I was able to get this circuit working and took out the two transistors since I found a schmitt trigger that could source enough current to operate the relay.

I didn't read the datasheet and just assumed the relay would be "regular" small sized, but it was surface mount and it sure is tiny!

Everything works and I can get it to reverse directions back and forth back and forth just fine.

But I was reading that a protection/flyback diode is needed for relays to prevent voltages spike that would damage your transistors/ICs.

I don't think you can use a diode in this circuit since the latching relay requires that current switches direction.

Is a diode not necessary in this circuit because because only a pulse is sent to operate the relay?

 

You don't need diodes because the capacitors slow down the current toward the end of their charging by having most of the 6 volts charged up in them (back EMF). Slow stop = no kick.

 

Thank you for your explanation. I'm glad I don't have to worry about that now.

 

I've noticed that the motor will switch directions at basically the same light level.

So in the evening the door will shut at a certain light level and then in the morning it will open at the same light level. I can use the potentiometer to get the door to move at a certain light level and this will be the same for evening and morning.

It can be adjusted two different ways:
1 Close later and open earlier
or
2 Close earlier and open later

I would like to adjust it independently so that I choose the light level it closes in the evening and then choose the light level it opens in the morning.

Is this something that would be relatively easy to do or am I entering into a whole new level of complexity?

 

#12, alec t any ideas?

I guess I could use 2 photoresistors (and pots) and put each one on a different circuit on the schmitt trigger ic?

 

I guess I could use 2 photoresistors (and pots) and put each one on a different circuit on the schmitt trigger ic?

That's one way. But the outputs of the two Schmitts would have to be logically combined to provide the correct relay operation.

 

Here's another option, using just one photoresistor. The 'dawn' and 'dusk' trimmers set the trigger points for increasing and decreasing light levels respectively, to set and reset a latch formed by two cross-coupled NAND gates. R8 provides positive-feedback to ensure a snap-action.

 

Great! Thanks Alec.

I understand the gist of this circuit, but need to study the logic a little more.

The output based on R8 changes based on the light intensity via the CDS cell, right? So does it move in sync with either the inputs of the dawn or dusk?

When one is high then it will be high and vice versa?

 

Consider dawn. As the light increases the CdS cell resistance falls so the voltage across R1 rises. This voltage is resistively summed with the voltage at the wiper of the dawn trimpot. When the sum reaches the trip point of gate U1d the gate output goes low, setting the latch formed by U1a/U1b. U1a output and R8 right end are now high, so the positive R8 current raises the voltage across R1, i.e. there is positive feedback via R8 to give the required snap action.
A similar situation arises at dusk. As the light level falls the input to U1b (and U1c, which is paralleled with U1b to boost the current available for driving R9) falls. When the trip point of U1b/c is reached the latch is reset, the right end of R8 goes low and the R1 voltage is thus pulled down by the negative current through R8.

 

Thanks Alex. This is very clever.

After reading your explanation and looking at the schematic several times and reading about NAND gates I understand it a little better now, but need to read up on SR latches more.

This is probably obvious to most, but why is it when you connect the wiper for either dusk or dawn going to U1c/b and U1d it changes the input voltage? I understand it when it is like an "inline" variable resistor, but not like this when only the wiper is connected to that input.

 

The wiper is connected via upper and lower parts of the trimpot resistance track and by fixed resistors to +6V and ground, so it has a voltage which is intermediate between 0 and 6V depending on the trimpot setting. A CMOS NAND gate changes output state when its input voltage moves past ~ half the supply voltage (i.e. about 3V here).

 

Thanks. The light is starting to come on in my head.

Why a fixed resistor both above and below the trimmers? Could not one just be used?

 

Why a fixed resistor both above and below the trimmers?

So that the adjustment range provided by the trimmer between its min amd max settings is optimum for use with the expected change in CdS cell resistance.

 

I'm trying something a little different for the first circuit Alec_t provided
Just for fun I hooked up an Odroid C1 to it (similar to a Raspberry Pi). I removed the Schmitt IC, the LDR sensor and the pot.
So all I have left are the two transistors, relay and two capacitors.

The Odroid output switches either to High 3.3v and 14ma or Low about 5mv and no ma. The negative of the Odroid is plugged into the negative rail and the positive is plugged into the input side of the relay. I can switch the relay direction when it needs the "high" voltage most of the time but I can't switch it back with "low".

The circuit is powered at about 5.6v. When in High mode the voltage at the input side of the relay is about 2.8v and when it is Low it is about 0.5v. The weird thing is that I could initially get it to switch back and forth both ways but now it only switches when High.

Is the voltage only 2.8v because it is coming through the transistor? But it is less than the .7v drop.

The relay is 3v 90ohm coil, 3ms switching time. Coil is rated from 2.25v to 9v.

Do you know why I can't get it to switch back and forth using the Odroid? Is the Low not low enough? Do I need transistors with different specs? Different capacitor specs?

I'm at a loss right now what to do and have been trying for over a week to figure it out. I have a hunch it is something simple but I'm stuck.

Thanks for any help.
Matt

 

It's bad form to hijack someone else's thread.
Please start you own.

 

I agree that would be in bad form.

Thankfully however I started this one. ; )

 

Sorry about that. Should have seen that it said you were the Thread Starter.

 

No problem. What do you think about my situation?

 

I'm not very good at visualizing what you have.
Can you post a circuit diagram?

 

Here it is. I removed the IC the LDR and the pot.

Sometimes it will switch over with High and sometimes not. I've tried doubling and then halving the capacitors and it does not matter.