Short-protected H-bridge

[iNfraNe]

Hello,

For my project I'm in need of a H-bridge design to run a 12V motor, controlled by 5V PIC outputs.The circuit schematic is displayed below. What I want to know is if the circuit is really protected from shorts this way. I made the circuit in a simulator, but before fry some transistors I would like to know if I did this the correct way!

View attachment 65474

 

[colin55]

The only problem with the circuit is the buffers are emitter-followers and don't provide any current-gain for the output transistors.

 

[colin55]

To get any current out of the circuit you will need to use the following resistor values.
View attachment 65475

 

[MOR_AL]

00 - All transistors are OFF
11 - Up transistors are ON and down (inside) transistors are OFF.
01 - Current on load in one direction.
10 – Current on load in other direction.
A good sequence to change load current direction is 01 - 00 - 10, or 10 - 00- 01
If your load has inductive component (motor) you may include four diodes in reverse position. From gnd to load, and from load to +12V. Include also two capacitors one 100nF and one electrolytic.
MOR_AL

 

[iNfraNe]

Thank you for your replies.

Colin: So, this means I cannot use my PIC directly to drive the transistors, correct? How can I solve this? Seems like a lot of current is wasted not going trough the center in this way (from the A/B points trough 1 diode (base->emitter), a 47ohm resistor and another diode).

Mor_Al: Yes I figured that. But changing the direction like that doesnt let the motor brake. The diodes/caps will be added, but are right now not essential for my testing.

Maybe I should just buy a DC motor driver, if there is no simple solution like this

 

[ericgibbs]

hi iN,
The circuit is unsuitable, especially with those suggested very low resistor values.

Where did the design come form.??

E.

 

[iNfraNe]

I designed it myself. In the circuit simulator java app (https://www.falstad.com/circuit/) it works like a charm. How would I need to change it to function properly?

 

[colin55]

Here are the values you should use:
**broken link removed**

 

[dougy83]

What about changing it slightly so we're just disabling the low-current control part instead of the higher-current bits? e.g. attached cct. Resistors sized according to collin55's calc'd values

 

[colin55]

How do you get BRAKE out of the circuit above?

 

[MOR_AL]

iNfraNe.
Please, specify the value of maximum (peak) current on your motor. 0,1Apk, 1Apk, 5Apk, ...?
Your circuit may have a minimum current gain of 150. 50 from external transistors and 3 from power transistors.
MOR_AL

 

[MrAl]

Hello,


Really though a circuit like these may not work very well in practice because there is no dead time designed into the circuit. Good quality circuits like this have dead time built in because a transistor does not turn 'off' immediately but takes some time to do so. Meanwhile, the other transistor on that same side of the bridge is turning on which means a current spike through both transistors occurs which can take out the two transistors or heat them up too much. The time it takes a transistor to fully turn off depends on the transistor and if it was in saturation just before turn off or not. It can take significant time to turn off so a little dead time is usually built into the circuits. If the power supply is not low impedance it may pull down thus protecting the transistors but then it could interfere with other parts of the circuit. It really depends on how sensitive the other parts of the circuit are.

The dead time is usually generated logically not with the bridge itself. This usually involves a couple of logic chips but could be done other ways too.

If you look at switching power supply controller chips you'll see a dead time spec which you can sometimes change by design.

Transistors that are not saturated turn off much faster. Look up 'storage time'.

 

[dougy83]

Really though a circuit like these may not work very well in practice because there is no dead time designed into the circuit.

There would only be shoot-through when changing direction instantaneously. Just follow what MOR_AL said above regarding this and there's no problem.

MOR_AL >> A good sequence to change load current direction is 01 - 00 - 10, or 10 - 00- 01

 

[MrAl]

Hi,

Oh ok when i looked at the circuits it didnt look like there was a way to turn off all four transistors at the same time. My 'bad' if there is
And if so, then yes turning them all off at the same time and keeping them off for a short while would be good enough, you're right and i agree fully.

 

[iNfraNe]

Hello guys,

Thanks again for the input.

My motor isnt a very demanding one. It draws 150 mA at 12V (the website from which I bought the device www.voti.nl stated so). The "datasheet" can be found here: https://www.shokaifareast.com/Products/Pots/motorized.htm

doughy83: Your circuit works fine too in my simulator. I am wondering about the resistor values tho. Why should they be this low (the 47R)? In my simulator 1k works perfectly fine there. Else there's too much wasted current for my 150 mA motor.
I'd like to have almost no wasted current to be able to just use normal transistors. Is this possible?

Is there a cheap IC that will do this job just as nice?

 

[colin55]

A 150mA motor takes 500mA to start and when heavily loaded. The driver transistor has a gain of 20 under these conditions. You need a base current of 20 - 30Ma The base resistor can be changed.

 

[dougy83]

doughy83: Your circuit works fine too in my simulator. I am wondering about the resistor values tho. Why should they be this low (the 47R)? In my simulator 1k works perfectly fine there. Else there's too much wasted current for my 150 mA motor.

I just used the ones collin55 used. They can can be increased somewhat.

I'd like to have almost no wasted current to be able to just use normal transistors. Is this possible?

I notice the motor you linked to is 3-6 VDC, not the 12V assumed in some of the posts above. You can use a simple emitter-follower configuration (scribble attached). This is as "short-protected" as any of the above diagrams.

You should also put a snubber of some sort across the motor terminals to protect the transistors

Is there a cheap IC that will do this job just as nice?

The BEAM robotics guys like using CMOS logic chips as motor drivers, e.g. **broken link removed** the 74AC240 & 74AC244 could be used in the same manner. Just parallel 4 drivers for each side terminal of the motor. Note these ICs are not suitable for > 6V operation.

 

[iNfraNe]

The motor is indeed 3-6V. However, voti.nl stated that the voltage can be increased to as high as 12 V without any problems. I'd like this as the motor is incredibly slow at 5 V.

The increased simplicity is of course appealing for a 5V setup... Heck, I'll just go for that.

Thanks again.

Bart

 

[dougy83]

I'd like this as the motor is incredibly slow at 5 V.

The emitter-follower drops the voltage to less than 5V - so it will be even slower than "incredibly slow". Make sure you test that it's fast enough before committing to a PCB or anything...

 

[MOR_AL]

Try this circuit.
View attachment 65511
View attachment 65512
MOR_AL