I'm a relative newbie, and I'm trying to understand how this H bridge works. ( link below) I want to build it for a microelectronics project. With the current "push pull" configuration of the amplifiers, I'm wondering how the motor is supposed to receive any current. It looks as though the 12 V supply will simply short to ground... but my understanding of transistors is somewhat limited, so I'd appreciate any help at all with figuring this out. Thank You.
I'm not competent to comment on the internals of one of those things, but I'm curious as to why you want to build one?
Unless the assembly is in and of itself a useful learning experience for you, I'd be soooooo inclined to buy one off the shelf. There are a zillion h-bridge-on-a-chip solutions, catering for different voltage and current requirements.
(Not that I want to discourage you from the experience of building one, but the rest of the solar project looks far more interesting than getting held up soldering transistors together.)
It looks a fairly simple arrangement. T1 is an NPN and T2 is a PNP. These have a shared input to thier base. If this signal is HIGH, the NPN will conduct. If the signal is LOW then the PNP will conduct. At any time one of these will be turned on.
So the left hand side of the motor is either connected to the +ve rail of the GND (depending which of the two is conducting).
T3 and T4 use the same arrangement to supply the other side of the motor.
LDR1 and LDR2 form a voltage dividier. The output of this is compared against a separate voltage divider (created by P1-R3-P2-R4).
The voltage output from that second divider will remain stable, but the voltage at the junction of the two LDRs will vary, dependent on which of the LDRs is receiving the greater amount of light.
When the LDR divider outputs a different voltage to the fixed divider, the two opamps will slew their output. Since there's no feedback from their output to their input, they will slew to either the positive rail value or ground.
The +ve and -ve inputs to the op amps are not marked but it seems obvious that they are connected back to front (with reference to each other) so when one op amp is putting out the +ve rail, the other is putting out the gnd.
I'd imagine there is some problem getting this to be stable though. Because at the point where both LDRS are ballanced, the opamps will also recieve a ballanced input. This would leave them in their current state (because that's what opamps do). So this would lead to the motors overshooting their target. until the difference is enough to swing the opamps back. (and therefore overshooting the other way).
So I think you'll end up with some oscillation that can be adjusted (but not completely removed) by those potentiometers. Seems wasteful on energy to me.
I want to build it
Lots of failed DIY H-bridge projects on the forum. Probably best to just buy one that works.
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The result looks like Link
(cool circuit but the mechanics look challenging)
First get your head round what you are trying to do with a H-bridge. You can do it with switch relays like half way down this page:- http://www.thebox.myzen.co.uk/Workshop/Motors_1.html
Then you take that principle and use transistors to do the switching, in the form of a chip:- http://www.thebox.myzen.co.uk/Workshop/Motors_2.html
There is little point making one from transistors as there is a lot of design to it. Most "simple" circuits don't work and end up shorting the power rails briefly as the change over occurs, this is known a shoot through.
chenchiller: I'm a relative newbie, and I'm trying to understand how this H bridge works. ( link below) I want to build it for a microelectronics project. With the current "push pull" configuration of the amplifiers, I'm wondering how the motor is supposed to receive any current. It looks as though the 12 V supply will simply short to ground... but my understanding of transistors is somewhat limited, so I'd appreciate any help at all with figuring this out. Thank You.
That particular circuit is using analog class-B driver stages to approximate an H-bridge switch setup. Approximates as the transistors will not act as good switches and waste a lot of power (and thus need more heatsinking that if done properly). Transistors used as switches are driven in common-emitter mode, not emitter-follower (class B is two complementary emitter followers back-to-back, basically).
Google "emitter follower", "class B", "common emitter" and "transistor saturation" for more introduction to these circuits.
The advantage of the class-B stage is that it is free from shoot-through as the high-side and low-side devices cannot be driven at once.
Note that any modern H-bridge will use MOSFETs driven by MOSFET driver chips which encapsulate much protective circuitry - if you want to see this sort of design look at the datasheet for the HIP4081 perhaps http://www.intersil.com/data/fn/fn3659.pdf - in particular the "typical application" circuit on page 3. High power circuitry is mainly about protection from fault conditions, since otherwise MOSFETs explode (literally).
That looks way better than the L293 !