Am I losin' it?

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I've been simulating some stuff on Proteus 7 and have been struggling to get something to work that, as far as I'm concerned, works on paper - a simple H-bridge. The H-bridge itself isn't really the point of interest for me, but it's part of the circuit and it simply doesn't work as I expected it too and for my own life I can't see why not.

This is a screen shot from Proteus, mid-simulation. Now, if I'm going crazy and I'm missing something obvious then great - shout it at me, but why on earth are there the huge voltage drops across Q1 & Q4!? They should be switched on hard should they not?

 

Interestingly I tried swapping out the MOSFETs for BJTs and found the same thing. (note the change in how R7 & R8 are connected and their value)

So in this case, there is a huge voltage drop across Q2 (which is the top right transistor, label's hiding behind R2 unfortunately), but more current is flowing through the motor. If I turn off Q2 & Q3, and turn on Q1 & Q4 however, then the circuit behaves as I expected it too originally with the MOSFETs (note there is no R3 in this case).

So what exactly am I missing here? Am I wrong to think that the H-bridge with MOSFETs will work as expected? I've attached the .dsn files to this post so that anyone else with Proteus can experiment themselves.

Attachment(s): 

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Odd. I don't use Proteus so I dropped your circuit into Circuitmaker (the original one) and the top one works just fine. I used 2N2222A for the BJTs and IRF640 and IRF9640 for the FETs.

#1 This forum helps those that help themselves

#2 All grounds are not created equal

#3 How have you proved that your chip is running at xxMHz?

#4 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand." - Heater's ex-boss

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#1 This forum helps those that help themselves

#2 All grounds are not created equal

#3 How have you proved that your chip is running at xxMHz?

#4 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand." - Heater's ex-boss

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Thanks Brian, I can sleep tonight now ha! Perhaps I should look into using Circuitmaker for simulation from now...

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R1 and R3 make a voltage divider such that when Q5 turns on, the voltage at the gate of Q1 would be 12 volts.  Shouldn't the voltage be something much smaller to turn off Q1?

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Howard_Smith wrote:

Perhaps I should look into using Circuitmaker for simulation from now...

 

It's long been obsolete. It was a Protel product, then bought out by Altium for the PCB side of it and then it disappeared.

#1 This forum helps those that help themselves

#2 All grounds are not created equal

#3 How have you proved that your chip is running at xxMHz?

#4 "If you think you need floating point to solve the problem then you don't understand the problem. If you really do need floating point then you have a problem you do not understand." - Heater's ex-boss

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Proteus makes mistakes? NO!!

Jim

I would rather attempt something great and fail, than attempt nothing and succeed - Fortune Cookie

 

"The critical shortage here is not stuff, but time." - Johan Ekdahl

 

"Step N is required before you can do step N+1!" - ka7ehk

 

"If you want a career with a known path - become an undertaker. Dead people don't sue!" - Kartman

"Why is there a "Highway to Hell" and only a "Stairway to Heaven"? A prediction of the expected traffic load?"  - Lee "theusch"

 

Speak sweetly. It makes your words easier to digest when at a later date you have to eat them ;-)  - Source Unknown

Please Read: Code-of-Conduct

Atmel Studio6.2/AS7, DipTrace, Quartus, MPLAB, RSLogix user

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When Q5 is on Q1 shouldn't be off, that is by design.

 

When Q5 is off R1 pulls Q1's gate to +24V. When Q5 is on, yes R1 & R3 form a voltage divider that pulls Q1's gate down to +12V, which is 12V negative with respect to the voltage at Q1's source, thus turning it hard on.

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Why are you simulating this(what do you hope to learn)....it's pretty straightforward, turning some switches on/off.

Add a 1 meg or so resistor to the motor terminal(s) to gnd.  when all fets are off the motor state is floating & undefined.  Add in a resistor  helps to constrain the calculations.

Floating nodes could give the solver a headache.

When in the dark remember-the future looks brighter than ever.   I look forward to being able to predict the future!

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Is there a reason R5 & R6 are mismatched ?

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As I said in my OP, the H-bridge is not what I was interested in here but still part of the circuit nonetheless, and ultimately was the reason that the part of the circuit that I am interested in wasn't working.

Thanks for the tip with the resistor.

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No I simply forgot to change it's value blush and 10k is the default value.