I don't see any advantage.

 

I am thinking that the CD 4049 has the greater pull-down capability rating.

 

Or is there an advantage that you think that IC might provide?

You only need one CD40107 dual-gate chip since it has a high sink-current capability (68mA @ 1V out with 10V supply).


.

 

The CD40107 is the same as using a discrete mosfet.

 

OK, and it seems that my guess was incorrect. I tend to use the 4049 because we always had them on hand, and they cost less.

 

The CD40107 is the same as using a discrete mosfet.

You're correct, it is the same as using two MOSFETs, but it comes in a single 8 pin package.

 

You're correct, it is the same as using two MOSFETs, but it comes in a single 8 pin package.

HUGE real estate savings compared to two 4049s!

However, if I'm understanding the datasheet correctly, the 40107s would require a pull-up resistor on each output and a diode parallel to each parallel coil set coil in each relay. So, 6 4 2 additional components. Is that correct?

 

No pull up resistors required.
Not sure about the diodes.

 

No pull up resistors required.

Yeah, just caught that: "no external resistor is required if [...] VDD < 15V".

That datasheet leaves much to be desired. :shrug:

 

a diode parallel to each coil in each relay.

Only requires one diode for the two coils driven in parallel, so two diodes total.
The diodes can be placed at each 40107 output (diode anode) to V+.

 

Only requires one diode for the two coils driven in parallel, so two diodes total.

We need a face-palm smilie... I should have known that.

 

No diodes needed if using discrete mosfets.

 

The diodes can be placed at each 40107 output (diode anode) to V+.

Just to remove any ambiguity, that means a diode between pin 3 (cathode) and pin 8 (anode) and another between pin 5 (cathode) and pin 8 (anode). Is that correct?

Forgive my noobness. I'm still learning the nomenclature. "V+" in this instance is the same as "VDD."

 

The turn off is still not instant and so there will be no generation of any spike due to the instant cutoff of coil current, because the current decays slowly as the capacitor charges.

 

The turn off is still not instant and so there will be no generation of any spike due to the instant cutoff of coil current, because the current decays slowly as the capacitor charges.

Not using a CD4017.

 

Still using a series capacitor to provide a short pulse to operate the relay. And the IC is a CD40107, an inverter, not a counter.

 

there will be no generation of any spike due to the instant cutoff of coil current, because the current decays slowly as the capacitor charges.

That may be true for a single MOSFET, but not necessarily for a CD4000B series device, which has a much higher gain due to having at least two stages from input to output.

Below is the I/O transfer function for 5V and 15V Vdd from an RCA app report
Note that it takes a only a few hundred mV change on the input for the output to go from Vcc to 0V.

On that basis I think the output could transition fast enough to cause a transient spike from the relay coil inductance, even with the slow input RC fall time.
A couple added diodes is cheap insurance against a possible damaging spike.

 

Schematic from post #21.
View attachment 336464

Here is a simulation of Sghioto's circuit.
I added the diodes because they are recommended by the manufacturer because the coils are wound on the same core.

Also, be aware there is a state where both Q and Qbar are high at the same time.

 

I added the diodes

Those diodes do not suppress the inductive transient, which generates a positive spike at the MOSFET's drain.
You need only two diodes, one connected from each MOSFET drain (diode anode) to VDD.

 

Those diodes do not suppress the inductive transient, which generates a positive spike at the MOSFET's drain.

They are meant to suppress inducing a current in the opposite coil.
The app note excerpt is below.