5
\$\begingroup\$

I have an old RC car but without the transmitter so I built my own Rx/Tx using MCU and 2.4Ghz transceiver, however, I want to use the car's original H-Bridge but not sure if I understood how that circuit works.

Here is the schematic for the bridge: enter image description here

My understanding is that P1 and P2 are for forward direction and provide different speeds!?, while P3 is the reverse direction. Correct me if I am wrong.

My questions are:
1- Can I use logic level to drive P1-3 (directly from MCU pin) ?
2- When setting one of them high should the other two be low? Can P1 and P2 be high at the same time?
3- What is the purpose of R1?, Why only 1Ω?, Why 10W? and Why it is only connected to Q1 and Q2 and not T9 (since all three are PNP high side transistors)
4- Why does it have a diode D4 for only one of the inputs (P1)?
5- What is the purpose of D1, D2 and D3?
6- How does T1 drive T5 and Q2? Since its collector and emitter are connected to the base of the other transistors! no ground or vcc!? I am confused.
7- Can I use PWM to control the speed of this bridge? (default frequency for Atmel Atmega328 is 32K)

(Note: C1, C2 and C3 are actually polarized electrolytic capacitors)

----- Updates ------

Here is the modified design after fixing the swapped transistors pins, removing the slow forward circuit and adding flyback diodes:

enter image description here

So how does it look? anything that need to be changed? removed? added?

\$\endgroup\$
  • \$\begingroup\$ Something is seriously messed up with that schematic. Did you get this from some documentation with the unit, or did you guess it and then enter it yourself? I'm figuring the latter. Some of the NPN transistors are used in ways that just don't make sense. \$\endgroup\$ – Olin Lathrop Sep 23 '12 at 23:17
  • 1
    \$\begingroup\$ You have the collector and emitter swapped on the S9013 transistors (T1, T2, T3). If you redraw the schematic with this fixed, it'll make a lot more sense. \$\endgroup\$ – Dave Tweed Sep 23 '12 at 23:18
  • \$\begingroup\$ @Olin Lathrop, I made this based on the actual board, tracing the lines and matching the pins with the datasheets for the transistors. I even checked it twice, but going to check it again. \$\endgroup\$ – Fahad Alduraibi Sep 24 '12 at 12:30
  • \$\begingroup\$ @Dave Tweed, You could be right, the drawing in the datasheet is confusing, I am not sure if what they have is top or bottom view or the NPN, here is the datasheet could you please verify, I assumed it is top view, if it is bottom then they are swapped: link \$\endgroup\$ – Fahad Alduraibi Sep 24 '12 at 13:18
  • 1
    \$\begingroup\$ @FAD: The new drawing looks good; however, there's really no need to keep R1 (the power resistor) or D4. The original functions of both of them are now going to be handled by your PWM firmware. \$\endgroup\$ – Dave Tweed Sep 25 '12 at 20:18
3
\$\begingroup\$
  1. Yes.
  2. a) Yes. b) Possibly; see answer #4.
  3. R1 limits the current for low-speed forward and reverse operation. T9 provides high-speed forward operation.
  4. It raises the switching threshold for P1 by one diode drop. My guess would be that if you drive P1 and P2 high simultaneously (from a dead stop), this allows the low-speed circuit to get the motor started before the high-speed circuit kicks in, providing smoother acceleration.
  5. The resistor-capacitor-diode combination provides a soft-start, fast-stop characteristic for the motor. When starting, the resistor and capacitor have a time constant of about 0.224 seconds, which helps prevent the motor from abruptly spinning the wheels. When stopping, the diode dischages the capacitor immediately for faster response.
  6. See my comment on your question. When hooked up correctly, T1 feeds base current from the high-side PNP (Q2) directly to the corresponding low-side NPN (T5).
  7. Only if you remove the three capacitors. You'll also need to add freewheeling diodes across each of the five power transistors.
\$\endgroup\$
  • \$\begingroup\$ Thanks for these answers, I will update the schematics and post it again here. BTW, since I am planing to use PWM, I think there is no need for the components that connects to P2 including Q1, so I should I remove them from the circuit or just leave it unconnected? \$\endgroup\$ – Fahad Alduraibi Sep 24 '12 at 21:12
  • \$\begingroup\$ The resistor-diode-cap arrangement are actually providing a hardware deadtime by delaying the rising edge but not the falling edge. Deadtime prevents both top and bottom transistors from being on at the same time. This is useful when using a single PWM output and its direct inverse to control the bridge. \$\endgroup\$ – BullBoyShoes Sep 25 '12 at 11:41
  • \$\begingroup\$ @BullBoyShoes: Do you understand that a time constant of 0.224 sec. is a ridiculous value for dead-time control? Besides, you don't do PWM the way you describe it in this sort of motor driver; you PWM each direction individually. \$\endgroup\$ – Dave Tweed Sep 25 '12 at 12:03
2
\$\begingroup\$

This design looks like it may have been mis-drawn. If you tried to draw it from inspecting the device itself, make sure your transistor pin-outs are correct. If it is drawn correctly, it must make use of some strange base/collector properties of the transistors, because it appears that the base of some of those transistors are used as current sources for other ones, which makes no sense.

Q1, Q2, T5, and T8 make up the H-bridge, and it looks like that input P1 and P2 drives the top-left and bottom-right quadrant of the H-bridge, and P3 drives the bottom-left and top-right quadrant. So, yes, P1 and P2 would be for "forward" and P3 for "reverse"

To answer your questions:

  1. Probably not -- you'll need 12V logic, and those RC circuits attached to the inputs are going to stress the microcontroller out a lot.
  2. Yes, but you'll probably get weird things happen
  3. Reduces the current through the H-bridge for low-speed operation
  4. It reduces the base current entering T2
  5. The 6.8k/33uF RC circuits cause the input to the transistor to change slowly, instead of instantly. By adding the diodes across the resistors, they're instant-off when the inputs go to 0 (since the capacitor discharges immediately through the diode). This is as long as the inputs are pulled to ground with a strong transistor. A microcontroller pin can only source/sink 20 or 30 mA, so that's going to defeat the purpose of the diodes.

  6. I'm also confused. They might be exploiting some strange property of the device.

  7. Not with that giant RC circuit attached to the inputs.

Other comments

This is a very inefficient circuit, and considering how cheap and easy monolithic H-bridges are to use, I'd recommend just going that route, especially since they have TTL-level inputs.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.