Discrete H-Bridge very low current output

Question

I have built two parallel H-bridges from "Design 1" on this page as you can see in the attached schematic.

However when I apply the external power (VCC1 in the center of the diagram) the two motors do not spin.

Using my multimeter, I can see that the entire circuit is working: the motor is getting a 'forward' and 'backward' current as the Arduino sets the pins high. The multimeter is reading +/- 5.54V with the motor replaced by the multimeter. (Across the motor this drops to 0.13V)

So then I measure the current by putting my multimeter both in series with the motors and replacing the motors. There's only 40mA available! Looking around the web I found the suggestion to reduce the 220Ω resistors at R2, R3, R5 and R6. So I switched in (randomly) 27Ω resistors and the output current is still only 40mA.

What else should I try to bring up the current enough to drive the motors?

Edit: So my uneducated guess is that it's likely that the limitation is coming from the transistors. The Ns are BC549 and the Ps are BC559. The datasheets list "low current" as a feature, but I figured that was the collector.

Edit 2: The motor is this one which requires 70mA when running (and presumably more during startup)

Schematic of two H-bridges controlled by Arduino

Edit 3: Code driving the Arduino:

// Pin setup
int pinLeftMotorFwd = 5;
int pinLeftMotorBck = 6;
int pinRightMotorFwd = 10;
int pinRightMotorBck = 11;
// Directions for the {left, right} motors to achieve the four directions
int FORWARD[] = { 1, 1 };
int BACKWARD[] = { -1, -1 };
int RIGHT[] = { -1, 1 };
int LEFT[] = { 1, -1 };
void setup() {
 Serial.begin(9600);
 pinMode(pinLeftMotorFwd, OUTPUT);
 pinMode(pinLeftMotorBck, OUTPUT);
 pinMode(pinRightMotorFwd, OUTPUT);
 pinMode(pinRightMotorBck, OUTPUT);
}
void loop() {
 go( FORWARD, 10 );
 go( RIGHT, 3 );
 go( FORWARD, 10 );
 go( LEFT, 6 ); 
}
// Move the bot in a particular direction from the array above for a given number of seconds
void go ( int dir[], int dur ){
 leftMotor( dir[0] );
 rightMotor( dir[1] ); 
 delay( dur * 1000 );
}
// Sill both motors (OK, so stop both motors. But stop is a keyword in C)
void still(){
 leftMotor( 0 );
 rightMotor( 0 );
}
// Set the left motor to Backward (-1), Forward (1), or Stop (0)
void leftMotor( int dir ){
 // Pull both pins low
 digitalWrite(pinLeftMotorFwd, LOW );
 Serial.println("pinLeftMotorFwd LOW");
 digitalWrite(pinLeftMotorBck, LOW );
 Serial.println("pinLeftMotorBck LOW");
 // If we're going forward, pull the Fwd pin high
 if( dir == 1 ){
 digitalWrite(pinLeftMotorFwd, HIGH);
 Serial.println("pinLeftMotorFwd HIGH");
 }
 // otherwise, pull the Bck pin high
 else if( dir == -1 ) {
 digitalWrite(pinLeftMotorBck, HIGH);
 Serial.println("pinLeftMotorBck HIGH");
 }
} 
// Set the right motor to Backward (-1), Forward (1), or Stop (0)
void rightMotor( int dir ){
 // Pull both pins low
 digitalWrite(pinRightMotorFwd, LOW );
 Serial.println("pinRightMotorFwd LOW");
 digitalWrite(pinRightMotorBck, LOW );
 Serial.println("pinRightMotorBck LOW");
 // If we're going forward, pull the Fwd pin high
 if( dir == 1 ){
 digitalWrite(pinRightMotorFwd, HIGH);
 Serial.println("pinRightMotorFwd HIGH");
 }
 // otherwise, pull the Bck pin high
 else if( dir == -1 ) {
 digitalWrite(pinRightMotorBck, HIGH);
 Serial.println("pinRightMotorBck HIGH");
 }
} 

• Ignacio Vazquez-Abrams: Design 1 would not have been my first choice.
• Gerben: What is the VCC1's power source. Is it a battery? By the way, 220 Ohm is pretty low value. That will result in 22mA of current (at 5v) going into the base. With a gain of 270 that would result in a collector current of around 6A. I'd try using increasing the resistors to 1 kOhm to 5 kOhm.
• RickMeasham: @Gerben: Ahh .. I see I may have messed up. I reduced the resistance. I'll try increasing it before I switch to a different circuit per Mark Williams below

Answer 1

Design 1 wouldn't be my first thought on how to do this either! The transistors aren't a great choice, but are cheap & common.

The layout here is more conventional, and you can physically link 1-2, & 3-4 to connect to your arduino without the risk of short circuits.

I suspect your code may be setting both lines high together - the only advantage I can see to the posted circuit is that this does not necessarily cause smoke (but may well do as you describe). This is why posting the code is recommended!

• RickMeasham: Thanks for the suggestion. The transistors were on-hand. Ultimately once I understand the concepts from "first principles" / discrete components, then I'd just replace the whole lot with either a motor driver circuit or a dedicated H-bridge. The table at the bottom of the circuit you suggest indicates I need to switch pins between VCC/GND/Disconnected. How can linking 1-2 attached to a pin and 3-4 attached to another pin give me full control? (Code isn't setting both high, but I'll attach it anyway)
• Mark Williams: The point of the H-Bridge is that the transistors work in pairs - so you would have 1&2 or 3&4 on together, but never 1&3 or 2&4 - if you go back to 'Design 1' you will observe that a single input has been put through an inverter stage to ensure this doesn't happen - your re-design replaces that with software - OK 'til a bug sets up a short.. So with the standard arrangement on that link, the truth table at the bottom (disregard the not-connected options) shows you the combinations. You have off/brake/forward/backward and possibly PWM as well..
• RickMeasham: No diodes at home, so I built it on circuit.io: 123d.circuits.io/circuits/511807-h-bridge where I had the exact same problem as my original circuit. Until I realised the NPNs were in backwards. Then everything came good. Going to check my original circuit tomorrow and see if I did the same there before I rebuild.
• RickMeasham: Turns out my actual problem was having the transistors around the wrong way. I couldn't get my circuit to work so I built yours and got exactly the same thing. The above circuits.io result gave me a clue so I played with them and got them around the right way. Turns out that the side that appears flat in Fritzing is the round side and vice-versa.