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24

It's not advisable unless the entire project is low voltage and battery operated. Anything mains operated could be dangerous even if transformer isolated. There are specific safety requirements for mains operated patient attached equipment to protect against excessive leakage currents and potential safety hazards due to equipment or component failures. ...


7

If the only problem is that the amplifier stage you made inverts the signal, just flip the inputs of the opamp around to compensate. For another approach, use a different opamp, or do the whole thing with discrete parts in the first place. You said in your first sentence "I've chosen to work with an High Frequency Opamp that ...". You chose it, you can ...


7

When doing an ECG test, the electrodes are much more conductive than dry skin, so the possibility of triggering fibrillation or other arrhythmia from small voltages is present and needs to be taken very seriously. Most pacemaker pulses for implanted pacemakers are from 2mV to 250mV. That's not a whole lot of voltage, and if your device accidentally drives ...


6

It might be the probe attenuation. Some probes have a switch on them that let you select an attenuation of 2x, 5x, 10x. Other probes are simply built to attenuate and don't have a select switch, but simply have a, say, 10x written on them. To account for this, oscilloscopes have an option for each channel to specify what kind of probe is connected to it. ...


6

You might need to look for rail to Rail op-amps if the output of op-amp is expected to be close to supply voltage. Below spec is from datasheet of LM358 op-amp which tells that the difference can be upto 1.5 V from the VCC (3 V, in OP case) Suggestion: MCP6001 from Microchip. Datasheet here


6

The transfer function is $$H(s)=\frac{-sR_1C_2}{1+s(R_1C_1+R_2C_2)+s^2R_1R_2C_1C_2}$$ and the maximum gain is $$A_{\text{max}}=\frac{R_1C_2}{R_1C_1+R_2C_2}=2.04$$


6

If nothing is said about it in the data sheet and application notes (as does occur sometimes), leave them floating. If there is direction in the data sheet or application notes, follow it precisely, as @IC_designer_Rimpelbekkie says. I can't recall any exceptions, but that doesn't mean that there are not one or two types of op-amps somewhere that require ...


5

All you need to convert 0-24 V to 0-5 volts is two resistors. The first resistor (R1) is in series with the input, and the second (R2) connects after that to ground. This is such a common construct it has it's own name, which is "voltage divider". Look that up, even here on this site, and you should find lots of info. If the first resistor has 4x the ...


4

Common mode range of the LMP2231 on a 3.3V supply is up to 2.5V. You have 2.77V on the non-inverting input. If you decrease R3 (to get say 2.4V) and increase R1 it might work, but the MOSFET is not guaranteed to give you even 250uA at 2.4V Vgs, so it would be a random non-design kind of working.


4

No, voltage gain is 20 log the ratio and not 10 log the ratio. 20dB = voltage gain of 10 and a power gain of 100. For the op-amp circuit it only makes sense to set it to have a voltage gain ratio of ten Voltage gain = 20\$log_{10}(\dfrac{V_O}{V_I})\$ or 10\$log_{10}(\dfrac{V_O}{V_I})^2\$ = 10\$log_{10}(\dfrac{P_O}{P_I})\$ Just think about the power in a ...


4

Both circuits are used and it depends on the application which of the two you should choose. There are two important differences between the modes the photodiode is used in those circuits: The first mode is called photoconductive. It has a relative large reverse voltage across the diode which results in low junction capacitance (good for high speed ...


4

Calculations which show "show Vin should be 1.02V as no current will be drawn towards negative terminal of opamp" are badly in error. As Peter Bennett, has pointed out, your power supply must be grounded. So the op amp portion would look like simulate this circuit – Schematic created using CircuitLab However, this will also not work. Since the ...


3

I did a quick simulation and maximum gain is 6.007 dB at 2.407kHz: - 6.007 dB is a voltage gain of 1.997 so I guess your mathematical approach is flawed in some respects.


3

It's a small-signal analysis, so ignore the fixed ideal current source (it has infinite impedance) and fixed output voltage offset that results. We're just looking for changes that result from en. Beta is just the ratio of the impedances and the output is just ~en/beta if the amplifier gain is high.


3

why not just measure the current by letting Vos be the reference voltage in our voltmeter? It's not really clear what you mean by this, but Vos is entirely internal to the op-amp, and it might be a combination of different errors in different stages of the amplifier. We only model it with a voltage source in series with the input. But there's no one ...


3

This would be better (output left open): simulate this circuit – Schematic created using CircuitLab Both inputs are within the common-mode range of the amplifer- see the datasheet which gives the values with a +/-6V supply and translate by adding 6V to the numbers.


3

You can usually find a fabrication hint on the front page of a device's data sheet - it might say something like: - Single Supply, Rail-to-Rail, Low Power, FET-Input Op Amp This is for the AD824 and it does mentions "FETs". However, I'd be more interested in it being low-power and rail-to-rail rather than the technology used. But, "FET" input usually ...


3

Decibels technically always measure power gain according to the formula $$\mathrm{Gain(dB)}=10\log_{10}{\frac{P_o}{P_i}}$$ When the input and load impedances are the same we have $$ \frac{P_o}{P_i}=\frac{V_o^2}{V_i^2}$$ so $$\mathrm{Gain(dB)}=10\log_{10}{\frac{V_o^2}{V_i^2}}=20\log_{10}{\frac{V_o}{V_i}}$$ Sometimes we use decibels imprecisely when the ...


3

I don't see much reason to have it. Perhaps it was once used for "filters" as the annotation suggests. Nor do I see a reason for R13. This circuit could probably be reduced to a single op-amp (with some improvement to the stability).


3

Your op-amp is a vital clog in the wheel in this circuit. Since the op-amp has infinite resistance, the current I entirely flows to the alternate path through the resistor R. Note that since the + end of the op-amp is at ground and because of the infinite gain of the op-amp, the - end is also held to ground. So the drop across the resistor is V = IR. So ...


3

Your first equation is not correct. It should be identical to the second equation, assuming the bias point of the output is really halfway between the supplies. (In practice, there's a small DC offset.) An op amp is a differential amplifier with high gain. It doesn't "know" what kind of supplies it's hooked up to, and in fact there is no way to know. The ...


3

Rules for most op-amps: - Output can usually get to within a couple of volts of the power rails and then will hard-limit unless it is a rail-to-rail opamp (then it might get to within 50mV) Inputs can't normally be taken to within a couple of volts of the power rails but there are some notable exceptions. An op-amp has no-idea that the negative supply to ...


3

I don't know what it says precisely in sedra smith but you can't do anything about input offset current. However, if you mean input bias current then that is eradicated by making the dc resistance seen by both inputs the same value. So, if there is a 100k feedback and a 100k input resistor to the inverting terminal then, ideally, you should have 50k seen by ...


2

The only possible reason for the series of op-amps in this set-up could be "filters" in the unity-gain region of the op-amps, but that would not be a very smart way to filter. It's also possible the U2D has a capacitor somewhere, because the composite maths of filtering in the post-amplifier became too complex for the designer around U2C and it is simply ...


2

Re drawing the schematic to I can add component references etc. simulate this circuit – Schematic created using CircuitLab I don't recognise this as a standard circuit but its got negative feedback so will try to keep both the inverting and non-inverting inputs the same The non-inverting input is simply \$ V_2 \$ And the inverting input: \$ ...


2

According to the docs the efficiency is reasonably high, I believe 89% for 8Ω is mentioned. If you want to be able to drive the speakers at full power continuously: Total power [Watt] = number of channels × power per channel P = 2 × 2.8 = 5.6 W Supply current drawn for the speakers alone would be: Current [Ampère] = Power [Watt] / Voltage[Volt] I = P ...


2

Among concerns posted in other answers, by other cool and froody people, I feel remiss if I do not point out a few other very important things. You selected a set-point for 1.06-ish ampere with 528mV over your 500mOhm resistor. Your MOST cannot regulate currents like that. The MOSFET you use is positively tiny. It might do for 500mA, but if you want good ...


2

Your power supply voltage is too low for the opamp to be able to drive that MOSFET. Look at the output characteristics of the NDS356: To be able to provide 1A of current it should be driven with at least 3V of Vgs (more or less). You have very little margin in your design and any spread in the curves could make it non-working. You could try to increase ...


2

If we take this 20dB as "power amplification", then The power gain of a voltage amplifier depends on the input resistance and the load resistance. For a non-ideal voltage amplifier, the power gain is related to the voltage gain as follows: $$G_P = \frac{P_L}{P_{in}} = \frac{V_LI_L}{V_{in}I_{in}}=\frac{V^2_LR_{in}}{V^2_{in}R_L}{} = ...


2

Deci-Bels (dB) always represent a power ratio. Specifically, the power ratio of P2 with respect to P1 expressed in dB is:   dB = 10Log10(P2/P1). Sometimes, as in the question you show, we use dB as a short form for voltage ratios. However, since power is proportional to the square of the voltage, a voltage ratio expressed in dB is: ...



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