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 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.
karesz said:Hi,
Is it possible for you to upload a schema pls?
I think in your system is some thing wrong...
K.
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.
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.
As we are writing it w. Betwixt too, but "the diode" can be any semiconductor junction in some component too...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.
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.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?