Op Amp with slew rate of 10,000 V/micro second exist?

Thread Starter

ping222

Joined Nov 7, 2013
13
Hello,

I would like to amplify a 2 GHz sinusoidal voltage signal (0.22 V peak to peak) using an operational amplifier. The gain I would like from the operational amplifier is about 5, that is, I would like a voltage output from the amplifier to be approximately 1 V peak to peak given the input. The problem is I can't seem to find an op. amp. with a slew rate that can give me the desired performance. The slew rate I require can be calculated for a sinusoid to be 2*pi*f*Vpp where,

f is the frequency of the signal (2 GHz in my case)
Vpp is the peak to peak output voltage required (1 V in my case).

Using this formula I require a slew rate of 12,566 V/micro second. I can't seem to find anything out there that exists that can go this fast. I don't have much experience with op. amps. Do they currently exist to meet my specifications? Thanks in advance for any feedback,

Regards,

Ping.
 

LvW

Joined Jun 13, 2013
1,760
Ping22, try to find a suitable amplifier in current-feedback topology (CFA).
I know they exhibit slew rates of several thousands of volts/µs.
Such amplifiers can be used and wired like conventional opamps.
The only point to be considered is that the feedback resistor must be larger than a specified minimum.
 

alfacliff

Joined Dec 13, 2013
2,458
does it need to be an op amp? if not, there are sevral MMIC's that will amplify with more gain than tha at 2 ghz. check out MINICIRCUITS LABS site for info.
 

#12

Joined Nov 30, 2010
18,224
Basic rule: The more parts in a stage, the slower they are, even if for no other reason than parasitic capacitance and inductance. One or two transistors stands a better chance of going fast than 50 or 100 transistor on an integrated chip.
 

bertus

Joined Apr 5, 2008
22,278

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Thread Starter

ping222

Joined Nov 7, 2013
13
Thanks guys for all of that, I will look into it and post what I find. With regards to 'RichardO' post, yeah I am pretty sure it is sinusoid peak to peak voltage that is required in the 2*pi*f*Vpp calculation. Therefore would not be off by a factor of two in that case.
 

crutschow

Joined Mar 14, 2008
34,468
I have to weigh in. It's 2*pi*f*Vp as shown here. It's calculated as the peak value of the derivative of the sinewave or d/dt (Vp*sin(2\(\omega\)t)) = 2*pi*f*Vp * cos(0) = 2*pi*f*Vp.
 
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