New H-Bridge Design

ScottWang

Joined Aug 23, 2012
7,400
About the all Bjt H-Bridge circuit.
If you like then you can combine Sw1,Sw2 become only one Sw1, you just add another PNP circuit as Q1 structure, connect the input of PNP circuit to forward pin, connect the C of PNP to backward pin, connect E of PNP to 5V.

And then you just use one Sw1 to control the motor to move forward or backward.
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
If you look at the datasheets for the darlington transistors, they already have R9, R10, R11 and R12 built-in.
Thanks. I checked and on the first page of the datasheet I saw this. Less four resistors.. nice.

@ScottWang

I tried to do what you said, but i do not understand very well.

May you draw it for me? Can be on paint :D.

thanks
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
ow ScottWang, thank's a lot! :D

I do not know why, but the mod you posted do not worked _on the simulation_. But I changed the PNP with a NPN transistor and folowed your tip. It worked.

With this circuit I can use a very small uC. Just 2 pins. Really nice.

The final circuito with BJTs:



Now, i wil try to adapt tis bridge to use with FET's..

EDIT:

Just make the mod here. Now it seems that work with FETs.

Nice?



bye
 
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ScottWang

Joined Aug 23, 2012
7,400
In the beginning, I want to use one PNP and one NPN, PNP for Hi, NPN for Lo, and I think only use PNP maybe will OK, but I forgot the Q8 still need a Low level as GND, the 10K is too much, if use 300 ohm, probably it will work, but it still need to try, like you used NPN, that is what I want to use in the beginning, it just replace the original Sw2 function and connect to GND, the R10(10K) just replace the PNP function.
 
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ScottWang

Joined Aug 23, 2012
7,400
Many Rb(1K) are too small, you need to increase the value. (4.7K~10K)
You lost some RB resistor at some B poles of transistors.
The C resistors of Q5,Q7,Q9,Q11, I think you can use them up to 10K.
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
Thanks again ScottWang, I changed the values and it seems that is working.

First I will mount the BJT bridge and when I got the FET's, I wil try the other.

Ahhhh... before I forgot, look that I have +24V to feed the gate of the FET's. I wil show I little trick to transform +12V into +24V:



With the doubler circuit this is easily accomplished.

bye
 

Audioguru

Joined Dec 20, 2007
11,248
IRF3205 datasheet.
page 2
VGS(th) Gate Threshold Voltage 2.0~4.0 V ← This is the smallest Voltage for Vgs.
The threshold voltage for a Mosfet should never be used for its input voltage because that is when it is almost turned off (0.25mA), not when it is turned on.

The datasheet says it is fully turned on when its Vgs is 10V.[/quote]
 

ScottWang

Joined Aug 23, 2012
7,400
IRF3205 datasheet.
page 2
VGS(th) Gate Threshold Voltage 2.0~4.0 V ← This is the smallest Voltage for Vgs.
The threshold voltage for a Mosfet should never be used for its input voltage because that is when it is almost turned off (0.25mA), not when it is turned on.

The datasheet says it is fully turned on when its Vgs is 10V.
Thanks for your Supplement.
I posted the input voltage range of MOSFET, that just to tell MatheusLPS the Vgs is too high, and it's unnecessary.

If you see that #8 I told MatheusLPS to reduce the input voltage, the voltage almost equal to the volage of motor(12V).
 

help_mum

Joined Sep 13, 2012
6
If you use I^RF389 HEXFETS you would only need 2, as you drive neg and pos. If I remember they are 30v 4to5 amps neg and up to 7.5 pos as gate drivers ROHM make twin NPN Transistors the last time I used them I tryed Hseries that have 4.5k internal resistors on the base pins but MX1 worked better. I would need to go back through my old Schematics. I think they were running at 1k
Just a thought if you try this post the spice Specs on them please I would like to see how well they worked.

HM
Beginner
 

Thread Starter

MatheusLPS

Joined Jul 16, 2011
34
The datasheet says it is fully turned on when its Vgs is 10V.
Agreed. To solve this problem with the +24V I can feed the 555 with +6V and in the end it will be +12V.

I am a bit confused with mosfets. I know that in the saturation mode:

VGS > Vth and VDS > ( VGS – Vth )

If I have this:

VCC --> LED --> resistor --> Drain --> Sorce --> VSS.

If VCC = 12V, Resistor = 470 ohms. And VGS is 10V.

So 10V > 4V and VDS (?) > 6.

In this case, how much is VDS?

bye
 

Audioguru

Joined Dec 20, 2007
11,248
VDS is almost 0V (a turned on switch) when the VGS is 10V.

EDIT: Don't even think about Vth since that is when the Mosfet is barely on and is almost turned off.
 
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