New answers tagged gain
0
The user guide referenced in the document you linked explains it fairly clearly.
Traditional velocity sensors consist of either a moving wire coil surrounding a fixed magnet or a fixed wire coil surrounding a moving magnet. The Velomitor® Sensor is more accurate than traditional velocity sensors. Because the Velomitor® Sensor contains no moving parts, it ...
3
The determination of the open-loop gain of an op-amp in a SPICE simulator can be done in various ways. However, in any case, to reveal the ac response, you need to maintain adequate bias on the output to prevent the op-amp from railing up or down. For instance, if you have a 12-V supply, you may want to maintain the output at 6 V so that enough dynamics ...
3
Well - here is the explanation for the unexpected behaviour.
You are not simulating open-loop but with feedback. I suppose that you are aware of this. For open-loop simulation you must take some actions to ensure a correct bias point within the linear amplification range
More interesting: What happens when the frequency is increasing up to pretty large ...
0
Obviously, since the hFE of the transistor is 20, you cannot expect a large gain. You wrongly deduced that the gain is R1/((re1+R2)+(R4||(R3+re2)), which you could have formatted in a proper manner. But I am digressing. The gain, in fact, is R1/((re1+R4)+(R2||(R3+re2)). If you take a look at it from the right side of the base, it becomes painfully obvious.
...
2
I guess your teacher did not ..... never mind.
I made your +10V and ground as straight lines and added some notes.
2
Your input source impedance is \$10\:\text{k}\Omega\$. This means your first stage is likely to attenuate the input, unless you work on the input impedance design. There are a couple of immediate-to-mind options to consider:
Boostrapped version of the standard CE BJT amplifier stage.
Long-tailed pair differential amplifier (diff-amp) using Darlingtons.
You ...
7
Why is there a resistor in the feedback section of this buffer
circuit?
If the op-amp has significant input bias currents (rather than offset currents) then making \$R_F = R_1||R_2\$ has some DC accuracy benefit.
If the op-amp has an unstable unity gain bandwidth, then using \$R_F\$ might just make the difference between the circuit oscillating or not ...
2
It depends on the type of opamp you use, whether the error due to the input bias current and different impedances seen by the "+" and the "-" inputs matter or not. If you use a modern opamp, you most likely can set the feedback resistor to zero ohms.
See also the summary of this application note, where one can read:
"With respect to ...
3
An OpAmp is never ideal. When the feed impedance seen by the two inputs (+ and -) is equal, certain non-ideal aspects cancel out. Hence the value of Rf should be the parallel of R1 and R2.
2
Most integrated Darlington pair power transistors have resistors so you would need a realistic current at the base and collector.
You could measure the resistance from base to emitter with an ohmmeter (the higher value of the two polarities, if one is higher, will be the correct one).
0
I wanted to add the derivation for the intrinsic gain, that might help to understand it.
$$A = \frac {dV_{out}} {d_{V_in}} = \frac {dI} {dV_g} \frac {dV_d} {dI} = g_m \frac {1} {g_d} = g_m r_0$$
6
The possible option I see for this circuit is to use the extra-element theorem or EET forged by Dr. Middlebrook some decades ago. You first identify an element that complicates the analysis. Here, it would be the series connection of \$R_3\$ and \$R_4\$. You have the choice to either bring it to infinity (temporarily remove the resistors from the circuit) or ...
0
Replace the feedback circuit from \$V_{out}\$ to \$V^-\$ with its Thevenin equivalent.
You could also replace the series combination of \$R_2\$ and \$R_4\$ with a single resistor, to simplify the algebra a little.
Top 50 recent answers are included
