Hello,
In the image below (left), there is a push-pull style transistor arrangement being used to control a motor. Contrast that to the h-bridge on the right. I realize that these are serving different purposes, the motor driver is sourcing and sinking current to run the motor in a single direction whereas the h-bridge is doing something similar but allows bi-directional rotation. What I don't quite understand is why the transistors in the motor driver case (left) are connected via their emitters, whereas the h-bridge (right) are connected their collectors. I wired up an h-bridge with the emitters connected instead of the collectors (push-pull style) and it seemed to work fine. However, all of the h-bridge diagrams I have found show the transistors connected via their collectors. Can someone please explain why these two cases should be connected differently? Thanks.
For a bipolar transistor to conduct at all, the base voltage must be about 0.7V higher than the emitter voltage.
The circuit shown on the left won't work with an Arduino output. The "direction control" (base) voltage has to be higher than 6.7 V for the upper transistor to be fully on.
Hi,
That's true for this specific example (I revised the diagram to eliminate the specific values). My burning question is why the motor driver uses a standard push-pull configuration vs. why the h-bridge doesn't. My hunch is that in an h-bridge, the PNP acts as a high side switch whereas the NPN acts as a low-side switch and that's pretty typical. So I could turn the question around. For the motor driver on the left, why didn't they instead use PNP high side and NPN low side (as in the h-bridge) rather than the standard push-pull amplifier configuration?
Common collector ("emitter follower") configuration as in the left diagram, have a lower output impedance due to the emitter feedback from the load. Also it has a voltage gain of almost unity, hence the "follower" part. That makes it a favourite for output stages, but it suffers from the junction drop issue mentioned above. The maximum output voltage is limited to about Vcc-0.7V, even when the transistor is saturated. Common emitter configuration doesn't have that problem, it can go almost to Vcc depending on base drive and collector load.
It's kind of moot, now that MOSFETs are available.
Often, a "why did they do that" question is pointless, as a circuit behaves according to its design, not what was in the mind of the designer on design day. We only hope that the two are closely similar and that the designer was competent. Here there is no context, either... it's just an orphaned diagram. It's better to just analyze and talk about what you see. The device in the diagram on the left would almost never appear in a real world product. It looks like a sketch from an engineering textbook that was intended to only show some basic aspect of operation.
Don't overthink those diagrams, if you want to know about drivers and bridges, keep looking.
Here is another reference to a "push-pull" motor driver. It does use a double ended supply to allow the motor to run in either direction. I hope this provides further context for my question.
That circuit uses analog control, not PWM. It has nothing to do with your question, except that it shows how a common collector output stage easily amplifies the output current of an op amp output. Easily, as I said before, because it is a "voltage follower", or unity voltage gain current amplifier.
My burning question is why the motor driver uses a standard push-pull configuration vs. why the h-bridge doesn't.
Both circuits are motor drivers, but in the circuit on the left, the PNP transistor acts only as a brake.
Either arm of an H-bridge is a "push-pull" configuration.
Maybe a better way to frame my question is this. "This works, but why shouldn't I make an h-bridge this way?" (push-pull amplifier style)
I already mostly answered that question. The output voltage can't cover the entire supply range, as it almost can with a common emitter configuration. That means, lower voltage to the load and higher losses (hence heat) in the transistors.
"I already mostly answered that question."
Yes you did, and I thank you for that. As you can tell, I can be a bit stubborn at times.
So for the common collector case, the "good" is that the output impedance is low, but the BE "diode drop" prevents it from going rail to rail making it less efficient. What makes the common emitter "better" is that it can deliver higher output current and is more efficient due to it being "rail to rail".
What confused me is that a push-pull (class B) amplifier appears to be so commonly used that I wouldn't have thought it "bad" for this purpose (to the point where I did find a non h-bridge example of it), although I suppose it really depends on the application. For example, it may be "better" for certain multi-stage audio amplification applications than it is for driving loads such as motors for the reasons mentioned. Does that seem like a reasonable way to summarize it?
Firstly the circuit on the left will only drive the motor in one direction. It will never be reversed.
With regards to the common collector the BE "diode drop" has nothing to do with it. A normal transistor being used as a switch will have a BE voltage drop of around 0.6 to 0.7V when saturated. The EC voltage drop will be about 0.2V. So in this circuit you will always be 0.4V at least less than rail to rail.
Also these transistors are being used as a switch and a class B amp is not. They are being used in their active zone.
AxiomGreg:
Maybe a better way to frame my question is this. "This works, but why shouldn't I make an h-bridge this way?" (push-pull amplifier style)
Because the modern designs are literally 100's of times more efficient (don't need big heatsinks, or even
any heatsink).
Firstly MOSFETs are far more efficient, secondly using BJTs as emitter followers will force several
volts loss in the transistors, rather than a fraction of a volt with BJT's used in common-emitter
(switching configuration). With MOSFETs you can reduce the voltage loss across the switching
devices to millivolts and use higher PWM rates.
windoze_killa:
With regards to the common collector the BE "diode drop" has nothing to do with it. A normal transistor being used as a switch will have a BE voltage drop of around 0.6 to 0.7V when saturated. The EC voltage drop will be about 0.2V. So in this circuit you will always be 0.4V at least less than rail to rail.
Under load, A transistor that is configured as common collector and passing the same current as a common emitter configuration, is also saturated and the EC voltage drop will be more than 0.6 to 0.7V. So, in fact, that disadvantage is real.
The EC voltage drop
Read as "the VCE drop" ( I believe that's what he meant)
raschemmel:
Read as "the VCE drop" ( I believe that's what he meant)
Correct.
It doesn't matter if they are common collector or common emitter a saturated transistor will have the same Vce, generally 0.2V.
windoze_killa:
Correct.It doesn't matter if they are common collector or common emitter a saturated transistor will have the same Vce, generally 0.2V.
Common collector can't saturate unless there's a separate extra supply rail to drive the base
above the collector voltage. We say "emitter follower" usually, "common collector" is not normally
used in practice.
MarkT:
Common collector can't saturate unless there's a separate extra supply rail to drive the base
above the collector voltage. We say "emitter follower" usually, "common collector" is not normally
used in practice.
WHAT????
To saturate a transistor you just have to increase the voltage on the base above the Emitter.
windoze_killa:
I addition I think this is what the OP meant by common emitter and common collector.
OP only used the term, "common collector" once, and didn't reference any particular diagram. I hope they didn't mean that, because of course, it's backwards.
windoze_killa:
WHAT????To saturate a transistor you just have to increase the voltage on the base above the Emitter.
No,
Saturation region is one in which both Emitter Base and Base Collector junctions of the transistor are forward biased. In this region high currents flows through the transistor, as both junctions of the transistor are forward biased and bulk resistance offered is very much less.Transistor in saturation region is considered as on state in digital logic.
A transistor is said to be in saturation if and only if
β > Ic/Ib
This is due to the fact that as both junctions of transistor are forward biased along with electron current flowing from emitter to base in active region there will be additional component of electron current flowing from collector to base.Small changes in Collector to base forward voltage leads to large variations in collector currents(variations in currents will be exponential as mentioned before through diode current equation).
https://ecetutorials.com/analog-electronics/operation-of-bjt/