V design limit with a 18V inductive load leaves no margin at all, and inductive loads are well-known to generate up to 2x the rated voltage as kickback! I'd use a 30V rated MOSFET at least. Also, use one with a lower Rdson (thus higher amp rating) to have some margin.

The reason you use P-channels for the high switcher is that it will conduct when the gate is more negative than the source (which is tied to the positive voltage.) An N-channel conducts when the gate is more positive than the source (which is tied to ground.) Tying the drain of an N-channel to positive, and the source to the load, using the N-channel as high switch, means that the gate needs to be more positive than the top end of your load -- you need to boost the gate voltage above the positive supply voltage! There are circuits that can do this (high-side N-channel MOSFET drivers) but that's why home-built H-bridges generally use P-channels instead.
Also, if your supply voltage is 18V, and your MCU output is 5V or 3.3V, then that MCU output is not enough to turn off a P-channel. Typically, you will use a pull-up resistor for the P-channel gate, and have the MCU controller drive a small N-channel device that, when open, pulls the gate of the P-channel to ground. This means you waste some current through the pull-up resistor while the P-channel is conducting. Again, with MOSFET driver chips, you can get around this -- look at something like the IR2183 or IR2301 for example (make sure you pay attention to the required supply voltage!)

Finally, again, if the driving current is AC, an H-bridge won't work. Make sure you check the output of your transformer. Ideally, using a scope, but if you don't have one, a multimeter can detect AC simply by measuring the output both in AC mode and DC mode. If it's the same value in both modes, it's likely DC, else it's likely AC.

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