A light-sensitive motor

[David Young]

This is a sketch of a circuit I have built from a suggestion from another site.

The idea is that the motor (is the circuit symbol correct?) should turn on when the light-sensitive resistor is exposed to light.

The motor will not turn when I rig this circuit up on a breadboard. On the other hand, when I replace the motor with an LED, it lights up when the LDR is exposed to light. Also, a voltage test with a multimeter shows that over 8 volts is running through the LED (the motor will turn with as little as 5v).

Can anyone tell me either what modification I should make to this circuit or if there is a different circuit that does what this one is supposed to do?

Any help would be greatly appreciated.

 

[(*steve*)]

The issue is current not voltage.

The bas current is amplified by the transistor to give the collector current. The amount of current required to turn on a LED is probably a lot smaller than for your motor.

At the very least, you need to reduce (or remove) the 10k resistor in series with the LDR.

In addition to thei, you should place a reverse biased diode across the motor to protect the transistor.

Once you have done these things, you *may* get the motor to turn when the LDR is exposed to light.

Note that, depending on the required motor current, the transistor may get quite hot. And that is bad. Too hot to touch is too, too hot.

 

[David Young]

Thanks. I've tried both: taking out the resistor in series and adding the reverse-biased diode. I can't get the motor to turn.

I do also have a relay with the specifications
5V DC
13x8C1 5A250V AC.

Is there anyway I could incorporate this into a circuit that has the desired effect? I am prepared to use two separate power supplies (one for the light-sensitive part and one for the motor) if I can find some way to get the circuit to work.

I have a large supply of resistors, capacitors and both NPN and PNP transistors, but very little knowledge of how this type of circuit should work (I got a fairly bad mark in A-level physics over 20 years ago and electronics was one of my weaker suits).

 

[(*steve*)]

A relay is just an on-off switch . It won't make the motor turn at different speeds.

The circuit you've shown might work, just to turn a motor, but it's not very practical.

Perhaps you need to describe what you want to do.

 

[wingnut]

Have you tried taking out all the resistors and connecting the photoresistor directly to the base?

 

[David Young]

I've tried it without the resistors, but nothing happens.

I need to complete a circuit consisting of a 9V battery and a motor when, and only when, light falls on the LDR. Ideally I would run the light-sensitive part off the same 9V battery, but if that is not possible, or it could be made simpler with a second battery for that part, then that would also do what I need it to do.

The less space the circuit takes up, the better.

 

[(*steve*)]

Perhaps you need to describe what you want to do.

*exactly*

What is the motor going to do? What are the specs for the motor? What load is it under? How much light? etc., etc., etc.

 

[Iron_man]

What is the current and voltage ratings of your motor? Perhaps a proper isolation circuit is in order. You can then use the LDR to drive the transistor into saturation allowing the relay to switch the motor on or off.

A relay is an on/off switch, configured with an isolation circuit, should allow you to make the motor work

 

[David Young]

If the circuit is placed in a drawer, in a room that is in daylight, then when the drawer is opened the equivalent of wiring the motor directly to the battery should happen. It has to be light-triggered, and not by breaking or making a physical contact. I don't have the motor to hand right now, so I can't check its resistance.

That said, if anyone can tell me how to rig up the relay I mentioned so that it switches when light falls on the LDR (it's an SPST relay), that will do exactly what I need the circuit to do, and the addition of a second power supply won't be a problem.

 

[wingnut]

I set up your original circuit but with these modifications...
I used a 6V power supply.
I used a TIP122 power transistor and using a 1k resistor instead of the pot, leaving out the 10k resistor completely. The motor turned on fine.

Then I tried it with a lowish power 2N222 transistor and this got hot - but worked, and the motor turned.

So I added a 500 ohm resistor where you had your 10k resistor and it still worked but the motor turned reluctantly and the 2N222 got quite warm. It would have turned a relay on.

With no resistors at all, the photoresistor turned the TIP122 on, but one needed the 1k to turn it off.

My photoresistor reads 160 ohm in full sunlight - what does yours read?
Photoresistors always have some resistance which usually is enough to protect the base while you are trying things out. But maybe your transistor got hot and blew.??

 

[David Young]

I haven't got the daylight resistance at the moment as it's the evening. In the meantime, with the lights in my living room, the LDR has a resistance of about 30 kilo ohms.

The resistance of the motor is tricky to ascertain. The figure on the multimeter bounces around so much when I apply the probes to the terminals, but the figure appears to be 1.8 ohms.

The two NPN transistors that I have ready to hand are 2N3904 and BC547.
I may have found one possible problem: the diagram requires a BC337 and mine turns out to have a BC547 for some reason. Would I be right in assuming that this would make a significant difference to the results?

 

[BobK]

At 30K the LDR is not going to supply enough base current to run the motor. You might try a darlington. A better circuit would use a comparator to discriminate light and dark and use its output to drive the base of the transistor.

Bob

 

[duke37]

The resistance of the motor does not mean much. A 100% efficient motor would have zero resistance.
The current taken by the motor will be dependant on the applied voltage and the speed. A spinning motor produces a back emf and the current will be the difference between the back emf and the applied voltage, divided by the resistance.

I am sure you can look up the data on the transistors. Why should I do it?

 

[David Young]

In reply to duke37: 1) I wouldn't know what information to look for on their datasheets (all of which I have) and 2) Nobody is forcing you to reply to this thread.

 

[David Young]

BobK: Is a darlington a type of LDR?

 

[(*steve*)]

No, it's a pair of transistors (sometimes in a single package).

 

[wingnut]

The TIP122 I recommended in an earlier post is a Darlington high power transistor.
When driving motors I don't like to mess around with low power transistors.

The BC547 is a low power transistor as Ic is 100mA and its power is max at 500mW

The transistor on your original diagram was a 800mA rated transistor if I remember correctly.

 

[KrisBlueNZ]

The problems with your circuit at the moment are:

You need a clean switching behaviour in response to a gradual, poorly defined change in the resistance of the LDR. This can be done with a circuit that uses a threshold - a comparator such as the LM393 (dual comparator) or a circuit built from transistors. Using the LDR to directly drive the base of a transistor is a bad idea because the transistor will operate in its "linear region" where it will dissipate a lot of power. I recommend using a Schmitt trigger, which is a comparator with positive feedback, which gives it a very clean switching action and a deadband, to prevent its output from wittering when the light level is near the threshold.

You need your switching transistor to be rated to carry the full motor current. The BC547 is only rated for a maximum collector current of 100 mA. This is sure to be insufficient for your motor. Power transistors can handle more current but need to be driven with a clean signal with plenty of current capability, to saturate the transistor. A Darlington transistor doesn't need much drive current, but it will drop around 1.5V which leaves only around 7.5V available for the motor. The best solution here is a MOSFET. These are ideal for switching heavy currents.

I will draw you up a suggested circuit. Can you tell me where you're located (put it in your profile so it shows up on all your posts - it's a common courtesy) so I know what supplier to direct you to, for the components.

It would also be useful to know how much current your motor draws. Connect an electrolytic (1000 uF or more) across the motor, and connect it to the battery with an ammeter in series, and tell us what the ammeter reads. (The ammeter would be a multimeter set to the 10A range.)

What kind of battery are you using? You should use something bigger than the little 9V batteries with the clip-on connectors, unless your motor is quite small.

 

[David Young]

Thanks for the advice. I'll rig up the circuit with the capacitor tonight and post the results.

Incidentally, I get nearly all of my components from eBay (one or two I can pick up from a local supplier when they're in stock).

The battery is a PP3 rechargeable, but I also use a combination of AA cells to reach the same voltage. The only other power sources are a mains adaptor (with everything from about 3V to 12V) and a Hornby R900 power unit (which goes up to 14V).

 

[KrisBlueNZ]

OK, here's my suggested circuit.

Q1 responds to the voltage between its base and emitter, which is the product of the current flowing through the LDR and the VR1 resistance. As the light on LDR1 increases, this current increases, until the voltage across VR1 reaches about 0.7V, and Q1 starts to conduct, and pulls its collector towards the positive supply rail.

This positive voltage on Q2's gate causes Q2 to start to conduct, and it pulls its drain down towards the 0V rail (bottom rail of the schematic). This causes extra current to flow through R4, reinforcing the change and causing Q1 to conduct harder. This is called positive feedback, and it makes the circuit "snap" quickly into the active state, where Q1 and Q2 are turned ON, and the 9V supply voltage appears across the motor.

Because of the extra current through R4, the brightness threshold to cause the circuit to turn OFF is now lower than the ON threshold. This gives a clean response to varying brightness and prevents wittering and indecision around the ON/OFF threshold, like a thermostat.

VR1 sets the brightness thresholds, and R4 determines the distance between the ON and OFF thesholds - that is, the amount of hysteresis. If the thresholds are too far apart, increase R4.

This circuit is designed for low quiescent current in the dark, to prolong battery life. When the motor is not running, the only current flowing will be through VR1, R1 and the LDR. If the LDR is in the dark, its resistance will be at least 100 kilohms, so this current will be less than 100 microamps (uA) and the battery should last most of its shelf life.

Q2 is a standard N-channel MOSFET that is rated to comfortably switch the amount of current required by the motor. I've specified an NTP3055 but there are MANY alternatives. The important specifications are:

Id(max) - the maximum drain current - must be comfortably greater than the current required by the motor.
Vds(max) - the maximum voltage across the MOSFET - must be comfortably more than the supply voltage; I suggest using something rated for at least 40V Vds.
Vgs(th) - the gate threshold voltage. In this application, the MOSFET will see 9V on the gate, which is plenty high enough for any normal MOSFET, so you don't need to worry about that specification.
Package: I would use a TO220 package (same as a 7805 regulator) for convenience.

There's not much penalty in choosing a device with much higher ratings than you need, as they're only a few dollars and you're only making one unit. MOSFETs are static sensitive so you need to follow the recommended anti-static handling procedures.

I'd suggest seeing what N-channel MOSFETs are available locally first. Post a list of options if you like, and we can tell you what's suitable.

Edit: When you're measuring your motor's current consumption, use a decent power source, not the little 9V battery. The motor will draw a significant amount of current and this could cause the battery voltage to drop, which in turn will reduce the current you measure. Unless your motor draws less than 50 mA or so, I would not use the PP3-style battery with this circuit.