Decreasing voltage being delivered to motor with H-bridge over time.
I am using an H-bridge to control current from a 12-V car battery to a drill motor. The H-bridge design is similar to the attached schematic. I have successfully gotten the motor to go in forward and reverse, but if I run the motor for more than about 5 seconds it begins to go slower and slower until it finally stops. Other evidence is the PNP transistors (TIP42) get very hot.
Here an example of my code for the attached circuit.
DO
HIGH 11
LOW 12
PAUSE 5000
LOW 11
LOW 12
PAUSE 5000
HIGH 12
LOW 11
PAUSE 5000
LOW 11
LOW 12
PAUSE 5000
LOOP
Any explanations or solutions to why my motor isn't running at full power?
Here an example of my code for the attached circuit.
DO
HIGH 11
LOW 12
PAUSE 5000
LOW 11
LOW 12
PAUSE 5000
HIGH 12
LOW 11
PAUSE 5000
LOW 11
LOW 12
PAUSE 5000
LOOP
Any explanations or solutions to why my motor isn't running at full power?
Comments
BJT's like the TIP42 are incredibly lossy. As they will heat up, you do have a heatsink on each one, right?
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- Stephen
Any recommendations on better transistor choices?
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.
Also I heard that car batteries are meant to supply a very large current for a short amount of time because their main purpose is turning over the engine. Could this be the root of my problem?
5 seconds, letting the motor coast for 5 seconds and then commanding full
reverse motion for 5 seconds.
If the motor isn't driving a load, it probably won't slow down much during
those coast times. That means that the back emf it is generating will be added
to the battery voltage when the reverse direction is selected.
I'm guessing that the back emf will be somewhere around 9v. Adding
this to the 10v battery voltage, the H-bridge will see a 19v supply
at the moment of reversal.
With only 1.1 ohms resistance, you are asking the H-bridge to deliver 17 amps
until the motor actually changes direction and comes up to speed.
The 2n6107 transistors are only rated for 7 amps continuous and 10 amps peak .
You need larger transistors capable of higher current plus good heat sinking.
You could also try increasing the 5 second delay to whatever is needed for
the motor to completely stop on its own while coasting. That may take 30
seconds with no load. This will put much less stress on the H-bridge when
the reverse direction is commanded.
phil
Post Edited (phil kenny) : 3/14/2008 6:28:47 PM GMT
The supply current is not your problem.... the required current from the motor is about 7 Amps. ((9V - (1.2Vdiode drop across 2-transistors) / 1.1Ohms = 7Amps)
You loose 8.5 Watts of heat across the PNP and NPN transistors (P = I * V) = 7 Amps * 1.2V = 8.5Watts.
The DC current gain or hfe is really low for that much current... less than 3! so if you figure in the 49.9 Ohm resistors, the drive current is only about 433mA !!
((9V - (1.8Vdiode drop across 3-transistors)) / 49.9) * 3(<-Hfe) = 433mA ... Simply put, your transistor does not saturate (turn “ON”[noparse];)[/noparse] enough to adequately drive the motor that you have selected.
My suggestion would be to find a transistor (both NPN and PNP) that has a better DC gain characteristic, or better yet switch your design over to using MOSFET transistors.
If you stay with Bipolar transistors, you are still going to need to dissipate 8.5Watts of heat even if they are completely saturated due to the diode drop across the transistor.
A MOSFET version would still dissipate heat, but at a fraction of the BiPolar counter part.
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Beau Schwabe
IC Layout Engineer
Parallax, Inc.