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DT800 and delay after digital output ON

Hi all.

This forums worth to spend time to learn by the experience of other users. I want to log data from sensors like flow meters and pressure transducers that will have to work on batteries (24 VDC).

So to save my batteries life, I want to sent a command to a Digital Output of the DT800, then wait for a couple of seconds to let the sensors stabilize and the log the measurements and go back to sleep.

The DT800 will have to shut down the Digital Output before going to sleep of course. I would like to know how to program this in my DT800. Is there a delay command? Any code or example would be appreciated.

Thanks a lot for any help.
Have a nice day.

Hi all. This forums worth to spend time to learn by the experience of other users. I want to log data from sensors like flow meters and pressure transducers that will have to work on batteries (24 VDC). So to save my batteries life, I want to sent a command to a Digital Output of the DT800, then wait for a couple of seconds to let the sensors stabilize and the log the measurements and go back to sleep. The DT800 will have to shut down the Digital Output before going to sleep of course. I would like to know how to program this in my DT800. Is there a delay command? Any code or example would be appreciated. Thanks a lot for any help. Have a nice day.

Good morning peterfreedom,

The DT800 has a delay command just for this.

DELAY=1000 will pause the operation of the D80/800 for 1 Sec.
In DeLogger4 the Delay is set by entering a number as the channel factor.

1DSO=1 'Turn on digital output 1
DELAY=1000 'deLay for 1 second
1V Take a reading
1DSO=0 'Turn off the output

Glad you like the forum

Cheers,
Roger

Good morning peterfreedom, The DT800 has a delay command just for this. DELAY=1000 will pause the operation of the D80/800 for 1 Sec. In DeLogger4 the Delay is set by entering a number as the channel factor. ```` 1DSO=1 'Turn on digital output 1 DELAY=1000 'deLay for 1 second 1V Take a reading 1DSO=0 'Turn off the output ```` Glad you like the forum Cheers, Roger

Hi!

I have connected a 24 VDC (60 mA coil) relay between a 24 VDC source and the Digital Output 1 of the DT800. I have wired the (-) of the power supply to the Gd terminal of the DT800 as explained in the manual. So the relay is activated by the Digital Output 1.

The problem is when 1 use:

RA10S LOGONA GA
1DSO=1
DELAY=1000
1V
1DSO=0

In reality the relay turns OFF with 1DSO=1 for 1 second and turns ON when 1DS0=0 for the rest of the time. So it is doing the contrary of what I expected. There is something I don't understand about the ACTIVE WHEN HIGH definition. Can you explain it to me?

I think that what happens is the FET turns ON (Conduct) when 1DSO=0, meaning there is O VDC at the Output. The FET turns OFF when 1DSO=1, meaning that there is 3.3 V DC thru a 47K Pull-Up resistor an d a 100 Ohms resistor.

I also tried to use a Solid State Relay (DC60S3 by Crydom). You can control them with only 3.5 to 32 VDC. But it never turns OFF. it takes 1 Vdc and below to turn OFF and the DT800 keeps a minimum of 2.7 V at the Crydom SS Realy. The DS60S3 has a 1500 ohms impedance and that could be to high.

I finally found how to make the SS Relay work. I put a 10K resistor in parallel with the input of the SS Relay. That keeps a minimum current and voltage drop. The bad thing is that you don't take advantage of the high impedance and low current consumption of a Solid State Relay.

I also tried to drive it directly with the output but the voltage drop across the 47 kOhms pull-up takes it all and leave me only 0.6 V to try to drive the SS Relay.

WHAT CAN YOU DRIVE THOUGH A 47K PULLUP? it gives you less than 70 uA for usage...

Hi! I have connected a 24 VDC (60 mA coil) relay between a 24 VDC source and the Digital Output 1 of the DT800. I have wired the (-) of the power supply to the Gd terminal of the DT800 as explained in the manual. So the relay is activated by the Digital Output 1. The problem is when 1 use: ```` RA10S LOGONA GA 1DSO=1 DELAY=1000 1V 1DSO=0 ```` In reality the relay turns OFF with 1DSO=1 for 1 second and turns ON when 1DS0=0 for the rest of the time. So it is doing the contrary of what I expected. There is something I don't understand about the ACTIVE WHEN HIGH definition. Can you explain it to me? I think that what happens is the FET turns ON (Conduct) when 1DSO=0, meaning there is O VDC at the Output. The FET turns OFF when 1DSO=1, meaning that there is 3.3 V DC thru a 47K Pull-Up resistor an d a 100 Ohms resistor. I also tried to use a Solid State Relay (DC60S3 by Crydom). You can control them with only 3.5 to 32 VDC. But it never turns OFF. it takes 1 Vdc and below to turn OFF and the DT800 keeps a minimum of 2.7 V at the Crydom SS Realy. The DS60S3 has a 1500 ohms impedance and that could be to high. I finally found how to make the SS Relay work. I put a 10K resistor in parallel with the input of the SS Relay. That keeps a minimum current and voltage drop. The bad thing is that you don't take advantage of the high impedance and low current consumption of a Solid State Relay. I also tried to drive it directly with the output but the voltage drop across the 47 kOhms pull-up takes it all and leave me only 0.6 V to try to drive the SS Relay. WHAT CAN YOU DRIVE THOUGH A 47K PULLUP? it gives you less than 70 uA for usage...

Good morning Peter,

  1. Active high.
    With most electronic circuits a high signal is a logic 1 (True) and a low is logic 0 (False) Because we have the system pulled high by default it is active when high. Your explanation is quite correct the FET conducts when DSO=0 and provides a path to ground for the power. With relays the state will depend on if the relay is normally open or normally closed. A normally closed relay will invert the logic.

  2. Wiring your Solid State Relay (SSR).
    Try connecting the SSR in parallel with the DT800 digital channel.
    From your power supply to drive the SSR use a 2200-Ohm resister to limit the current. Then connect to both the power input to 1 (+) on the SSR and to the digital channel on the DT800. Connect the Ground of the power supply to the digital ground on the DT800 and the 2 (-) on the SSR

Cheers,
Roger

Good morning Peter, 1. Active high. With most electronic circuits a high signal is a logic 1 (True) and a low is logic 0 (False) Because we have the system pulled high by default it is active when high. Your explanation is quite correct the FET conducts when DSO=0 and provides a path to ground for the power. With relays the state will depend on if the relay is normally open or normally closed. A normally closed relay will invert the logic. 2. Wiring your Solid State Relay (SSR). Try connecting the SSR in parallel with the DT800 digital channel. From your power supply to drive the SSR use a 2200-Ohm resister to limit the current. Then connect to both the power input to 1 (+) on the SSR and to the digital channel on the DT800. Connect the Ground of the power supply to the digital ground on the DT800 and the 2 (-) on the SSR Cheers, Roger
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