Relay switch to detect power source

[vikum123]

I need to find a switch that can detect whether it should go from 9v source or 5v Source. I am making a tool that has a battery power of 9v and it can also get powered from a USB device, which is 5V.
If the customer need to connect the tool to a computer while its running from the battery, will it harm the system? i was asked to use a relay switch. Please help.

 

[uart]

Is the 9V an internal source (eg battery) or an external source (eg "wall wart")? I might make a difference as to what choices you'd make.

 

[vikum123]

9V is coming from the Battery and 5V from USB

 

[uart]

BTW. The solution can be as simple as a schottkey diode in series with the USB +5V, depending upon the exact configuration.

 

[uart]

9V is coming from the Battery and 5V from USB

Ok, then the simple diode soln works (protects the usb), but has the disadvantage of continuing to drain the battery after the usb is connected. So no good. (that's the reason I needed to know how the 9V was sourced BTW).

 

[uart]

You can certainly detect the USB powersupply and use it to switch a relay, but there still might be some simple "relayless" options.

If your device is fairly low power (should be so if running from USB) then you could consider a simple voltage boost circuit, say 5V to 9.8V, and then a simple diode solution would select the highest available voltage automatically (without a relay).

Will the circuit that normally runs on 9V run satisfactorily on 5V anyway?

 

[vikum123]

Yes, its a Low power device. The system runs at 3.6V. I had to drop 5V to 3.6V and 9V to 3.6V.

 

[vikum123]

So what you are suggesting is to boost the 5V to 9V, so then there won't be a system damage and i don't need to use a relay switch?

 

[uart]

So what you are suggesting is to boost the 5V to 9V, so then there won't be a system damage and i don't need to use a relay switch?

No the 3.6V information adds new light to the problem. What circuitry are you using to drop the voltage, and is it common circuitry for both 5V and 9V? If not common then the solution is trivial, drop each to 4.3/4.4 volts and add series diodes (or 4.0/4.1 volts and add series Schotkey diodes).

 

[uart]

If not common then the solution is trivial, drop each to 4.3/4.4 volts and add series diodes

Sorry scrub that idea, it doesn't necessarily remove the loading from the battery (or only partially removes it). Unless it's feasible to boost the USB +5 to slightly higher than the battery then you may need to use a relay.

One other idea. Have you thought about powering it from 3x AAA 1.5V cells? This makes the USB voltage slightly higher than the battery voltage and passive diode switching/selection really does become trivial.

 

[vikum123]

I am using a voltage regulator to drop 5V to 3.6V and a Flyback to drop 9v to 3.6V. So its not a common circuit for both.

 

[vikum123]

I was thinking 3xAAA batteries too. But the requirement is to use a 9V alkaline battery.

 

[uart]

I am using a voltage regulator to drop 5V to 3.6V and a Flyback to drop 9v to 3.6V. So its not a common circuit for both.

Well you could replace the 5V to 3.6 regulator with a crude voltage boost circuit (probably doesn't need to be very well regulated as I assume your flyback has some regulation) and then just use the one common power supply (your 9V to 3.6V flyback) to power the device. Diodes can automatically select the higher of the two voltages (ie the boosted usb pwr if it's there) as the input to the flyback.

Take a serious look at my previous comment about using AAA batteries though (if space permits), because it might be a much simpler solution (depending upon how loose are the requirement for the 3.6 volts though).

 

[uart]

I was thinking 3xAAA batteries too. But the requirement is to use a 9V alkaline battery.

OK missed that before (posting at the same time).

 

[vikum123]

Thank you for your help. I will try what you suggest.

 

[saifnobel]

Can somebody recommend a relay for me?

Hi there,

I have to drive a 6V DC mini air pump motor for a portable blood pressure monitor from my PC. I am not sure about the current rating of this motor but I imagine it should be within 500 mA.

I want to drive the above motor using National instruments DAQ system (NI 9263). My DAQBook can give me +/- 10V supply. However, the current drive (output current load) of the DAQBook is only 1 mA.

Therefore, I will need a relay.

Can someone please suggest the right one for me? The smaller the better.

The many relays that I have seen online are bulky looking and have higher than 1 mA switching current requirements.

Thanks in advance!



Saif

 

[davenn]

Hi Saifnobel
welcome to the PF forums
its not a good thing to hijack some one elses thread

mods would you like to move post to his own thread please :)

Dave

 

[vikum123]

Well you could replace the 5V to 3.6 regulator with a crude voltage boost circuit (probably doesn't need to be very well regulated as I assume your flyback has some regulation) and then just use the one common power supply (your 9V to 3.6V flyback) to power the device. Diodes can automatically select the higher of the two voltages (ie the boosted usb pwr if it's there) as the input to the flyback.

Take a serious look at my previous comment about using AAA batteries though (if space permits), because it might be a much simpler solution (depending upon how loose are the requirement for the 3.6 volts though).

I tried to find a Crude voltage boost circuit but could not come up with that. Also i am wondering what do you mean by use a common power supply. the issue is, i have to use a USB to connect the device to the computer and the USB gives 5V. At the same time the system is running at 9V with a alkaline battery. I was asked to use a relay so then it will detect which power to use. Please Help me with this issue. Thank you