Amplifiers

Err, a current amplifier amplifies current and a voltage amplifier amplifies voltage?

The simple answer is yes.

If I recall correctly there was a nice on-demand online seminar about them somewhere on either the Analog Devices website or National Semiconductor's website. I don't recall all of the details, and I haven't used one since I watched the seminar, but here is what I remember off the top of my head. Current Feedback Op Amps (That is what you were referring to, yes?) have a completely different internal configuration than Voltage Feedback Op Amps. The two inputs are not high impedance "voltage measuring terminals", but rather are part of a current path to ground. The output stage "looks at" how much current is flowing between the two terminals and either increases or decreases its voltage to try to keep the current flowing between the two input terminals at a constant level. As you can see, this is wildly different from a voltage feedback op amp. However, surprisingly enough, many of the standard op amp circuit configurations will actually work very well with this setup. The important thing to remember had to do with the purpose of the feedback resistor. It seems to me that it set the gain independently of the other resistors in the circuit, where as a resistive divider sets the gain in a voltage feedback configuration. I could be wrong about the FB resistor setting the gain, but I can't remember for certain right now. The reason for using a Current Feedback op amp had to do with its high bandwidth and the fact that it doesn't suffer from the gain-bandwidth product limitation of voltage feedback op amps. If you recall, as you increase the gain on a VFBOA circuit, you begin to lose bandwidth. The main drawback stemmed from the CFBOA not having the high DC accuracy that the VFBOAs typically enjoy.

Well, there you go – nice and simple. Get the details for yourself for the full story and to make sure I didn't get some of it wrong. I hope this post was of more help than the last two.

Regards,

Scott

P.S. If you were not referring to op amp feedback configuration, but rather to the more fundamental topic of current-to-current, voltage-to-voltage, current-to-voltage and voltage-to-current amplification, then you may want to refer to a good general text on electronics. I highly recommend The Art of Electronics by Horowitz & Hill.

P.P.S. to my previous post above. I couldn't resist checking myself, and sure enough I was wrong about the feedback resistor setting the gain. The R1/R2 ratio still sets the gain, the FB resistor (R2) sets the cut-off frequency. The manufacturer usually specifies a relatively narrow range for the feedback resistor. I am pretty sure I remember reading somewhere that VFB amps have gotten so good (read "fast") that CFB amps aren't as big of a deal as they were when they first appeared.

Regards,

Scott

People are use to dealing with voltage feedback amps. For a normal non-inverting amp with two resistors the transfer function is vout/vin=(1+R2/R1), and for an inverting the transfer function is vout/vin=(-R2/R1). For a current feedback amp the transfer functions are the exact same.

There are many differences between the two amps in how they work. The normal voltage amp takes the voltages on the input (V+ and V-) subtracts the difference and then it multiplies the difference or voltage error by the open loop gain. So for a voltage feedback amp the output voltage is Vout=A*(V+ - V-). When you use feedback then you get the transfer function vout/vin=(1+R2/R1) assuming the open loop gain is very High.

For a current feedback amp the amp doesn't do anything to the input voltages, it only cares about the input current so the amp tries to make the input current into the V- equal to zero. The amp open loop gain looks something like vout=A*(I). When you do the math after adding feedback resistors you get the same transfer functions as the voltage feedback amps. One thing to note is it's not the exact same as the voltage feedback amp. If you put a cap across the feedback resistor, it can make the amplifier unstable. So you have to do the math.