Circuit quest use of LM324/555 as motor driver

I'm planning to build this light following microbot:

https://www.winnipegrobotics.com/Projects/How It Works - Coke Can Robot.pdf

basically it uses 4 photocells. These are fed into two comparators, the outputs fed into two 556's which then drive the motors.

According to the author the 556 is used for for its comparator, flip-flop, and its ability to source or sink up to 200mA of motor current.

I'm very new at this and I don't quite understand why I can't drive the motors directly from the comparator outputs. From what I understand the outputs will be either 7.5V or 0V. Enough to drive a mall hobby motor?

I guess I'm not sure what s meant by "source or sink 200ma".

Thanks
Marc

marchache said:

I'm very new at this and I don't quite understand why I can't drive the motors directly from the comparator outputs. From what I understand the outputs will be either 7.5V or 0V. Enough to drive a mall hobby motor?

Click to expand...

The outputs of the comparator can't handle enough power to feed the motor, it's not the volts which are the problem, but the current capacity which isn't high enough.

I guess I'm not sure what s meant by "source or sink 200ma".

Click to expand...

200mA refers to the current it can handle, it would probably be enough for very small motors. Source or sink refer to the actual method of motor connection - for 'source', the power flows out of the 555, through the motor, and down to ground - so the 555 is the actual source of power for the motor. For 'sink', power is connected to the top of the motor, and the 555 connects the bottom of the motor to ground - 'sinking' the current to ground. Basically it's where the 555 switches the motor, either at the top (positive) or bottom (negative).

Thanks Nigel. Appreciate the quick reply.

So when designing these circuits, if an an IC is the source of power for a load circuit, I need to determinw how much currrent the load circuit will draw and make sure the sink/source of the IC will handle it in addition to making sure the output voltage is sufficient. Correct?

Secondly- Is there an easy way to determine how much current a motor draws ? These hobby motors typically only give a supply voltage range.

Thanks again, you guys are great.

Marc

Motor driver

Hello,

The current taken, under load, must be less than 200mA due to the capacity fo the 556 driver. If you put in series a 1 Ohm, 1 W resistor with the motor and feed it with 9 volts, the resistor will drop: I xR= max .2Ax9V= max. 1.8V at full load. This can be checked by applying step by step a friction brake, and reading the voltage drop in the resistor.

One word about the comparator: never let the unused inputs floating, tie them to ground or +V. This IC can sink only 20mA and source 40mA and are very sensitive to spikes and noise in general, so it would not be a good idea to use it as a motor driver, unless using a very good designed suppressor circuit.

marchache said:

So when designing these circuits, if an an IC is the source of power for a load circuit, I need to determinw how much currrent the load circuit will draw and make sure the sink/source of the IC will handle it in addition to making sure the output voltage is sufficient. Correct?

Click to expand...

Yes, that's correct.

Secondly- Is there an easy way to determine how much current a motor draws ? These hobby motors typically only give a supply voltage range.

Click to expand...

The only way is to measure them - it will vary a great deal under different load conditions, you should check the current off load (which should be fairly low), and check it with the motor stalled (which will be a great deal higher, in fact many times higher). In a robot, if it runs into something it's likely to stall the motors - this can blow the drivers, so you need to either limit the current available, or make sure the drivers can handle the load.

The only way is to measure them - it will vary a great deal under different load conditions, you should check the current off load (which should be fairly low), and check it with the motor stalled (which will be a great deal higher, in fact many times higher).

Click to expand...

Thanks again. Is this as simple as putting a multimeter (set to current of course) in series with the motor ?

Assuming this is the case, a further question: my multimeter can handle a maximum input current of 200ma. Although the current is not likely to be this high, would it be safer to instead place a resistor in series with the motor and then measure the voltage drop across the resistor ?

Thanks
Marc

marchache said:

The only way is to measure them - it will vary a great deal under different load conditions, you should check the current off load (which should be fairly low), and check it with the motor stalled (which will be a great deal higher, in fact many times higher).

Click to expand...

Thanks again. Is this as simple as putting a multimeter (set to current of course) in series with the motor ?

Yes.

Click to expand...

Assuming this is the case, a further question: my multimeter can handle a maximum input current of 200ma. Although the current is not likely to be this high, would it be safer to instead place a resistor in series with the motor and then measure the voltage drop across the resistor ?

Click to expand...

I would expect the stall current to be far higher than that!, I measured the stall current of the Cybot motors (a small magazine robot) at more than 6A - and I suspect that was probably limited by the battery I used, a PJ996, rather than the motor. But the Cybot motors are fairly powerful for a small robot.

Yes, you could measure the voltage drop across a small value resistor (which is how the current ranges in your meter work anyway).

Re: Motor driver

IRQ57 said:

Hello,

The current taken, under load, must be less than 200mA due to the capacity fo the 556 driver. If you put in series a 1 ohm, 1 W resistor with the motor and feed it with 9 volts, the resistor will drop: I xR= max .2Ax9V= max. 1.8V at full load. This can be checked by applying step by step a friction brake, and reading the voltage drop in the resistor.

Click to expand...

0.2Ax9V=1.8watts (not volts) is the power dissipated by the motor. The voltage drop across the resistor will be 0.2v.

Motor driver

Ron,

You are completely right, sorry for the mistake.