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'[EE] fast differential in-out amplifiers'
2005\12\20@090409
by
Vasile Surducan
Hi all,
I'm looking for a high frequency differential amplifier (> 400MHz BW
at -3dBm) optimized for high input impedance and very large slew rate
(1000 to 5000 V/us) running at 3V or 3.3V single supply.
I found some differential in-out amplifiers (AD8132, LMH6551) but all
are using small input/gain resistors to keep high the bandwidth, and
this is reducing the differential input impedance. Requirement for
differential input impedance is 10K at a differential gain of 5.
I don't want to use any buffers between the input signal and
differential amplifier.
I do not want to design my own differential in-differential out
amplifier using two or three high frequency OA.
If there is any RF master here with experience in high frequency
differential amplifiers, I'm searching for a clue.
cheers,
Vasile
2005\12\21@005727
by
Mike Singer
Vasile Surducan wrote:
> I'm looking for a high frequency differential amplifier (> 400MHz
> ... Requirement for differential input impedance is 10K
...
> If there is any RF master here with experience in high frequency
> differential amplifiers, I'm searching for a clue.
Vasile,
There is a formula F = 1 / (R * C)
C = 1 / (R * F) = 1 / (1 000 000 000 Hz * 10 000 Ohm) = 0.1 pF
pretty low input capacitance, isn't it?
Mike.
2005\12\21@012433
by
Dmitriy Kiryashov
Hi Mike.
What 10K _differential_input_impedance_ has to do with RC filter ?
Dmitriy.
Mike Singer wrote:
{Quote hidden}>
> Vasile Surducan wrote:
> > I'm looking for a high frequency differential amplifier (> 400MHz
> > ... Requirement for differential input impedance is 10K
> ...
> > If there is any RF master here with experience in high frequency
> > differential amplifiers, I'm searching for a clue.
>
> Vasile,
>
> There is a formula F = 1 / (R * C)
>
> C = 1 / (R * F) = 1 / (1 000 000 000 Hz * 10 000 Ohm) = 0.1 pF
>
> pretty low input capacitance, isn't it?
>
> Mike.
>
> -
2005\12\21@044958
by
Vasile Surducan
10 K input impedance with a gain of 5 means 50K feedback resistor,
with 2pF input impedance, voilla RC filter.
Vasile
On 12/21/05, Dmitriy Kiryashov <spam_OUTvze27bymTakeThisOuT
verizon.net> wrote:
{Quote hidden}> Hi Mike.
>
> What 10K _differential_input_impedance_ has to do with RC filter ?
>
>
> Dmitriy.
>
>
>
> Mike Singer wrote:
> >
> > Vasile Surducan wrote:
> > > I'm looking for a high frequency differential amplifier (> 400MHz
> > > ... Requirement for differential input impedance is 10K
> > ...
> > > If there is any RF master here with experience in high frequency
> > > differential amplifiers, I'm searching for a clue.
> >
> > Vasile,
> >
> > There is a formula F = 1 / (R * C)
> >
> > C = 1 / (R * F) = 1 / (1 000 000 000 Hz * 10 000 Ohm) = 0.1 pF
> >
> > pretty low input capacitance, isn't it?
> >
> > Mike.
> >
> > --
2005\12\21@070659
by
Dmitriy Kiryashov
|
Just an idea.
Active element ( buffer ? ) in feedback net.
Providing required attenuation and removing
capacitance effect the same time.
Vasile Surducan wrote:
{Quote hidden}>
> 10 K input impedance with a gain of 5 means 50K feedback resistor,
> with 2pF input impedance, voilla RC filter.
>
> Vasile
>
> On 12/21/05, Dmitriy Kiryashov <
.....vze27bymKILLspam@spam@verizon.net> wrote:
> > Hi Mike.
> >
> > What 10K _differential_input_impedance_ has to do with RC filter ?
> >
> >
> > Dmitriy.
> >
> >
> >
> > Mike Singer wrote:
> > >
> > > Vasile Surducan wrote:
> > > > I'm looking for a high frequency differential amplifier (> 400MHz
> > > > ... Requirement for differential input impedance is 10K
> > > ...
> > > > If there is any RF master here with experience in high frequency
> > > > differential amplifiers, I'm searching for a clue.
> > >
> > > Vasile,
> > >
> > > There is a formula F = 1 / (R * C)
> > >
> > > C = 1 / (R * F) = 1 / (1 000 000 000 Hz * 10 000 Ohm) = 0.1 pF
> > >
> > > pretty low input capacitance, isn't it?
> > >
> > > Mike.
> > >
> > > --
>
2005\12\21@073252
by
olin piclist
Mike Singer wrote:
> There is a formula F = 1 / (R * C)
There is, but if you're looking for the 3dB rolloff frequency then it's
F = 1 / (2 * Pi * R * C)
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2005\12\21@082007
by
Gerhard Fiedler
Mike Singer wrote:
> There is a formula F = 1 / (R * C)
Actually, it's
f = 1 / (2*pi * R * C)
(R * C) by itself it the time constant of the filter. 1/RC is the "angular
speed", so to speak, so it has to be divided by the angle of a full circle
(2*pi) to get the frequency.
Not that it changes the basic thought a lot... just the numbers a bit.
Gerhard
2005\12\21@131420
by
Vasile Surducan
On 12/21/05, Dmitriy Kiryashov <vze27bymKILLspamverizon.net> wrote:
> Just an idea.
>
> Active element ( buffer ? ) in feedback net.
> Providing required attenuation and removing
> capacitance effect the same time.
No, it's easyest to add buffers on the input signal and low down with
the input impedance at about 150 ohm, that's tipical for this
frequency. The better choice is to use a transimpedance amplifier, but
I didn't found any at 3V single supply.
At +/-5V there are plenty differential in /single out or differential
in/differential out
buffers with 5000V/uS or more but power hungry devices or
transimpedance amplifiers with fully differential outputs. But I can't
afford to spend energy for 3/+-5V conversion.
Vasile
{Quote hidden}>
> Vasile Surducan wrote:
> >
> > 10 K input impedance with a gain of 5 means 50K feedback resistor,
> > with 2pF input impedance, voilla RC filter.
> >
> > Vasile
> >
> > On 12/21/05, Dmitriy Kiryashov <
.....vze27bymKILLspam.....verizon.net> wrote:
> > > Hi Mike.
> > >
> > > What 10K _differential_input_impedance_ has to do with RC filter ?
> > >
> > >
> > > Dmitriy.
> > >
> > >
> > >
> > > Mike Singer wrote:
> > > >
> > > > Vasile Surducan wrote:
> > > > > I'm looking for a high frequency differential amplifier (> 400MHz
> > > > > ... Requirement for differential input impedance is 10K
> > > > ...
> > > > > If there is any RF master here with experience in high frequency
> > > > > differential amplifiers, I'm searching for a clue.
> > > >
> > > > Vasile,
> > > >
> > > > There is a formula F = 1 / (R * C)
> > > >
> > > > C = 1 / (R * F) = 1 / (1 000 000 000 Hz * 10 000 Ohm) = 0.1 pF
> > > >
> > > > pretty low input capacitance, isn't it?
> > > >
> > > > Mike.
> > > >
> > > > --
> >
2005\12\21@134353
by
Peter
On Wed, 21 Dec 2005, Dmitriy Kiryashov wrote:
> Just an idea.
>
> Active element ( buffer ? ) in feedback net.
> Providing required attenuation and removing
> capacitance effect the same time.
Or a 'driven shield' ;-)
Peter
2005\12\21@195943
by
Mike Singer
Olin Lathrop wrote:
> > There is a formula F = 1 / (R * C)
>
> There is, but if you're looking for the 3dB rolloff frequency then it's
>
> F = 1 / (2 * Pi * R * C)
But Vasya did say nothing about the signal being sinusoidal, If I'm
not mistaken.
Mike
2005\12\22@180739
by
Mark Rages
On 12/21/05, Mike Singer <EraseMEznatokspam_OUTTakeThisOuTgmail.com> wrote:
> Olin Lathrop wrote:
> > > There is a formula F = 1 / (R * C)
> >
> > There is, but if you're looking for the 3dB rolloff frequency then it's
> >
> > F = 1 / (2 * Pi * R * C)
>
> But Vasya did say nothing about the signal being sinusoidal, If I'm
> not mistaken.
>
> Mike
Vasile only said 400MHz bandwidth. Joe Fourier added the part about sinusoids.
Regards,
Mark
markrages@gmail
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