12VDC to 9VDC converter with 12VDC output added

Good morning,
 
I used the following schematic to make a 12VDC to 9VDC converter, which worked perfect (see attachment).

 
Since I need both 12VDC and 9VDC for my project (12V to run a motor, 9V to run WC8.0), I could simply tie off the 12VDC input of the converter (see second attachment).
This, however, caused a periodic flip flop of the 12V OUT as if the caps had an impact; which shouldn't happen since I tapped off the 12V before the circuit using the caps.
 

 
Where am I going wrong?

Rainer
 
 
 

rfeyer said:

I used the following schematic to make a 12VDC to 9VDC converter, which worked perfect (see attachment).
Since I need both 12VDC and 9VDC for my project (12V to run a motor, 9V to run WC8.0), I could simply tie off the 12VDC input of the converter (see second attachment).

This, however, caused a periodic flip flop of the 12V OUT as if the caps had an impact; which shouldn't happen since I tapped off the 12V before the circuit using the caps.

Click to expand...

 
I'm not sure that a 7809 will work with 2.3V differential. (12V - 700mV across the diode only leaves 11.3V)
 
I've been running WC8 off 12V supplies 24/7 for years. As long as they have adequate ventilation, they're quite happy.
If you're concerned, just whack 3 or 4 diodes in series, that'll drop a couple of volts, a lot easier than a linear regulator.

I will try that on next board - TY!
 
Is there any reason why the 12VDC OUT, which is directly connected to the 12VDC supply to the drop down circuit should be flip flopping? Am I correct that, drawing 12VDC prior to entrance of the circuit should shortcut the capacitor effects?
You're up very early, RossW!
Rainer

rfeyer said:

I will try that on next board - TY!
 
Is there any reason why the 12VDC OUT, which is directly connected to the 12VDC supply to the drop down circuit should be flip flopping?

Click to expand...

 
Are you sure that it is? Is it the 9V that's dropping? How are you measuring it and what makes you think it's the 12V side?
 
 

rfeyer said:

Am I correct that, drawing 12VDC prior to entrance of the circuit should shortcut the capacitor effects?

Click to expand...

 
Depends where your isolation or decoupling is. If it's before the diode, and the capacitor is after it, then anything after the diode should have minimal effect.
 

rfeyer said:

You're up very early, RossW!

Click to expand...

 
Firstly, I work all sorts of hours. Getting more than 3 hours sleep in one hit is a rare thing.
Second, I've done this for 25 years or more, so it's second nature.
Finally, and this is the real kicker... I'm in Australia, so it's not "early", it's "early evening".  (UTC+11 at the moment, soon to be UTC+10)

You might want to try eliminating the protection diode. Its function is to protect the regulator if you
hook up up the power supply backwards, but as Rossw posted it costs you 700mv of headroom.

In you second picture there is just a wire between the diode and the regulator. Anywhere along that
wire is the same as far as voltage drop is concerned.

I'd try doing the following:

1) Move the 12V motor takeoff to in front of the protection diode.

2) If that does not work eliminate the diode.

3) If neither of these experiments work try eliminating the 9V regulator as Rossw suggested or
substitute a bigger 12V supply that will not droop as much when the motors are turned on.

/tom

TY both!!!
I will do the experimentation when I get home later (when RossW is starting his day again
 
To answer how I know about the voltage drop: As mentioned initially, if hooked up as schematic, the motor flip flops a few times; but the WC board light does not seem to flicker. So, I am guessing it is the 12V OUT.
 
I will remove the diode and see what happens then - it is a 4A power supply, should be good.\
 
RossW: the last person who helped me with my surveilance camera / PC recording dilemma was also from Australia, an auther, great books, his name is Geoff Worboys (close to Sidney)
 
Rainer

I would vote for Ross' idea that to have couple diodes to drop the voltage. That actually also protect any negative going surge spike from motor, sometimes inductive load can produce spikes that are negative going cause damage.

TY support, I will do that.
So, there is no need to have any capacitors to eliminate spikes?
 
so:
 
in:  12V +  -----------  >|--->|----->|----9V + ----
                  |                                                        to WC 8.0
                  |  in 12V -  --------------------  -  ---
                  |                                        |
                  |                                        |
                  12V + to Motor                  12V neg to motor
 
 
As simple as this?

Rainer

rfeyer said:

As simple as this?

Click to expand...

 
Pretty much, yes.

Love it! TY
 
And I bougth 10 of those 7809's

rfeyer said:

Love it! TY
 
And I bougth 10 of those 7809's

Click to expand...

 
They're cheap, and you'll use 'em on something else, one day.
 
I'm a great fan of keeping things simple. The diodes don't do the same job as a 7809, but since they will drop (roughly) the same voltage all the time, and since your input is regulated, 12V-(3*0.7) is always about 9.9 or 12-(4*0.7) is always about 9.2V, if you use 4 discrete diodes, it'll achieve very close to the same thing.

Especially for me, simple should be best!!
 
I did make a little board with 3 diodes, dropped voltage to 10.1 volts, input 12VDC 5A and a direct 12V output as above diagram.
 
Weird thing, it would not drive the motor again. Yet, when I hooked up a small 9v 1.2A power supply directly to the motor, it worked (not as fast, but it worked).
 
I am sure running the WC8.0 off the 12V 5A PS will not drain it.
May have to work with 2 PS's - though a pain.
 
Rainer
 
Pic labesl probably too small, but: 12VDC 5A coming in on Left, at right is 3 diode reduced 10VDC, at mid bottom is 12VDC straight connected to 12VDC in - yet, the 9DVC works, the 12V at botttom does not drive the motor - even though says 12V on multimeter.

How do you drive the 12V motor, through a relay?

Yes, WC adtivates a transistor which allows the 12v relays to be activated. This relay circuit uses the 12v power supply from the above schemwtic at the bottom labeled 12v OUT

with a volt meter, active the motor, there should be 12V on the motor terminal, deactivate the motor, 0V across the motor.
Do you see 12V across the motor, but motor does not turn?