Hello,
I need to use about 60 CD4051s in my circuit for analogue signal. The main supply is -12V +12V that is used for other ICs. The problem is that CD4051 can't accept more that 20v between negative and positive supplies. So I was thinking about using a voltage regulator but I can't see on the datasheet how much current the CD4051 needs (the datasheet mention only the quiescient current). Then I would be able to know output current specs for the regulator to supply the 60 CD4051s.
Other "solution" would be using a resistive divider on the main supplies to get something like -7V +7V. Is there any inconvenient ? impedance problems or other?
So how can I know the current supply needed for the CD4051 (if on the datasheet I didn't find it).
And is the resistive divider a good alternative for so much ics ?
Thanks for your help!
datasheet:
https://www.ti.com/general/docs/sup...=26&gotoUrl=http://www.ti.com/lit/gpn/cd4051b

 

Hello,

Have a look at figure 6 at page 9 for the power dissipation of the chip.

Bertus

 

It is Cmos with a very low quiescent current of 10uA max.
You cannot simply use one resistor to +12 to make +7V and another resistor to -12V to make -7V because you do not know their minimum current which will cause their supply voltages to be too high.

Use a voltage divider to make +7V and another voltage divider to make -7V. Calculate the resistors for ten times the ICs max current (10 x 60 if them x 10uA=) 6mA for each divider. The +7V and -7V will each need a decoupling capacitor to ground.

 

Have you considered the DG408 instead of the 4051? It will manage ±12V supplies. It also has lower distortion than the 4051.

 

Thank you all for your replies!
The DG408 seems to have better specs but dissipates too much power and not sure sure it can manage -12+12V , maximum V- to V+ is 18V. It is more expensive too. So for the moment I prefer to stick with the CD4051 at least for testing especially If the 60 CD4051 needs only 6mA.

 

You can't draw appreciable current from a voltage divider, and certainly not a varying current. The first picture shows a voltage divider to make 7 volts from 12 volts. R1 = 13k and R2 = 18k to make ~7 volts at the junction, with no load being drawn.



What happens if we try to supply 10mA from this divider? R = V/I = 7/0.01 = 700 ohms. This picture shows the result. The output voltage has dropped to 0.6v!



Ok, what if we drop R1 & R2 by a factor of 10? The output voltage is closer to where we want it, but still far from the desired 7 volts.



Hmmm. Let's try cutting R1 and R2 by another factor of 10, making R1 =130 ohms and R2 180 ohms? Ok, that's better. The output voltage will be 6.3 volts with a 10mA load.



So this could work. Or could it? The picture below shows the total current drawn by the voltage divider – 44mA to get 10mA out. And over half a watt power dissipation.



Voltage dividers are not all-together practical which appreciable current will be drawn; a lot of power is wasted as heat.

 

I would want the safest solution that works. For that a 79L05 might be the best answer.

 

Thank you all for your replies!
The DG408 seems to have better specs but dissipates too much power and not sure sure it can manage -12+12V , maximum V- to V+ is 18V. It is more expensive too. So for the moment I prefer to stick with the CD4051 at least for testing especially If the 60 CD4051 needs only 6mA.

To quote from the datasheet:
"Wide supply ranges - Single supply: +5 V to 36 V - Dual supplies: ± 5 V to ± 20 V"
Current requirements <500uA at room temperature. 60 off DG408 is 30mA. Your resistive dividers use more than that.

Or alternatively, TI do some low noise, low-quiescent-current regulators with part numbers that start TPS7A. They need only microamps.

 

Thank you so much, I am very grateful for all your answers.
@Ian0 sorry I must have read wrong datasheet. I think I 'll use DG408 in future projects that don't need that much ICs.
@JonChandler, thanks for your time and explanation, it was very pedagogical and helpfull, I really appreciate.
So I think I will take the voltage regulator solution (thanks to @DickCappels to point at the 79L05)

 

If low current is important, the there are many voltage regulators with lower quiescent current that the 78L/79L series.

 

I said to use a divider that uses 6mA. The 13k and 18k divider uses only 0.39mA so of course its voltage drops when loaded with 0.6mA.

What is the very low resistance of 700 ohms? The CD4051 draws only 10uA and all of its output current comes from its input, not from its power supply.

 

My post was an example to explain why voltage dividers don't make effective power supplies. The idea that significant current can be drawn from a voltage divider seems to be a common fallacy here (i.e., "I want to make 5 volts from 12 volts to charge my phone using a voltage divider.").

 

How about using a 7808 and a 7908 (for +/- 8v) or a 7809 and a 7909 (for +/- 9v), both within spec for the CD4051?