Current Feedback Amplifier doubts - Page 1
Forum section: electronics / beginners

author: jimmyWalton
author_rank: Newbie
timestamp: May 23, 2024, 11:01:47 am
content: Hello!I am a newbie as regards the Current Feedback Amplifier CFA. Its implementation may be different according to the producer, but consider the model used for the numerical example in Figure 2 (b) (the one on the right) here:https://www.allaboutcircuits.com/technical-articles/current-feedback-amplifiers-vs.-voltage-feedback-amplifiers/The same model is used in some pdfs likehttps://www.ti.com/lit/an/sboa081/sboa081.pdf, with non-inverting configuration.I am struggling in understanding some aspects of its operation.A traditional opamp (Voltage FeedBack, VFB device) connected the same way would provide an output voltage v_o such that its input v_d = v_+ - v_- = 0. Here, instead, it should be the current generated by the output that causes I_n = 0.In that Figure 2 (b), itseemsthat 7.18 mA out of the total 7.20 mA flowing through resistor R_G come from the output of the CFA, while instead the voltage buffer between v_+ and v_+ needs only to provide about 0.02 mA, so it's almost the condition I_n = 0 (it's not exactly 0 due to some non-idealities). But the output of the voltage buffer is a low impedance, so: the current coming from V_o, when reaching the node at 1.0 V, shouldn't divide between R_G and the internal low impedance to the buffer, which is a much favourable path to ground? Why insteadallthe current from R_F flows through R_G? What can guarantee this?Also: this CFA seems to be conceived to be only used with feedback. Excluding a feedback network, or putting the v_- node to gnd, would prevent it from working (because the buffer would become useless). Is this correct?

author: magic
author_rank: Super Contributor
timestamp: May 23, 2024, 11:58:10 am
content: Quote from: jimmyWalton on May 23, 2024, 11:01:47 amIt isn't due to non-idealities, it's the core principle of CF amplifier operation: output voltage is proportional to IN- pin current. The only case when IN- current is zero is when OUT is also zero*. A small current must flow out of IN- (as shown on figure 2b) for the output to be positive.*Almost, because the model is incomplete. Z is not a simple resistance like 500kΩ, it necessarily includes capacitance too. It is possible that output voltage is non-zero due to charge/voltage stored on the internal capacitor while IN- current is zero, in such case the internal capacitor simply discharges through the internal resistance and the output moves towards ground but it isn't at ground just yet.Quote from: jimmyWalton on May 23, 2024, 11:01:47 amBehavior is determined by gain of the amplifier and feedback action. If too much current flows in/out of IN-, the output very quickly slews in the opposite direction and Rf current is changed to match Rg current (minus the small current controlling the amp).Quote from: jimmyWalton on May 23, 2024, 11:01:47 amShorting IN- of a CFA to any fixed voltage makes no theoretical sense because it's a voltage source too. But if you put a resistor between IN- and GND then it could work open loop. With the usual caveats about stupidly high gain, offset voltage and so on.

author: jimmyWalton
author_rank: Newbie
timestamp: May 23, 2024, 02:52:42 pm
content: First of all, thanks for your observations.Quote from: magic on May 23, 2024, 11:58:10 amSorry, I can't understand this point. Assume the condition in the figure (a stationary condition): why the current coming from V_o doesn't divide between R_G and the internal low impedance of the buffer? Why such current only flows through R_G?And what happens if current I_n raises?Quote from: magic on May 23, 2024, 11:58:10 amBut if I_n increases, V_o = Z I_n is proportional to it and it has the same sign, so V_o would increase too.

author: magic
author_rank: Super Contributor
timestamp: May 23, 2024, 03:24:45 pm
content: Yes, V_o increases if more I_n flows out of the pin. In your example this happens when Rf current is lower than Rg current, i.e. the output is "too low". And it causes the output to move up.Conversely, if the output is "too high" then I_n decreases or reverses polarity and the output moves down. That's how negative feedback works in these circuits.