Delay On/Off Setup

[KatieandDad]

How long are these Power Outages ?

You might need a huge Capacitor bank if its for too long.

Anything more than a few seconds might need a battery.

 

[AndrewT]

a few seconds

Just long enough to allow a controlled and quiet shutdown.
And similarly a delay just long enough to allow a controlled and quiet start up.

It might turn out that the DSP only needs 100,000us to start up, or to shut down quietly.

 

[KatieandDad]

Do we know how much current the DSP requires to keep going ?

Batteries are cheaper than huge caps ?

I've got 4 x 10000uF 16V caps going spare here that I can't use.

Any offers about £10 the lot would do me.

 

[DQ828]

Just long enough to allow a controlled and quiet shutdown.
And similarly a delay just long enough to allow a controlled and quiet start up.

It might turn out that the DSP only needs 100,000us to start up, or to shut down quietly.

Correct

Whats 100,000us? Just sounds like a lot of money to me

 

[Mooly]

Correct

Whats 100,000us? Just sounds like a lot of money to me

1/10 of a second

 

[DQ828]

Thank you All Again

I went for the higher voltage diode (D2) because that is what I get from the local supplier. I have rotated D3

I'd like to put this together in a few days so I will have to pass on the caps, thanks anyway.

I didn't notice the cap was changed to 47uf, I have made the adjustment.

I'm sure this electronics business is part black magic

 

[Mooly]

Any problems, just ask

 

[DQ828]

I did forget to ask, will this type of resistor be suitable 0.6W 100k Metal Film?

 

[KatieandDad]

I did forget to ask, will this type of resistor be suitable 0.6W 100k Metal Film?

That'll be fine - you don't need anything special here.

 

[DQ828]

That'll be fine - you don't need anything special here.

It's always the wattage that I'm not sure about with resistors

 

[KatieandDad]

Wattage is:

Either V x V / R
Or I x I x R

You know that the MAXIMUM voltage that can appear across the resistor is 12V

Therefore The MAXIMUM power that could ever be expected of it would be

12 x 12 / 100000 = 0.001 Watts

That's 1/1000 of a watt.

Your 0.6W resistor will be absolutely fine.

 

[AndrewT]

P = I^2 * R

If you know the current, you can calculate the power to be dissipated.
I generally choose 600mW and that allows for a nominal 150mW of maximum dissipation for non critical applications.
Or, allows for 60mW of maximum dissipation for more critical applications.

 

[sofaspud]

And the only DC path to ground is through the 680k resistor. The only significant current in that circuit is through the relay coil.

 

[Mooly]

Another useful way to look at it, and understanding that this applies only when you know the absolute max voltage the circuit can ever see.

You know the supply is 12 volts and you know W=(V*V)/R

So knowing that and taking a 0.6 watt rating we can say this,

(V*V)/0.6 gives us the "lowest" value of resistor that can safely be used in a design running on 12 volts without having to "think" about it.

So we get 144/0.6 which is 240 ohms. So any resistor anywhere in that circuit that is over 240 ohms can be a 0.6 watt. Anything less than 240 ohms and you need to look and see what the resistor will really see across it under all conditions.

And factor in a safety margin too. If buying 0.6 watt resistors lets call them 0.4 watt and do the calculation again which gives 360 ohm.

 

[MCPete]

I'm attaching the schematic of a power-on delay based on the 555 timer that I believe would be the simplest solution. Given the values of C1 and R2 shown in the schematic, the delay of the output pin 3 of U1 (the timer IC) going from ground to +12V is about 2 seconds. If the connection of +12V to the circuit is disrupted, it takes about 1/10 of a second in theory for C1 to become completely discharged and ready to delay power again.

I built and tested this circuit. No matter how quickly I (manually) disconnected from and reconnected the circuit to the +12V power supply, the output pin was at ground when reconnecting and took about 2 seconds to go high to +12V.

Note that all the resistors are 1/4W and diode D2 is a small signal diode (1N914).

Now given that delayed power on is effected with this circuit no matter how short or long the disruption of the power supply, I would think that this would mean that there is no need to power the miniDSP through the power outage. That is, the miniDSP is always going to receive power before the power amp.

Regards,
Pete

 

[DQ828]

I'm attaching the schematic of a power-on delay based on the 555 timer that I believe would be the simplest solution. Given the values of C1 and R2 shown in the schematic, the delay of the output pin 3 of U1 (the timer IC) going from ground to +12V is about 2 seconds. If the connection of +12V to the circuit is disrupted, it takes about 1/10 of a second in theory for C1 to become completely discharged and ready to delay power again.

I built and tested this circuit. No matter how quickly I (manually) disconnected from and reconnected the circuit to the +12V power supply, the output pin was at ground when reconnecting and took about 2 seconds to go high to +12V.

Note that all the resistors are 1/4W and diode D2 is a small signal diode (1N914).

Now given that delayed power on is effected with this circuit no matter how short or long the disruption of the power supply, I would think that this would mean that there is no need to power the miniDSP through the power outage. That is, the miniDSP is always going to receive power before the power amp.

Regards,
Pete

Thanks for the help, I have already ordered the part for the other design but you can be assured I will keep yours up my sleeve for another time.

 

[Mooly]

Interesting circuit there MCPete. It's great to see novel circuits using the ever versatile 555

It will certainly work and the 555 can easily drive small loads directly (to source or sink current) so an external transistor may not be needed (depending on relay).

Perhaps a .1uf across pins 1 and 8 maybe if using the "original" version of the 555 ?

 

[MCPete]

Perhaps a .1uf across pins 1 and 8 maybe if using the "original" version of the 555 ?

I haven't seen that used. What purpose would the 0.1uF capacitor serve?

A concern that I have about applying my circuit to DQ828's application that I didn't mention is that most versions of the 555 aren't designed to tolerate a supply voltage greater than 15 VDC. If his 12V PSU is regulated, then there is no problem. However if it is unregulated, then possibly the output voltage of the nominally 12V supply could go above 15V when relatively unloaded.

I know DQ828 says that he doesn't plan to use a 555 IC this time around, but I thought that I would mention my concern now anyway.

-Pete

 

[MCPete]

Interesting circuit there MCPete. It's great to see novel circuits using the ever versatile 555

Of course this is mostly straight out of a textbook. My contribution is the loop for rapidly discharging C1 including diode D2 connected in parallel with resistor R2 and resistor R3 connected from ground to the terminal of C1 connected to V+.

 

[Mooly]

I haven't seen that used. What purpose would the 0.1uF capacitor serve? -Pete

The original Signetics NE555 had a well known issue inherent in its design of "spiking" the supply as the output changed state. I think a value of 400ma for a few microseconds was somewhere near the mark. That could cause false triggering and erratic behaviour. The recommended solution was a cap as close as possible to the supply pins... and I have just looked at some old data sheets and some mention it and some don't. Some suggest a 0.1 in parallel with a 1uf.

We forget how old these devices are at 40 years and there have been many subtle changes and revisions. Later versions and the low power and CMOS versions pretty much eliminated the problem.

Page 8 on this one, additional information
http://www.datasheetcatalog.org/datasheet/nationalsemiconductor/DS007851.PDF