Driver Circuit Components

[BlackMelon]

Hello,

I have just bought a dc motor driver board. A load (motor) is connected to M+ and M-. The jumper J is a power source selector. connecting the two pins of J together means that you want to use 12V source while leaving the pins unconnected means 24V source.

Curiously, I want to know the function of components highlighted in red. I was trying to google this phrase: "Common base BJT and a motor driver" but found nothing related. Could you please clarify its mystery?

Thank you
BlackMelon

 

[Harald Kapp]

This is a simple linear voltage regulator for ~9.4V output voltage (V(ZD1)-Vbe(TR1)).
The jumper is set for 12V to give sufficient base drive current to TR1. At 24V input voltage the current through R11 is suffiient so the jumper is not set in this case.

 

[BlackMelon]

Oh! Thank you very much!

PS. Nowadays, there are many ICs available for this purpose, even with the full H-bridges. Could you suggest the most efficient one for a full-Hbridge DC motor driver?

 

[BlackMelon]

Oh! I have some more questions. It's clear that C3 and R11 in the circuit , together, can act like a low-pass filter. However, why doesn't the filter C4 require a resistor to form a low-pass filter?

Also, will I be OK if I remove the diode D1? (since the current can flow in just one direction from 12V to the collector of TR1)

 

[Harald Kapp]

Nowadays, there are many ICs available for this purpose

Which purpose? Regulating the voltage or driving the motor?

• For voltage regulation the most efficient circuit will use a switch mode regulator.
• For driving the motor several ICs are availabe. Which one is the most efficient depends on your definition of efficiency in terms of:
  - cost
  - PCB real estate
  - power loss
  - ...

It's clear that C3 and R11 in the circuit , together, can act like a low-pass filter. However, why doesn't the filter C4 require a resistor to form a low-pass filter?

These components serve different purposes.:

• C3/R11 indeed forma a low pass filter to keep noise away from the base of TR1 which in turn serves to create a clean output voltage at the emitter of TR1.
• C4 is a buffer capacitor which needs to be able to be charged quickly and to give of charge quickly in case of peak current demand. This capacitor is known as smoothing capacitor.

 

[Colin Mitchell]

This is a simple linear voltage regulator for -9.4V output voltage (V(ZD1)-Vbe(TR1)).
should be
This is a simple linear voltage regulator for +9.4V output voltage (V(ZD1)-Vbe(TR1)).

C3/R11 indeed form a a low pass filter
The more-accurate definition is:
R11 provides base current for the transistor and the transistor can deliver about 100 times the base current to the circuit connected to the collector.
C3 is a storage capacitor and the transistor effectively multiplies the capacitance by 100 times to produce an effective value of 1,000u.
The circuit is called a voltage regulator.
The circuit has 3 electrolytics across the power rails: 47u 1,000u and 100u.

 

[BlackMelon]

Thank you for completely clarifying every component.
I think the pin 2 of the comparator IC1/2 should be connected to pin 7 of IC1/1, cuz pin 7 is an output of the square wave oscillator, consisting of IC1/1 and its peripheral components. Also, I think the voltage divider R7 and R8 will provide excessive voltage to the base of TR2. To explain, when pin1 of the comparator equals to 9.4volt (high time), the divider will provide 2.2*9.4/(2.2+5) = 2.87volt to the base of TR2. However, the base of TR2 requires only about 0.7 volts when it turns on. Do you agree with my opinions?

PS. One thing, is it a good idea to replace the linear regulator with a switching regulator? I think this should add some efficiency to the circuit.

 

[Harald Kapp]

I think the pin 2 of the comparator IC1/2 should be connected to pin 7 of IC1/1,

No, it shouldn't. On pin 7 of IC 1/1 there is already a square wave. no need to put a comparator there. The idea of the circuit si that the square wave on IC 1/1 pin 7 is integrated by R4/C1 into an approximate triangular waveform. IC 1/2 compares this triangular waveform to a reference voltage adjusted by VR1. Whenever the triangular signal is higher than the reference voltage, the output IC 1/2 pin 1 goes low, otherwise high. Thereby you can adjust the puse/pulse ratio of the pwm by adjusting the pot. VR1. See here for more details.

 

[Colin Mitchell]

"when pin1 of the comparator equals to 9.4volt (high time), the divider will provide 2.2*9.4/(2.2+5) = 2.87volt to the base of TR2. However, the base of TR2 requires only about 0.7 volts when it turns on."

No. That's not the way to consider it.
Remove R8. R7 will deliver about 2mA to the base of the transistor. The transistor will deliver 200mA if needed.
The voltage across R8 will NEVER be higher than 0.7v and it will bleed (take away) less than one-quarter of a milliamp from the 2mA, so it will not change the conditions.
R7 is simply to take the base voltage below 0.5v when the output of the op-amp goes LOW. If it goes low to say 1v, the base will see 200mV and the transistor will definitely be turned OFF.

 

[Harald Kapp]

PS. One thing, is it a good idea to replace the linear regulator with a switching regulator? I think this should add some efficiency to the circuit.

The main portion of the power will be dissipated by the motor. Using a switch mode regulator may add more complexity to the circuit than necessary.