DC motor speed controller

[ant9985]

hi, iv been assigned to give a breif description of how the DC motor speed controller circuit below works. i understand some of the circuit but some parts baffle me, i dont suppose anyone would be kind enough to have a look at the circuit and tell would they? id appreciate it a lot

http://img175.exs.cx/img175/2246/dcmotorcontroller8bm.jpg

 

[audioguru]

Hi Ant,
That's a very complicated motor speed controller for its simple low performance. It doesn't even use pulse-width-modulation.

1) The motor is driven with a DC voltage from emitter-follower transistor T2.
2) IC3 amplifies the variable voltage from the VR1 control and feeds it to T2.
3) IC4 is a voltmeter and displays the DC output voltage of IC3.
4) The complicated arrangement of optocoupler OP1, transistor T1, opamp IC1 and the monostable timer IC2 simply force IC3 to its maximum voltage for a moment when power is 1st applied to the circuit.

 

[ant9985]

oh right, thnx, but how does the signal get from IC3 to OP1 and from there how does it get to T1, im not sure which route it takes, it seems like theres only the ground and the voltage rail connecting OP1 to the first part of the circuit..

 

[spuffock]

its an exercise in using a lot of components to perform a simple task. All you need is the pot and transistor. The 555 could be used to get a more efficient drive, but it seems that the "designer" has a very tenuous grasp on the basics.

 

[audioguru]

Hi Ant,
Have you seen the circuit and tried it? Maybe PO1 has a slotted disc in it that is driven by the motor for some kind of tachometer-controlled speed control, and maybe VR1 is just a preset set-up control and not a speed adjustment. That is about the only reason I can think of for having the opto and other pile of parts connected to it.

Originally I was thinking that when power is 1st applied to the circuit, the LED in the opto would light immediately, turning on its phototransistor. Since C1 wouldn't be charged yet, it couples the ground from the turned-on phototransistor to TR1 which also turns on for a moment until C1 charges.

Spuffock is correct, the 555's wiring could be simply changed to provide much more efficient and useful pulse-width-modulation speed control.

Without having a slotted disc, the circuit is simply a very inefficient amplified rheostat, with a maximum speed pulse when its power is 1st applied.

 

[ant9985]

well im not really sure, iv been doing a BTEC nation deploma in electrical and electronic engineering, this is an assignment iv been given from the fault finding module. i suspect they made it complicated intentionaly to give a challenge, but im not that great at reading circuits . i understand the basics of it such as u turn VR1 to increase or decrease speed, the LEDS connected to IC4 turn on when more speed is increased, IC3 comparator compares the two signals from VR1 and the lower circuit, and levels them out and outputs the results. this output decides the speed of the motor. i think IC2 gives some sort of pulsed signal, im not really sure. what i need is more detail really, and the bit that reall confuses me is how OP1 takes a signal from the first part of the circuit, i wasnt aware that a signal could travel through ground or power rails... i thought they only put power into a circuit, im pretty much a noob in electronics really what i really need is a fairly detailed, but not too much indepth, description of the circuit. where the signal goes really. thnx for the help so far

 

[Nigel Goodwin]

well im not really sure, iv been doing a BTEC nation deploma in electrical and electronic engineering, this is an assignment iv been given from the fault finding module. i suspect they made it complicated intentionaly to give a challenge, but im not that great at reading circuits . i understand the basics of it such as u turn VR1 to increase or decrease speed, the LEDS connected to IC4 turn on when more speed is increased, IC3 comparator compares the two signals from VR1 and the lower circuit, and levels them out and outputs the results. this output decides the speed of the motor. i think IC2 gives some sort of pulsed signal, im not really sure. what i need is more detail really, and the bit that reall confuses me is how OP1 takes a signal from the first part of the circuit, i wasnt aware that a signal could travel through ground or power rails... i thought they only put power into a circuit, im pretty much a noob in electronics really what i really need is a fairly detailed, but not too much indepth, description of the circuit. where the signal goes really. thnx for the help so far

As Audioguru suggested, you're missing an important piece of the puzzle!, OP1 is a slotted opto-switch - you need a slotted disk fed from the motor interrupting the beam as it rotates - this provides feedback to trigger the 555 timer. R12 and C6 form a triangle waveform on pin 2 of IC3, this is compared with the input on pin 3, and produces a PWM wave on the output, the pulse width depending on the voltage on pin 3.

T2 is a very badly designed driver for the motor, the motor should really be in the collector, not the emitter!.

I would suspect that at startup, with the motor not moving, the 555 isn't running, and full DC voltage will be applied to the motor. This will make the motor spin, and the resulting pulses from the slotted disk will trigger the 555 timer. This will reduce the power to the motor (via PWM), which will slow down, resulting in the 555 being triggered more slowly - which will cause the power to increase again. So the motor will stabilize at a constant speed - if the load increases, and the motor slows, the PWM will increase power again to maintain the set speed. Standard negative feedback! - a control signal fed back to maintain the speed.

 

[audioguru]

Hi Ant,
Yes, you really are a noob if you can't see how IC2 isn't an oscillator making many pulses, unless the optocoupler is with a spinning slotted disc in it.
You also don't understand how the optocoupler works: When its LED shines on its detector transistor's crystal, that transistor turns-on. If the optocoupler had a gap with a spinning slotted disc in the gap, then the LED's light and therefore the detector transistor would be gated on and off performing like a tachometer.
IC2 gives a constant output pulse width each time it is triggered by a pulse from the optocoupler, and IC2's output is a higher average voltage at higher RPM's. Therefore if the motor slows down from a heavy load, IC2's averaged output voltage drops lower than VR1's setting and the motor speeds up. The opposite occurs if the motor speeds up too much when its load is removed.
Do you fully understand now?

The problem with your schematic is that it doesn't show a gap in the optocoupler between its LED and its phototransistor.

 

[ant9985]

yes it is clear to me now, thnx a lot all, the help is greatly appreciated

 

[ant9985]

could i also just ask...

T2 is a very badly designed driver for the motor, the motor should really be in the collector, not the emitter!.

why should it be in the collector, not the emitter? :?

 

[Nigel Goodwin]

could i also just ask...

T2 is a very badly designed driver for the motor, the motor should really be in the collector, not the emitter!.

why should it be in the collector, not the emitter? :?

Because been in the collector the transistor will drop far less voltage, in the emitter it's got the Vbe drop as well - which is an extra 0.7V. The other reason is that you've got far more base current available to turn the transistor on - with the emitter to chassis (and motor in the collector) you've got 11.3V voltage to feed the base from (with a 12V supply), simple ohms law will give you the base current from the value of the base resistor. With the motor in the emitter the base has to go to 12V (or at least as close as it can), this means there's almost no voltage available to power the base current - again you can work the base current out from the resistor value.

As shown, with the 1.5K base resistor, if we assume the base only goes as high as 10.5v (leaving only 9.8V to feed the motor) the base current will be 1.5/1500=1mA. With the motor in the collector you will get something like 11.9V to the motor, and the base current will be 11.3/1500=7.5mA.

 

[ant9985]

k, cheers