Re: Voltage Gain Switch (Design Help)


Phil Hobbs ha escrito:

leo2100@xxxxxxxxx wrote:
Phil Hobbs ha escrito:


leo2100@xxxxxxxxx wrote:

The only problem is that the control signal comes in binary (from a
PIC), so I will have to use a decoder to drive the MOSFETs accordingly.
That adds another IC to the circuit, yet another thing that can go
wrong (but very unlikely). The decoder pops a question: what VGS
voltage shoul I use ?. From what I gather, more VGS voltage better
linearity and for a wider range of VDS. I was thinking about VGS=10V,
but I will need to raise up the voltage coming from the PIC with a BJT,
so I think VGS=5V is fine, the VDS voltage is already low enough so it
doesn`t matter.


I'd probably want to use CD4000 series logic, running from whatever your
analog supply is (+12 or +15, probably).

There are only 4 gain settings, so it ought to be possible to do this
with a few gates.

The gains are not in a binary sequence, so it isn't instantly obvious
how they're encoded---but one natural guess would be

A1 A0 Gain G2 G1 G0
0 0 100 1 0 0
0 1 500 0 1 0
1 0 1000 0 0 1
1 1 5000 0 0 0

where we assume positive logic, and calling the gates of the MOSFETs G0,
G1, G2 for /5, /10, and /50,

G0 = ~A0 & A1 = ~( A0 + ~A1)
G1 = A0 & ~A1 = ~(~A0 + A1)
G2 = ~AO & ~A1 = ~( A0 + A1)

The obvious solution needs two CD4002 quad 2-input NOR gate packages, or
one CD4002 and a couple of spare inverters. You'll want to use a really
slow RC lowpass between the logic and the MOSFETs to make sure you don't
couple anything off the supply rail into the signal path. If you're
imaginative about it, you can probably do it with one gate package, but
this is left as an exercise for the reader, along with the 5V->10V level
shifters you'll need. Alternatively you can run the gates at 5V, but
don't run them from the logic supply, or you'll be wishing for those
nicce well-behaved relays.

Cheers,

Phil Hobbs


I was thinking of maybe using CD4555 Binary to 1 of 4 Decoder instead
of simple gates (which might mean more ICs unless I complicate the
design for this one-off). I need to connect 3 outputs only, plus the
inputs from the PIC. The CD 4002 are actually Dual 4-Input NORs.

I could drive the Gate directly from the 4555's output at 5V, it says
it can drive or sink up to 6.8mA, more than enough I guess, but I do
not know why you say that's not advisable. If not I would use a PNP
transistor to drive the Gate with higher voltage, but I will need to
invert the logic from the 4555, or use 4556 Low on select.

I'm interested in knowing what kind RC lowpass filter I will need, I do
not have any experience with this kind of coupling, I just assume they
are rather isolated parts of the circuit.

Thanks.


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------| | <--t
| | | |
C |----*
C |
100 nF C |
C |
C |
| |
GNDGNDGNDGNDGNDGNDGNDGND

It isn't a complete solution because it only rolls off like 1/frequency
(20 dB per decade). That means just what it sounds like: when they get
to the MOSFET gate, signals at 160 Hz will be down by 10x in voltage,
signals at 1.6 kHz will be down by 100x, etc. Big signals will still
get through, so really, really, don't run that bit of the logic from the
'digital' +5V.


Cheers,

Phil Hobbs

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 data*** states
the drain current with VGS=0V and VDS=48V to be 1mA Max at Tj=125ºC,
don`t know about the impedance, but the current at much lower VDS
should be significantly smaller, right ?

Thanks.

.

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