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very simple question about grounds in circuits...

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laurah

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Hi, I have a development board powered by 12VDC that has a CMOS 5V output pin that I can control.

I want to use the 5V output to switch on a 12V relay. So I am using a transistor to avoid drawing too much current from the CMOS to switch on the relay. My problem is that the CMOS 0V is 0.7V higher than the negative terminal of the 12VDC supply.

How do I reconcile this? Do I connect the emittor of the transistor to the 0V of the CMOS or to the negative terminal of the 12VDC supply?

Many thanks, sorry if this is a trivial question.
 

betwixt

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The simplest solution is to add a diode in series with the base resistor so a further 0.6V is dropped from the CMOS output. That would ensure the transistor switched off when the CMOS output was low. Alternatively, you could use a darlington transistor which needs more B-E voltage anyway.

Be careful with split ground circuits though, they are not usually a good idea. Be sure the noise between them is kept as low as possible and be particularly careful if they can be isolated, for example through a power switch. You could get almost any voltage or polarity between them and that really would cause you problems!

Brian.
 

    laurah

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laurah

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Thanks, that sounds like a good solution. So I would then connect the transistor emitter to the ground of the supply rather than the "0V" of the CMOS?



betwixt said:
The simplest solution is to add a diode in series with the base resistor so a further 0.6V is dropped from the CMOS output. That would ensure the transistor switched off when the CMOS output was low. Alternatively, you could use a darlington transistor which needs more B-E voltage anyway.

Be careful with split ground circuits though, they are not usually a good idea. Be sure the noise between them is kept as low as possible and be particularly careful if they can be isolated, for example through a power switch. You could get almost any voltage or polarity between them and that really would cause you problems!

Brian.
 

matbob

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Hi laurah,

If you are connecting the CMOS output to a transistor driving the relay, then you should have the grounds of both the CMOS power supply and relay driving power supply connected together. This is because the output of the CMOS chip is with respect to its ground and if you connect this simply to the base of the transistor then with respect to the 12V supply this voltage is floating ie, it is not referenced to 12V supply's ground. So you need to connect the CMOS circuit's ground and 12V supply's ground and also the emitter of the transistor together.

I would prefer an optocoupler in between for isolation.
 

    laurah

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laurah

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Thanks for your reply. If I connect the two grounds together then won't there be a constant flow of current from the CMOS (the CMOS "0V" is 0.7V higher than the power supply 0V)



matbob said:
Hi laurah,

If you are connecting the CMOS output to a transistor driving the relay, then you should have the grounds of both the CMOS power supply and relay driving power supply connected together. This is because the output of the CMOS chip is with respect to its ground and if you connect this simply to the base of the transistor then with respect to the 12V supply this voltage is floating ie, it is not referenced to 12V supply's ground. So you need to connect the CMOS circuit's ground and 12V supply's ground and also the emitter of the transistor together.

I would prefer an optocoupler in between for isolation.
 

matbob

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Hi laurah,

You wrote : 'CMOS 0V is 0.7V higher than the negative terminal of the 12VDC supply' . Please explain.

Are you using the same 12VDC available in you development board for the relay? Is the power supply for the CMOS section obtained from the same 12VDC?

If answer to above questions are affirmative then you needn't connect both grounds together and you may use the technique posted by betwixt.
 

    laurah

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betwixt

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My solution will work if the CMOS ground is always 0.7V above power ground but I'm concerned why you have two different grounds anyway.

To turn the relay on you have to supply more than 0.6V between the transistor base and emitter. You can do that by connecting the emitter directly to the CMOS ground so it shares the same 0V reference. The trouble is that the relay current is now conducting from 12V to the CMOS ground rather than the power ground and that might cause you problems. For better advice, please explain why you have two grounds instead of one.

Brian.
 

dick_freebird

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It seems to me that you want to better understand the ground
situation, its consistency and whether the CMOS ground
is an appropriate return for the 12V current. Maybe your
dev board has separate ground domains, or maybe you just
forgot to cross-strap the power supplies and "CMOS ground"
is really an ESD diode pulling from some other source
(0.7V is just too convenient an offset).

If in doubt, an optoisolator might be a better relay interface.
Add current-gain if required.
 

    laurah

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laurah

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I am using a multi-tap transformer as a power supply. I have rectified and regulated oneof the taps to generate a 12vdc steady power supply. I did this because my dev kit takes 9-12vdc that it then rectifies/ regulates down to 5v to power a programmable IC. This IC has a bunch of input/output pins. In one block, there are 8 IO pins plus 2 other pins: a Vcc and a 0v. I use the multimeter to measure the voltage between Vcc and 0v and I get 5v as expected. I then used the multimeter to measure the voltage between the negative terminal of my little 12v power supply and the 0v pin expecting it go
be pretty much 0v however it was +0.7v... I didn't build the dev board so I don't know what is going on with it...

I wanted to use these IO pins as logical 1s and 0s to switch on/off a 12v relay. Because the relay pulls quite a bit of current to energies it's coil, I didn't want to burn out the programmable IC on the dev board. So I want to use a transistor to handle the current.

My question is: which ground do I use for the emittor of the transistor?

Hope that is clearer! Thanks to all who are helping me
 

karesz

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Hi,
Is it possible for you to upload a schema pls?
I think in your system is some thing wrong...
K.
 

    laurah

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laurah

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I don't have one sorry :(

karesz said:
Hi,
Is it possible for you to upload a schema pls?
I think in your system is some thing wrong...
K.
:cry::cry:
 

betwixt

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Is your board home-made or a built unit?

If it is built, does it have a part number or manufacturers name?
Which country are you in?

It might be possible to trace the information you need. I fear you might be returning your relay coil current through a digital output set to low rather than real ground.

Brian.
 

laurah

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betwixt said:
Is your board home-made or a built unit?

If it is built, does it have a part number or manufacturers name?
Which country are you in?

It might be possible to trace the information you need. I fear you might be returning your relay coil current through a digital output set to low rather than real ground.

Brian.

It doesn't unfortunately. If it is the case that it is a digital output set to low, what do I do about it?

Added after 22 minutes:

betwixt said:
Is your board home-made or a built unit?

If it is built, does it have a part number or manufacturers name?
Which country are you in?

It might be possible to trace the information you need. I fear you might be returning your relay coil current through a digital output set to low rather than real ground.

Brian.

ahh... So I have a case of analog ground (AGND) vs digital ground (DGND)! Which one do I use?
 

karesz

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Thes "different GND" potentials of special 0.7V is "too special" for me!!
Are your supply voltages all correct_with load too, can you check it with a oscilloscope pls?
Seems me as is a diode in the bridge(if you have it) is dead_or the GND-connections arnt OK!?
I think you have to check for a clean Groundig as first "to do"!
K.

Added after 5 minutes:

AGND & DGND are to arrange at a "Star point", usual at the ADC input/refvoltage GND, or at the most sensitive part...
K.
 

betwixt

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As Karesz says, if it really is a case of analog or digital grounds, they should still be connected together and therefore both be at the same DC potential. The reason for having two in this case is so that the 'noise' from digital switching currents is kept away from analog signals where it could appear as interference.

The 0.7V difference could come from several sources, it may be coincidence it is the same as the voltage dropped across a diode junction or it may be the result of current 'pulling' a digital low signal up slightly. We are being cautious about the solution because if you do have the current returned through a digital output, you are likely to cause damage to it.

Can you post a photograph of it?

Brian.
 

matbob

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Hi laurah,

I think you have a bridge rectifier at the DC input of the development board. This is to avoid the polarity reversal by mistake when you provide the DC input. So the bridge may be creating a drop of 0.7 volts due to the diode conduction and when you measure the CMOS ground wrt your DC source's ground, you see 0.7 volts.

To find whether my assumption is right, please check for any series diodes or bridge rectifiers at the DC input section.
 

kabiru

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I think it will be better for you to use opto isolator.
 

karesz

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matbob said:
... drop of 0.7 volts due to the diode conduction and when you measure the CMOS ground wrt your DC source's ground, you see 0.7 volts.

To find whether my assumption is right, please check for any series diodes or bridge rectifiers at the DC input section.
As we are writing it w. Betwixt too, but "the diode" can be any semiconductor junction in some component too...
Than:
its "A" possibility, but ALL THE GNDs ARE TO ARRANGE AT ONE "STAR POINT" IN THE SYSTEM_in other case you have anytimes such diodes they are -and will be- uncontrollable!:-(...
@kabiru,
Optoisolators are often a muss for good signal separations, but they need exactly a clear _and functioning_ GND-System too:)
K.
 

laurah

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matbob said:
Hi laurah,

I think you have a bridge rectifier at the DC input of the development board. This is to avoid the polarity reversal by mistake when you provide the DC input. So the bridge may be creating a drop of 0.7 volts due to the diode conduction and when you measure the CMOS ground wrt your DC source's ground, you see 0.7 volts.

To find whether my assumption is right, please check for any series diodes or bridge rectifiers at the DC input section.

ahh - yes there is a bridge rectifier - the circuit will accept AC!

OK - so this is the source of my 0.7V drop? If so, what do I do about it!!!

Thanks for all the help people, i really appreciate it
 

betwixt

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Assuming the bridge rectifier will carry the extra relay current, you should be OK to return the transistor ground to the CMOS ground. If your supply is DC it should be safe to short out the CMOS ground to the negative supply, effectively shorting the diode out, then both supplies will be really zero volts.

Brian.
 

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