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Old 19th January 2012, 10:05 AM   #1
DQ828 is offline DQ828  Australia
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Default Delay On/Off Setup

I have built small amps for my kids & incorporated a MiniDSP unit into them so I can use it to control a sub as well. The problem I would like to solve is how to turn the DSP on first (as it make horrible noises) and hold the DSP on for a few seconds even if we have a power outage (happens a bit around here).

My electronics knowledge is limited & would greatly appreciate a bit of help.

The DSP is powered by a separate 12VDC 1A Walwart PSU. To hold the DSP on I was planning on installing a capacitor bank to keep the unit on if the power failed or on normal turn off. I am not sure if it is ok to do this as per my drawing, can anyone advise me please.

The turn on circuit for the amp is copied from the WEB so I assume its kosher.

As you can see I dont have a lot of room to work with.
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File Type: jpg 12-01-19 Delay Circuit.jpg (411.3 KB, 494 views)
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File Type: pdf 12-01-19 Delay On Off Circuit.pdf (74.6 KB, 86 views)

Last edited by DQ828; 19th January 2012 at 10:13 AM.
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Old 19th January 2012, 11:05 AM   #2
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Someone might have a more sophisticated watchdog circuit, but your idea is correct. The caps have to hold enough charge for the load+time, but I'm not sure what that is. Put a Schottky diode to the left of the cap bank to force the voltage to the DSP.
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Old 19th January 2012, 11:18 AM   #3
Mooly is offline Mooly  United Kingdom
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So many approaches to this kind of issue.

Doing as sofaspud suggests and incorporating a diode is essential as without the DSP supply discharges back into the PSU circuitry. In practice it may or may not make much difference depending on the loading.

A diode across the 68K will help discharge the cap quicker on power off so that the "reset action" works straight away again.

These kind of circuits can be a bit tricky and indeterminate. The second transistor collector to go to ground of course. First transistor. A 470uF from base to ground is a "short" as it initially charges pulling all the current through the B-E junction. Not good imo. It's all very unpredictable in use.

How about a proper speaker delay instead ? One that gives a few seconds delay at power up and is able to detect short mains drop outs and so always giving the full delay time.
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Old 19th January 2012, 11:54 AM   #4
AndrewT is offline AndrewT  Scotland
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The Vbe of that left most PNP is enormous (12Vbe) at the moment of switch on.
The collector current is limited by the 10k load to -12V (~1.2mApk).
The base current has no current limiter !!!
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Old 19th January 2012, 09:12 PM   #5
DQ828 is offline DQ828  Australia
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Quote:
Originally Posted by AndrewT View Post
The Vbe of that left most PNP is enormous (12Vbe) at the moment of switch on.
The collector current is limited by the 10k load to -12V (~1.2mApk).
The base current has no current limiter !!!
Is that a bad thing?

Should I do something differently?
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Old 20th January 2012, 12:13 AM   #6
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Quote:
Originally Posted by DQ828 View Post
The DSP is powered by a separate 12VDC 1A Walwart PSU.
How about using a 12V battery.
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Old 20th January 2012, 01:17 AM   #7
DQ828 is offline DQ828  Australia
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Quote:
Originally Posted by Michael Chua View Post
How about using a 12V battery.
Then I'd need a charger, also I need to cram it into the existing case.
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Old 20th January 2012, 01:49 AM   #8
MCPete is offline MCPete  United States
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The 555 timer IC can easily be configured to give you a delayed turn-on, the duration of which is set by the resistance of a single resistor. Also, the 555 IC and associated resistors and capacitors draws negligible current.

Probably you can find a suitable schematic diagram by searching for "555 timer circuits".

At the present time I can't post graphics here as I don't have a web address for them. Otherwise I would post the circuit.

-Pete
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Old 20th January 2012, 08:09 AM   #9
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by DQ828 View Post
Is that a bad thing?

Should I do something differently?
It's a bad thing and yes.

Something like this is easy. I've just made this circuit up but all the values are in the right ball park and will work. The 47uF cap can be altered to give a different delay. The zener just means the voltage across the cap has to rise a bit more (4.7 volts more) before the FET conducts.

The 100K and 47uf charge at switch on. When the voltage across the cap exceeds 4.7 volts plus the turn on volts of the FET (around 4 volts) then the relay pulls in.

Using a FET means we need no equivalent of "base current" and so we can design using high value resistors and small caps.

The diode across the 100k pulls the voltage on the cap down quickly at power off.

Edit... and remember to include and try the diode sofaspud mentioned
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Old 20th January 2012, 09:33 AM   #10
DQ828 is offline DQ828  Australia
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Quote:
Originally Posted by sofaspud View Post
Someone might have a more sophisticated watchdog circuit, but your idea is correct. The caps have to hold enough charge for the load+time, but I'm not sure what that is. Put a Schottky diode to the left of the cap bank to force the voltage to the DSP.
Is there a reason it needs to be a Schottky diode?
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