6
What would be some of the benefits of doing this?
Power supply noise regulation is the main benefit - op-amps are good but they won't deal with power supply rail voltage variations without producing some distortion on the output signal. For an op-amp, it's called PSRR (power supply rejection ratio) and tells you (in the data sheet for the op-amp) what ...
3
As a bipolar-input amplifier, The input impedance of that part is relatively low, less than 1G ohm, typically as you can see from the change in input bias current with CM voltage. Eg. Fig.15.
Typically, to get the benefit of the front end which is run at relatively high current with super beta transistors, you’d want a source impedance less than 1K so that’s ...
3
The data sheet says this: -
The AD8597/AD8599 are the optimum choice for low noise performance if
the source resistance is kept < 1 kΩ
Clearly, this amplifier should only really be used when the source impedance is less than or equal to 1 kohm. Reason: it's a very low noise amplifier and this means that having a source impedance higher than 1 kohm is ...
2
Substrate is normally connected to the most negative voltage in the circuit (Vee in the case of the 741) since it is p-type. Since the emitter and base are built on top of the collector diffusion in a vertical NPN, the collector-substrate junction characteristics are of interest.
Lateral PNP transistors as you'd find in the 741 are going to be different, I ...
2
It's one of the ideal op-amp assumptions that the current into either input pin is zero.
In even a crummy real op-amp it's generally less than 1 uA. In FET-input op-amps it can be as low as femtoamps.
2
Your paragraph 1 is correct, as it is a feedback loop. MOSFET gate voltage cannot be calculated because it is in the middle of a feedback loop, and it depends on the MOSFETs uS value.
For the same reason the Vout of the op-amp is unknown without testing and recording its behavior. The MOSFET chosen will directly affect this value. Voltage-to-current tracking ...
2
simulate this circuit – Schematic created using CircuitLab
The way I do this is by taking the voltage you want to measure with the ADC, and voltage divide it down so that the maximum voltage is slightly less than your MCU supply voltage which in this case is 3.3v. In this case it works out to be around 3.2v if the signal is the maximum 10v. Then you ...
1
THe LM358 has an input impedance ~ 10M and output impedance ~ 1k @ 1V/1mA and 100mV /50uA ~2k so it is does not reach negative rail to 0V. But perhaps close enough if your load is < 50uA or >> 2k. That is due to saturation of BJT’s.
Consider matching input R to V+,V- to null input bias current induced Vio with ~1M on each input.
Consider output ...
1
The FET is just a nonlinear voltage-controlled resistor, so the gate voltage will also be nonlinear relative to the current waveform due to the resistance divider created with the current sense R.
You always want to scale your control voltage to match the current sense voltage which for low power must be 50~100mV max in typical situations. So choose Imax ...
1
The input impedance of AD8597 ca. 45MΩ @ 1kHz and 2MΩ @ 20kHz.
1
There are two basic considerations when solving this problem:
The first consideration is to expand the maximum voltage drop across the load aka "compliance voltage of the output current source". It is determined by the supply voltage of the transistor stage minus the voltage drop across the current-setting resistor. If the DAC directly controls the ...
1
Instead of the constant source that is fed into the non-inverting
input of the 2nd stage op-amp, if I feed in the voltage from the DAC
directly, why don’t I see a constant current through Rload?
It won't work as it stands - to make this work you need to make VM2 (the input demand signal voltage) referenced to the positive supply rail (PVDD) so that the op-...
1
A lot of instability comes from stray positive feedback. this may be conducted or radiated. This stems from the fact that wire leads can be both distributed capacitance and inductance depending on geometric size and direction relative to nearby conductors.
Inductors depend on length and also diameter being a small ratio 5~10nH/cm but don’t contribute much ...
answered Aug 6 at 19:44
Tony Stewart Sunnyskyguy EE75
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1
My answer
It is simple and short: There is no use of this arrangement (negative "resistor" connected to a voltage source). In order to be useful, there must be a load connected in parallel. Then the voltage source will supply the load and the negative resistor will "help" it by passing an additional current through the load. Depending on ...
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