Voltage Gain Switch (Design Help)

MOSFETs don't draw any dc gate current, and you aren't trying to get them to switch fast. A 100k resistor and 100 nF capacitor make a time constant of 10 ms, corresponding to a 3 dB corner frequency of 16 Hz.

If you put a 100k resistor from the logic gate's output to the MOSFET gate, with a 100 nF capacitor from the gate to ground, that 16-Hz rollof will get rid of most of the noise that otherwise might couple from the logic output to your analogue path.

|---- 100k | | Logic 0----RRRRRRRR---x------| |

Reply to
Phil Hobbs
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Pity they don't mention the most useful low-cost parts.

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 Thanks,
    - Win
Reply to
Winfield Hill

. Gain switching the feedback dividers eliminates effects . __ from the CMOS switch's series resistance. . ------|+ \\ . | >-----+----+----+--- . .--|-_/ R1 | | . | | R3 | non-inverting gain . '---o-->o----+ | R5 . o--- | ---+ | . o--- | -- | ---+ . | | | . R2 R4 R6 . | | | . GND --+----+----'

And if Gmin = 2, i.e., R1 => R2, etc., then a +/-5V switch can be used with +/-10V output signals.

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 Thanks,
    - Win
Reply to
Winfield Hill

This arrangement very nicely eliminates the effects of switch resistance, but output will swing to the rails when all switches are open (if that is allowed to happen) and some opamps might not like the high input voltage resulting when a feedback switch is then closed, so opamp input protection diodes may be advisable. Arrangements which switch parallel resistors in the input and/or feedback can be done which never open the feedback loop, but then the switch resistance shows up as an error term. So the designer seems to have a choice between a low glitch design with switch resistance errors or a potentially high glitch design without them - but a make before break switching arrangement could eliminate the switching glitches which will otherwise occur with the above ckt.

Reply to
Glen Walpert

In article , Winfield Hill wrote: [....]

Also:

You can use a series string of resistors to save on resistor count.

If you use a 4052, one divider can be on the input side and one in the feedback path. You hook two switch inputs to each resistor step. You get

4 levels with only 4 resistors but your swing ends up limited by one input connection.

Also also:

If you really have to have a large input swing and a large output swing in the inverting mode, you can use the 4051 to apply a positive feedback. This only works for narrow groups of gain values since the inaccuracy of the resistors gets amplified.

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Reply to
Ken Smith

See....

formatting link

Took awhile. I had to collect my works for On-Semi's analog switches into a "blind" part that didn't divulge what I designed for them at the device-level.

I am sure there are better 1% values, but I just "slung" it together.

The 16-Bit-Step is just my test part to generate digital words.

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

Phil Hobbs ha escrito:

it

So in short I shouldn`t use it. What do you mean when you say "Big Signals" ? I created a new post about this part of the circuit, the PNP switch I was bound to do drove me insane for an hour, thought it deserved it`s own post, the title is: Turning on a MOSFET from 0-5V.

I forgot to ask before about the Bandwidth change with the attenuator, why suddenly when one MOSFET is ON the other's capacitance doesn`t matter ?

I also wonder about the OFF-state of the MOSFETs. The datasheet states the drain current with VGS=3D0V and VDS=3D48V to be 1mA Max at Tj=3D125=BAC, don`t know about the impedance, but the current at much lower VDS should be significantly smaller, right ?

Thanks.

Reply to
leo2100

Winfield Hill ha escrito:

An interconnection of ASCII characters is worth a million words :)

Thanks.

Reply to
leo2100

Glen Walpert ha escrito:

So what you are trying to say is that, when the above circuit switches the feedback, there might be some time of inestability in the circuit. If that's what you meant there is no problem, there were relays before that took forever to change the feedback, there's no problem with that kind of inestability. How's this make before break circuit you mention ?

Reply to
leo2100

Ken Smith ha escrito:

I don`t understand the positive feedback arrangement. If you could work out an example that would be great.

Reply to
leo2100
[snip]

Did you miss my post...

formatting link

???

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.
Reply to
Jim Thompson

Jim Thompson ha escrito:

Thanks for the effort. The plot's are particulary useful, I need to get PSpice.

Reply to
leo2100

In article , wrote: [.....]

ASCII art: R1 R2 10K 10K ---/\\/\\/------+-------/\\/\\---------- ! ! ---!-\\ ! ! >------------+-+------- ---!+/ ! ! \\ ! / 10K ! \\ ! ! ! A O-----+ ------------> ! O \\ ! / 1K GND \\ ! GND

When the switch is at ground, the circuit is simple. Just ignore the extra stuff and you have a gain = -1

In the other case, if Vout = 1V, the voltage at A=0.1 so R2 has 0.9V on it and thus the input must be -0.8V

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

I think I'm about done with this, actually. With the amount of help you've had from some pretty high-class engineers, you have lots of things to choose from. Wire a few up and see how they work!

Cheers,

Phil Hobbs

Reply to
Phil Hobbs

That doesn't Glen's address transient situations, does it? But yes, I agree there isn't really a problem. The hc405x series in fact switches so fast that ordinary opamps don't have time to respond to open-switch situations, as they might with other slower high-voltage, breakdown-protected CMOS switches. While Glen's comments are on target for many if not most traditional analog switch families, the hc405x switch family was first and foremost a logic switch, with low-resistance low-capacitance MOSFETs, and it responds rapidly, appropriate to that task.

These are also readily-available low-cost parts, and attractive if they're buried deep inside a circuit, where static discharge and low maximum-operating voltages may not be an issue, like Leo's amplifier.

That said, there's still the issue of charge injection, where an output pulse appears from the transient current injected into the opamp's feedback loop from the MOSFET-switch gate-voltage swing. Using low-value feedback divider resistors minimizes this affect. Using low-capacitance switches also helps, but these are slow, and therefore can exacerbate the open-switch problem.

--
 Thanks,
    - Win
Reply to
Winfield Hill

Phil Hobbs ha escrito:

Well everyone thanks, I think I`m gonna follow Phil's suggestion and learn the good ol' way, blowing stuff up. I`m gonna limit myself to concrete questions, afterall from the experience I`ve got, all this might be too much information, which I`m not sure I`m ready for. But there sure is some top quality information, talk about information overload.

Reply to
leo2100

On 18 Sep 2006 16:23:29 -0700, snipped-for-privacy@gmail.com wrote in Msg.

No, you need to get LTSpice which is available for free from Linear Technology. Or if you're on a Unix environment, look around for a "vanilla" Spice3f5. I've heard that LTSpice runs under wine/Linux, but I've never tried that.

robert

Reply to
Robert Latest

Ken Smith a écrit :

Should be:

"In the other case, if Vout = 1V, the voltage at A=0.090909... so R2 has

0.9090909...V on it and thus the input must be -0.8181818...V"

or make the divider 9K/1K :-)

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Thanks,
Fred.
Reply to
Fred Bartoli

Make before break switching means that you close one switch before opening the previous one, to prevent ever opening the circuit entirely.

Reply to
Glen Walpert

You are right. I can only put the error down to hunger. My wife called me to dinner just as I typed the end of it so I didn't proof read it. I think you will forgive me when I tell you that it was roast beef, fresh green beans from the garden, fresh carrots from the garden and potatoes also from the garden.

This circuit can be improved somewhat by placing a small capacitor from A to GND. This eats up the switching glitches.

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kensmith@rahul.net   forging knowledge
Reply to
Ken Smith

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