Looking for a CD4053BE sub with low "on" resistance...

[Dave]

I am trying to build something that uses a digital switching mechanism, but
I believe the 4053 I am currently using is blowing me out of the water with
it's 120 ohm "on" resistance. Is there a sub for this part with little or
no "on" resistance? I found a referrence on the net to an unnamed Analog
Devices part with a mere .4 ohms "on" resistance. I guess I could work with
an analog switch instead of a digital, but how do I find this part?

Many thanks,

Dave

 

[John Popelish]

I am trying to build something that uses a digital switching mechanism, but
I believe the 4053 I am currently using is blowing me out of the water with
it's 120 ohm "on" resistance. Is there a sub for this part with little or
no "on" resistance? I found a referrence on the net to an unnamed Analog
Devices part with a mere .4 ohms "on" resistance. I guess I could work with
an analog switch instead of a digital, but how do I find this part?

You have lots of choices, and analog switches are fine
possibilities. The main concern that limits your choices is
the power supply voltage you are using.

For example, if you are using a 5 volt supply you might be
able to make use of the version of the 4053 in the 74HC
family. Though from its maximum resistance, that may be
what you have already tried (though I generally assume that
a 4000 number implied the CD series).
http://www.fairchildsemi.com/ds/MM/MM74HC4053.pdf

What part are you using now, and what power supply and
signal range do you need?

 

[Dave]

You have lots of choices, and analog switches are fine possibilities. The
main concern that limits your choices is the power supply voltage you are
using.

For example, if you are using a 5 volt supply you might be able to make
use of the version of the 4053 in the 74HC family. Though from its
maximum resistance, that may be what you have already tried (though I
generally assume that a 4000 number implied the CD series).
http://www.fairchildsemi.com/ds/MM/MM74HC4053.pdf

What part are you using now, and what power supply and signal range do you
need?

Well hello John,

Thank you for the reply. And your assumption would be correct, I am using
the CD4053. For a power supply I am using a 9V battery, and would like to
continue with either that or perhaps something with AA batteries. From a
brief glance at the datasheet you so kindly gave me the link for, I am
thinking that perhaps a +5V regulator and a -5V regulator would be in order
.. The 30 ohm "on" resistance (I think that's what I saw) sounds a lot
better than what I am currently dealing with. You don't know of anything
even lower, do you?

Oh, and for signal range I am using the complementary Q and not-Q outputs
from a 4027 JK flip-flop. Nothing too extreme there. I believe that is
just +5V and ground.

Thank you again for the reply. I am planning a trip to the parts depot Sat.
AM, and your info is most welcome.

Dave

 

[Bob Masta]

I am trying to build something that uses a digital switching mechanism, but
I believe the 4053 I am currently using is blowing me out of the water with
it's 120 ohm "on" resistance. Is there a sub for this part with little or
no "on" resistance? I found a referrence on the net to an unnamed Analog
Devices part with a mere .4 ohms "on" resistance. I guess I could work with
an analog switch instead of a digital, but how do I find this part?

Many thanks,

Dave

What exactly are you trying to do? Often it is much simpler
to redesign the circuit so the "on" resistance is not an issue.
"On" resistance tends to be nonlinear, so if you are putting
any kind of a signal through it, you can have problems.
The solution typically involves buffering the switch output
with a high input impedance op-amp using (say) 100k to ground.
So the switch is the top leg in a voltage divider and 100k is
the bottom, meaning there is not much voltage across the
switch itself.

Best regards,





Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[John Popelish]

Well hello John,

Thank you for the reply. And your assumption would be correct, I am using
the CD4053. For a power supply I am using a 9V battery, and would like to
continue with either that or perhaps something with AA batteries. From a
brief glance at the datasheet you so kindly gave me the link for, I am
thinking that perhaps a +5V regulator and a -5V regulator would be in order
. The 30 ohm "on" resistance (I think that's what I saw) sounds a lot
better than what I am currently dealing with. You don't know of anything
even lower, do you?

Oh, and for signal range I am using the complementary Q and not-Q outputs
from a 4027 JK flip-flop. Nothing too extreme there. I believe that is
just +5V and ground.

Thank you again for the reply. I am planning a trip to the parts depot Sat.
AM, and your info is most welcome.

If your signal swings only between +5 and zero, then you
also only need a switch supply with that range.

However, if your signal is truly digital, why not use gating
instead of a switch? That way, the gate reproduces the
logic signal with its output, instead of passing it through
resistance.

This one has 4 selectors (similar to analog switches, but
makes logical copies of one or the other logical inputs),
but all 4 selectors are switched by a single input:
http://focus.ti.com/lit/ds/symlink/sn74hc158.pdf

 

[Fred Bartoli]

John Popelish a écrit :

If your signal swings only between +5 and zero, then you also only need
a switch supply with that range.

However, if your signal is truly digital, why not use gating instead of
a switch? That way, the gate reproduces the logic signal with its
output, instead of passing it through resistance.

This one has 4 selectors (similar to analog switches, but makes logical
copies of one or the other logical inputs), but all 4 selectors are
switched by a single input:
http://focus.ti.com/lit/ds/symlink/sn74hc158.pdf

For a pure 0/5V operation maxim has the nice MAX4617-18-19 which is pin
compatible, 10 times faster, low Ron, but has no neg supply.

 

[John Larkin]

Well hello John,

Thank you for the reply. And your assumption would be correct, I am using
the CD4053. For a power supply I am using a 9V battery, and would like to
continue with either that or perhaps something with AA batteries. From a
brief glance at the datasheet you so kindly gave me the link for, I am
thinking that perhaps a +5V regulator and a -5V regulator would be in order
. The 30 ohm "on" resistance (I think that's what I saw) sounds a lot
better than what I am currently dealing with. You don't know of anything
even lower, do you?

Oh, and for signal range I am using the complementary Q and not-Q outputs
from a 4027 JK flip-flop. Nothing too extreme there. I believe that is
just +5V and ground.

Why not just use an xor gate? Or a true digital mux?

John

 

[Ben Jackson]

I am trying to build something that uses a digital switching mechanism, but
I believe the 4053 I am currently using is blowing me out of the water with
it's 120 ohm "on" resistance.

You got other good advice. If you're not sure about the on resistance
as the problem, you could always stack another 4053 on the first one
(piggy-back) and cut Ron in half (at the expense of doubling all the pin
capacitances, but that might not matter to you).

 

[Dave]

If your signal swings only between +5 and zero, then you also only need a
switch supply with that range.

However, if your signal is truly digital, why not use gating instead of a
switch? That way, the gate reproduces the logic signal with its output,
instead of passing it through resistance.

This one has 4 selectors (similar to analog switches, but makes logical
copies of one or the other logical inputs), but all 4 selectors are
switched by a single input:
http://focus.ti.com/lit/ds/symlink/sn74hc158.pdf

Hey John,

Not sure what to say. My "signal" (on the Ain/out and the A1/A0 etc.of the
4053) is actually analog (thus my concerns about Ron), and the 74HC4053
(according to that datasheet) requires both a +5V (Vdd) and a -5V (Vee).
And (BTW) what is Vss?

Someone told me that using a 7805 "backwards) to simulate -5V can cause
flaky problems, so I am thinking I need a 7805 *and* a 7905. (Can these
share a common ground?) Or should I use a real powersupply with a 12.6V CT
transformer and associated hardware to provide my voltages?

Sigh Sorry to be so obtuse, but CMOS is still pretty new to me. Hate to
admit it, but it's obvious. I'm not sure what I'm doing, or how I am doing
it.

Best regards,

Dave

 

[Dave]

John Popelish a écrit :

For a pure 0/5V operation maxim has the nice MAX4617-18-19 which is pin
compatible, 10 times faster, low Ron, but has no neg supply.

Hey Fred,

What do you mean, no neg supply? I'm lost with that statement. Can you
illuminate?

Thanks,

Dave

 

[Dave]

What exactly are you trying to do? Often it is much simpler
to redesign the circuit so the "on" resistance is not an issue.
"On" resistance tends to be nonlinear, so if you are putting
any kind of a signal through it, you can have problems.
The solution typically involves buffering the switch output
with a high input impedance op-amp using (say) 100k to ground.
So the switch is the top leg in a voltage divider and 100k is
the bottom, meaning there is not much voltage across the
switch itself.

Best regards,





Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

Hey, here we go. I may need to hear more about this type of operation.
What is this technique called, and whre can I find more info about it?

Many thanks for the interest,

Dave

 

[Dave]

You got other good advice. If you're not sure about the on resistance
as the problem, you could always stack another 4053 on the first one
(piggy-back) and cut Ron in half (at the expense of doubling all the pin
capacitances, but that might not matter to you).

Thank you for the idea, Ben. I may use that. Thank you very much...

Dave

 

[Fred Bartoli]

Dave a écrit :

Hey Fred,

What do you mean, no neg supply? I'm lost with that statement. Can you
illuminate?

Thanks,

Dave

The 405x series can handle positive and negative signals because some
portions of the internals are biased through VEE which you can either
tie to ground or make negative. The full voltage range is then from
VEE(or GND) to VDD.

For the max461x you don't have the VEE pin (not connected) and the
signal range is therefore limited to GND-VDD.

 

[Dave]

Dave a écrit :

The 405x series can handle positive and negative signals because some
portions of the internals are biased through VEE which you can either tie
to ground or make negative. The full voltage range is then from VEE(or
GND) to VDD.

For the max461x you don't have the VEE pin (not connected) and the signal
range is therefore limited to GND-VDD.

Oh how cool. Will have to check that out. Thank you.

Dave

 

[John Popelish]

Not sure what to say. My "signal" (on the Ain/out and the A1/A0 etc.of the
4053) is actually analog (thus my concerns about Ron), and the 74HC4053
(according to that datasheet) requires both a +5V (Vdd) and a -5V (Vee).
And (BTW) what is Vss?

Vss is the negative logic supply rail. The control logic
operates between Vss and Vdd. The analog signals can swing
between Vee and Vcc, though Vee can be connected to Vss if
the analog signals stay between the logic rails. But
reducing the total difference between Vdd and Vee from 10
volts to 5 volts almost doubles the switch on resistance
(from something like 4o ohms to something like 70 ohms).

Someone told me that using a 7805 "backwards) to simulate -5V can cause
flaky problems, so I am thinking I need a 7805 *and* a 7905.

Much better idea.

(Can these share a common ground?)
Yes.

Or should I use a real powersupply with a 12.6V CT
transformer and associated hardware to provide my voltages?

To use a 7805 and a 79o5 to produce positive and negative 5
volt rails, you will need a raw positive and negative 7 to 9
volts, referenced to the same ground. A 12.6 Vct secondary
will produce about positive and negative 8.8 volts, peak,
minus a diode drop or about 8 volts, so it may work fine
with the two regulators.

Sigh Sorry to be so obtuse, but CMOS is still pretty new to me. Hate to
admit it, but it's obvious. I'm not sure what I'm doing, or how I am doing
it.

You are getting there.

 

[Dave]

Vss is the negative logic supply rail. The control logic operates between
Vss and Vdd. The analog signals can swing between Vee and Vcc, though Vee
can be connected to Vss if the analog signals stay between the logic
rails. But reducing the total difference between Vdd and Vee from 10
volts to 5 volts almost doubles the switch on resistance (from something
like 4o ohms to something like 70 ohms).


Much better idea.


To use a 7805 and a 79o5 to produce positive and negative 5 volt rails,
you will need a raw positive and negative 7 to 9 volts, referenced to the
same ground. A 12.6 Vct secondary will produce about positive and
negative 8.8 volts, peak, minus a diode drop or about 8 volts, so it may
work fine with the two regulators.


You are getting there.

Okay, one more stupid question. Do you think I could run both the 7805 and
the 7905 off of the same 9V battery? Probably wouldn't last long if I did,
though. Still... I have to ask.

Thank you for your help. It is appreciated.

Regards,

Dave

 

[John Popelish]

Okay, one more stupid question. Do you think I could run both the 7805 and
the 7905 off of the same 9V battery? Probably wouldn't last long if I did,
though. Still... I have to ask.

Can't do it. you would need two in a stack, with the center
point as ground, and the two ends as plus and minus 9 volts
as inputs to the regulators. That would run for a few hours.

 

[Bob Masta]

Hey, here we go. I may need to hear more about this type of operation.
What is this technique called, and whre can I find more info about it?

Many thanks for the interest,

Dave

I don't know of a specific name, it's just one of the ways to deal
with nonlinear analog switches. If you will explain what it is you
are trying to do, we can probably give more specific advice.

BTW, in answer to your question about running positive and negative
regulators off one 9V battery, you may not need any regulator at all,
just a "ground splitter". This is a non-inverting op-amp buffer whose
positive input is tied to a voltage divider that provides 4.5V from
the 9V battery. The output of the amp (which is also connected to
the inverting input to make it a buffer) drives the circuit ground.
If your ground current needs are not too high, this may be all you
need, or else you may need some output transistors to provide some
extra current capability.

The above won't be regulated, but CMOS doesn't care about that as
long as the voltage is in a (fairly wide) useable range.

Best regards,



Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!