very simple question about grounds in circuits...

[laurah]

Thanks Brian. Could you please explain in more detail what connecting the transistor ground to the CMOS ground will exactly do? I am interested!

EDIT: sorry, i meant connecting the negative supply to the CMOS ground...

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.

Added after 4 hours 19 minutes:

I simulated a similar circuit (substituting a NAND gate for my PIC) in NI MultiSim and it is reporting that if I connect the negative terminal to VDD, I will get a constant drain from VDD on the CMOS to the negative terminal of about 4.5mA (plus any current from the load). Does this sound right??

 

[betwixt]

Normally, you would connect VSS to negative and VDD to positive but it depends on the exact design you have.

Before answering your main question, please tell me:

1. is the CMOS board powered from AC or DC?
2. is the same power source being used to power the relay or are you using two separate power supplies?

Which area are you in, if our time zones are similar it may be easier to 'live' chat.

Brian.

 

[laurah]

Normally, you would connect VSS to negative and VDD to positive but it depends on the exact design you have.

Before answering your main question, please tell me:

1. is the CMOS board powered from AC or DC?
2. is the same power source being used to power the relay or are you using two separate power supplies?

Which area are you in, if our time zones are similar it may be easier to 'live' chat.

Brian.

Sorry, I meant to say VSS

I am using a multi-tap transformer as the main power source.
- 15VAC from this is rectified/regulated down to 12VDC. This goes into the dev-board that includes the CMOS. This 12VDC is rectified/regulated on the dev-board down to 5VDC to power the PIC etc.
- The 12VDC is also used across the relay's coil.
- 24VAC is the voltage that the relay is switching.

Opposite time zones unfortunately! But thanks for the offer.

I have one more question if you don't mind... is there any problem with having two relays (controlled independently) which switch the same 24VAC source to two different loads?

 

[betwixt]

Thanks for the explanation, the situation is clearer now.

This is how I would tackle the problem:

Your Dev board obviously has on board rectifiers so it can accept an AC supply but you are feeding it DC anyway. This means the rectifiers are not really serving any useful purpose and the ones on the ground side of the bridge are responsible for the boards ground being 0.7V above the power supply ground. If possible, short out the bridge rectifier so the negative side of your 12V is connected directly to the boards ground. If that isn't possible, provide a wire link between supply ground and the board ground at the boards ground terminal. This effectively shorts the diode out but externally to the dev board.

With only one ground, you don't have to worry about the voltage drop as the transistors emitter pin and dev board ground will be at the same potential and both returned to the same point at the power supply.

Switching more than one load is not a problem. You can switch as many as you like until the power supply can handle no more. Bear in mind that as well as the current being drawn by your load, you have to add the current drawn by all the relay coils as well when you work out the total load. Use a transistor to drive each relay like you do at the moment but drive the base from different signals from the dev board if you want to control them individually.

Brian.

 

[laurah]

That's great, thanks for that. What do you make of the 4.5mA flow when the the two grounds are connected? Does that make sense and how would I have calculated that without the use of Multisim?

Cheers!

Thanks for the explanation, the situation is clearer now.

This is how I would tackle the problem:

Your Dev board obviously has on board rectifiers so it can accept an AC supply but you are feeding it DC anyway. This means the rectifiers are not really serving any useful purpose and the ones on the ground side of the bridge are responsible for the boards ground being 0.7V above the power supply ground. If possible, short out the bridge rectifier so the negative side of your 12V is connected directly to the boards ground. If that isn't possible, provide a wire link between supply ground and the board ground at the boards ground terminal. This effectively shorts the diode out but externally to the dev board.

With only one ground, you don't have to worry about the voltage drop as the transistors emitter pin and dev board ground will be at the same potential and both returned to the same point at the power supply.

Switching more than one load is not a problem. You can switch as many as you like until the power supply can handle no more. Bear in mind that as well as the current being drawn by your load, you have to add the current drawn by all the relay coils as well when you work out the total load. Use a transistor to drive each relay like you do at the moment but drive the base from different signals from the dev board if you want to control them individually.

Brian.

 

[betwixt]

I'm not sure where Multisim gets the current flow from. My guess would be that it assumes 0.7V across the rectifier and 0.6V across the transistor B-E junction so it calculated 0.1V across Rb (the base series resistor) and hence the current through it.

Without knowing the full simulation parameters it's difficult to be sure through. I have to honest, I rarely use simulators, I'm more of a pocket calculator person. The results are usually the same, at least in DC circuits where things tend to be very predictable.

Brian.

 

[laurah]

Maybe for someone with your knowledge!

Thanks again to everyone who helped with this.

I'm not sure where Multisim gets the current flow from. My guess would be that it assumes 0.7V across the rectifier and 0.6V across the transistor B-E junction so it calculated 0.1V across Rb (the base series resistor) and hence the current through it.

Without knowing the full simulation parameters it's difficult to be sure through. I have to honest, I rarely use simulators, I'm more of a pocket calculator person. The results are usually the same, at least in DC circuits where things tend to be very predictable.

Brian.

 

[VSMVDD]

cmos>>68R>>..|>1n4148... base

add a 1k res between 1n4148 cathode and ground

the 68R resistor limits any current from the cmos i/o more than desired

the diodes gets rid of .7v

the 1k to ground
allows for the right bias lift up from diode cathode

use a bd139 in this circuit

 

[laurah]

Thanks! That's an interesting solution...

cmos>>68R>>..|>1n4148... base

add a 1k res between 1n4148 cathode and ground

the 68R resistor limits any current from the cmos i/o more than desired

the diodes gets rid of .7v

the 1k to ground
allows for the right bias lift up from diode cathode

use a bd139 in this circuit