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H-bridge with TIP 122/127

Discussion in 'General Electronics Discussion' started by wingnut, Aug 10, 2012.

  1. wingnut

    wingnut

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    Aug 9, 2012
    Hi all

    I am trying to build an H-bridge using TIP 122 npn Darlington pair transistors and TIP 127 pnp's.

    The bridge got the 3V motor to turn one way but refused to reverse. The bases had 350 ohm resistors and I did not include 4 diodes since the data sheet shows that TIP's have their own protective diodes. My power source is a 6V battery pack.

    When it would not reverse, I tried each transistor separately with only the motor as load across the emitter/collector. The base was biased with a 2k to infinite ohm CdS photoresistor. The transistors turned the motor on individually but got hot in about 30 seconds of use. Is this normal? Surely if one is driving something as small as a 3V DC motor, the TIP should not overheat so?

    I copied the circuit found half way down the page at http://multyremotes.com/dc-motor-controller.htm

    Any suggestions on these two problems would be most welcomed.
     
  2. john monks

    john monks

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    Mar 9, 2012
    What are the voltages coming out of RL2 pins 1 and 4?
    Do you have the specifications of the motor?
     
    Last edited: Aug 10, 2012
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Let me start with problem no. 2: You do not control a motor by connecting it across emitter/collector of a transistor. You put the motor in series from collector to power supply.
    As I understand your setup the transistor shorted the motor and thus had to bear all the power. No wonder it got hot.

    Problem no. 1: I'm quite sure you either made an error building the H-bridge or you did not use the correct control signals. An H-bridge is a fairly simple and proven design. Check your connections. Also check your driving signals. They need to be complementary, meaning that if one driving signal is High (which is, I assume, 6V) the other driving signal needs to be low (0 V) and vice versa. If either both signals are high or low, the motor will not turn.

    Also a note on the schematic on the page you linked to: The use of darlington transistors in this configuration is a waste of energy.
    For example look at Q5: there are two base-emitter diodes in series from the base of Q5 to the emitter of Q5. This means a voltage difference of approx. 2*0.6V = 1.2V. If your base voltage is 6 V, this means that the emittter voltage is at least 1.2 V less than 6V -> 4.8 V .
    Make the same consideration for Q4 and you will find that the emitter voltage of Q4 is at least 1.2V.
    So the Motor will see only 4.8V-1.2V=3.6V. This may be suitable for your application using a 3V motor, but the voltage drop of 2.4V across the transistors (1.2V per transistor) will lead to waste power heating the transistors.

    A much better circuit is here: http://www.mcmanis.com/chuck/robotics/tutorial/h-bridge/bjt-circuit.html - you don't have to use the phototransistors.
     
  4. wingnut

    wingnut

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    Aug 9, 2012
    I "jumpered" my inputs straight from the battery + and - terminals but through 350 ohm resistors as shown on the schematic so I am sure the inputs are + and -. I am not using voltages from RL2 yet since I just want to get the H bridge to work.

    I have rebuilt the circuit twice, so I think everything is connected.

    Maybe the TIP needs more than 6V.

    When testing the TIP's I connected the TIP (c,e), battery and motor all in series with a separate bias to the base of the tip through a photoresistor. The DC motor itself has very little resistance (7 ohm when not turning to 500 ohm turning it by hand. I presume it gathers inductive resistance as it turns, but to start it is almost a short circuit). Is it wrong the way I tested it?

    Thanks for the help so far
     
  5. john monks

    john monks

    693
    1
    Mar 9, 2012
    I'm assuming that you have the motor connected to the emitter of your TIP.
    If that is the case you might connect a large, maybe a 1 Kohm resistor from the base to the collector just to see one transistor works.

    If your emitter is connected to the battery you might damage the TIP.

    How do you have it hooked up?
     
  6. wingnut

    wingnut

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    Aug 9, 2012
    I tried putting the motor on the collector side and then on the emitter side of TIP122, leaving the battery on the other side. In both cases motor turns on when light falls on photoresistor or if a 1k resistor applies a p voltage to base. In both cases TIP got hot in30s. If I left it longer, no doubt I could fry an egg on it. I did not think it makes a difference which side of the TIP one places the motor.
     
  7. Harald Kapp

    Harald Kapp Moderator Moderator

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    What current does the motor draw when operated in your test setup? Multiply that current by 1.2V and you get the power dissipated in one transistor.
    Do the transistors have heatsinks? From your explanation I'd guess they don't.

    Another thing that factors in is the resistance of the photoresistor. Is it low enough to fully turn on the transistors? If not, the base current may be too low to saturate the transistor. The transistor will operate in linear mode and consume even more power.

    The TIP need ~1.2 V Vbe, so 6 V can be used. But again: The H-bridge circuit from your reference may work but is far from optimal in terms of power dissipation. See my first answer.

    With reagrd to the motor resistance: The resistance is 7 Ohm as you have measured on the idle motor. When turning, the resistance doesn't change. What you observe is that the motor develops an internal counter electromagnetic force (essentially the rotating rotor acts as a generator). This counter EMF works opposite to the applied voltage, thus reducing the current. If you turn the rotor without applying an external voltage, you can measure this EMF as generated voltage at the motor's terminals.
     
  8. wingnut

    wingnut

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    Aug 9, 2012
    Thanks again for the feedback

    I measured the current through the TIP and motor as 0.8A. Multiply this by 1.2V drop you say occurs across the TIP and one gets 1W lost across the TIP, which I guess, explains why it gets so hot. As you said, there may be a more efficient non-Darlington transistor.

    The induced (reverse) V when turning the DC motor slowly by hand, was measured as 0.08V, which would rise as the motor speed increases.

    After many months of struggle I managed to build a solar tracker using a comparator and a NJM2670D2 dual H-bridge motor driver chip. It works well and does not overheat but is a bit pricey. I was hoping to use TIP's in place of the NJM. So far I have not had much luck with TIP's. But I guess the TIPs have had worse luck when, in the recent past I blew a few of these by forgetting to include base resistors.

    Thanks again.
     
  9. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    It's not a matter of the efficiency of the transistor, but of the circuit. look at the circuit I've linked. The emitters are connected to GND or VCC and the motor is connected to the collectors. Thus you have only the VcE voltage drop (0.1 V - 0.2 V) * motor current which in your case translates to 0.2V*0.8A=0.16W or 1/6 of the power dissipation of your circuit.
     
  10. wingnut

    wingnut

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    Aug 9, 2012
    Harald - I made the circuit you recommended using Tip122s and 127s and it works a treat. The motor flies both ways and the TIPs seems to stay reasonably cool. You made my day. :D

    I still don't understand why Vce is less (and efficiency much greater) when emitters are connected to the power supply rather than when collectors are attached to the power supply, and why the motor works better drawing current from collectors rather than emitters.
     
  11. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    It's a matter of Vbe, re-read my first answer. Come back if still unclear.
     
  12. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Wingnut, There are two basic categories of transistor usage. One is as an amplifier, where the output current is an amplified image of the input (Base-Emitter) current. In this mode (Linear) the transistor is never completely turned on or off.

    The other use category is called "Switching". In this mode the transistor (Vce) is ideally completely "ON" or completely "OFF". The idea being that the transistor should behave as closely to a "Switch" as possible . To do this the transistor must be "Saturated" (Google "Transistor Saturation"). Note: Emitter Followers can't be "Saturated".

    See the attached schematic and plots of DC Characteristic Curves for and "Common Collector" (Emitter Follower) vs a "Common Emitter" using the same transistor. You will clearly see why Common Emitter configuration is usually preferred for switching applications.
     

    Attached Files:

    Last edited: Aug 11, 2012
  13. wingnut

    wingnut

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    Aug 9, 2012
    Thanks for the graphs which illustrate how the same transistor can act as a switch or an amplifier. I must say that it is not intuitive how this works (meaning very difficult to understand) but having seen the result, I am trying to catch up on the theory.
     
  14. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    May 8, 2012
    Actually those plots don't demonstrate the same transistor being used as a switch or and amplifier. They're plots of base current and voltage vs output current and voltage. On the other hand the plots do demonstrate how the common emitter collector current suddenly snaps on as compared to a common collector.

    I'm heading out to the watering hole. When I get back I'll post CE & CC as a switch. This will be a Transit Plot as compared to DC "Transfer" Characteristics.
     
    Last edited: Aug 12, 2012
  15. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    These are plots of Common Emitter and Common Collector amplifiers being used as switches. Note that I intentionally made the source voltage (VS1) only 1.0V because it will give you a much better idea of how these two configurations will respond to a small control voltage.

    This is called a "transient plot" that starts at time zero and where VS1 switches from 0V to 1.0V one second later. Study this as common emitter - common collector circuits are the building blocks of all BJT designs. Take particular note that a common collector provides NO voltage gain but does provide current gain. The common emitter provides both.
     

    Attached Files:

  16. wingnut

    wingnut

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    Aug 9, 2012
    Thanks for taking the trouble to send these plots. I certainly hope to come to grips with this soon.

    I downloaded LTSPICE and spent time on a site called circuitlab.com which allows one to design common emiter and common collector circuits and then see changes in V etc. I am from the school of fiddle with components till the circuit works (kinda), but hope to join those of you who are not surprised when circuits work first time.
     
  17. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

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    Electronics Point has a tool bar. Scroll up to the top of the page and click on the "Tutorials" button and start reading. It's a structured electronics course.
     
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