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21

The pros about using a single resistor over either parallel or series: Less cost Less wiring Less space needed Less error prone (one component less to fail) The pros about using a combination of resistors: You can select resistors from the ones you have if you have a limited number of values When putting them parallel, you increase the power (Watts), see ...


15

This is not an amplifier. Q1 will clamp the signal at power-on to keep DC from propagating. If the input has +DC on it, the DC will propagate until C10 charges. Q1 will clamp the DC until C11 charges. After C11 charges, there is no base current path and Q1 is effectively out of the circuit. D1 keeps Q1's base from going negative on power-off.


7

If you are purchasing parts for an amplifier design that you will be tweaking resistor sizes to optimize performance I strongly suggest that you invest in a resistor assortment kit. Such a kit will prove useful for you for many future projects as well. I still regularly use resistors from an 0805 SMT kit that I purchased over 16 years ago. Kits such as ...


6

What they should have said, was that in practice you work with what you got and it often isn't necessary to use networks just to get a non-standard value. For that, you really want to get used to the E12/24/48/96 series. For example, if you calculated you need 18800 Ohm, a 18k7 would usually be good enough. Often enough 18k will do fine. Not even talking ...


5

why? Positive feedback seeks to maximize the voltage difference between Vin- and Vin+ therefore it cannot produce a virtual ground. Negative feedback seeks to minimize the voltage difference between Vin- and Vin+ therefore it can produce a virtual ground.


5

That's an old site I had maybe 20 years ago. I had forgotten about it... In retrospect I guess this memory distortion thing was just a red herring. This kind of stuff pop up periodically in the audio community, someone finds a magical reason that explains everything, it gets popular then is forgotten then it starts all over again with something else. Sure, ...


4

Generally speaking combining resistors is either irellevant or self defeating, but when prototyping one does what one needs to so of course it is done extensively. For example it is easy to solder an smt resistor on top of another and saves a step vs desoldering if I want to quickly try a value smaller than what I chose . Consider that resistors values ...


3

The pins with ADC input channels are used already. The analog mode for a GPIO pad is just a multiplexer setting. You still need the internal wiring to the ADC, which is only available for pins specified with an ADC channel marked ADC_INxx as additional function in table 16 of the datasheet.


3

He multiplies the value of the thermal voltage, \$V_T\$, by 17.2 and then adds the nominal base-emitter voltage, \$V_{BE}\$. Both \$V_T\$ and \$V_{BE}\$ will have been discussed at length in previous paragraphs.


2

Thanks a lot for your answers. Texas Instruments kindly explained the reason for those resistors: The in-line resistance acts as part of a low-pass filter given the input capacitance is 3pF at the START and STOP pins. This is to prevent false-positives otherwise created by high-frequency transients-glitches at these pins.


2

The proper way to measure ripple in a 1MHz SMPS is to use a 100nF ~ 1uF low ESR DC blocking cap and 50 Ohm coax with 50 Ohm termination . 0.1uF*50R =...30kHz HPF This may reject low frequency noise and pass the ripple you are trying to measure. The causes of Ripple are mainly the ESR of the caps, chokes and switches with some possibilities of resonance ...


2

My equation was -5*ia - 6V + 18*-6R = 0 The current through the unknown resistor is not \$6R\$, it's \$6/R\$. So you should have $$-5i_a - 6\ {\rm V} -\left(18\ \Omega\right)\left(\frac{6\ {\rm V}}{R}\right)=0$$ (I've assumed the blue probe of the voltmeter is the positive probe, because otherwise you're going to end up with a negative resistor value) (...


2

I don't think that's possible. Do you know what the aux port is there for? I assume it's there to give feedback to blind people using the ATM. If that's the case, the aux port probably is output only and hacking the atm this way would be pretty much like hacking the atm via its display using a video projector.


2

We only tend to need 'impedance matching' when dealing with RF, and an LF351 is not an RF amplifier. 50 ohms is quite a low load impedance for an opamp to drive, so the main part of driving a 50 ohm load is to make sure it will source enough current. With a +/- 15v supply, the 351 is specified to drive a minimum of 10v, and typically 12v, into a 2k load, ...


2

There is no universal answer on your question. Normally in low power electronics you try to use available space as efficient as possible by taking in consideration possible interaction of closely located components. For example in radio circuits you would try to shield oscillators from amplifiers, make traces as short as possible to avoid antenna effect ...


2

The LM324 datasheet specifies the output voltage max as Vsupply - 1.5V, so given your 5V power supply, 3.5V out (max) is exactly what you would expect. The LM324 should be able to drive the MOSFET at low frequencies, although the gate is a capacitive load so stability can be an issue. You will do better using a rail-to-rail output opamp, although even ...


2

Assume the output impedance of U2 is 0 ohms across the frequency band of interest. If so, then the impedance seen by the U3 circuit is 50 ohms at one end of the FREQ1 pot rotation, 0 ohms at the other end, and approx. 12.5K ohms in the center. Same for the U3 circuit. This introduces a non-linearity into the circuit's response to pot rotation. Both pots ...


2

Frequency is proportional to 1/(R*C), so it won't be linear with R.


1

In the circuit of an op-amp non-inverting comparator (Schmitt trigger), a virtual ground appears during the transition, in the moment when Vout/Vin = -R2/R1.


1

OP Amps, by circuit design, are linear ONLY when the output is not saturated which means if the input differential is >>0 there is zero linear gain. ( ie no virtual ground, Vin NOT =0 ). As soon as the input >> 0 OR the output goes to the rail, the closed loop gain must (by definition ) = ZERO (0) Av(f) = linear gain = ΔVo/ΔVin ( at any frequency ...


1

The easiest modification you can make to your single rail photodiode amplifier is to reverse the photodiode position and have the anode grounded like this: - I've also marked on the direction of photodiode current with a red arrow. This circuit would now satisfy the feedback constraints of a single rail op-amp; the output will rise positively from close to ...


1

I did it, guys. The gpioAddGeneric function is called with parameters i and ppm_factory.waves[ppm_factory.next_wave_id]. If you can see, in my for loop, i is incremented up to *ppm_factory.channel count. Then, incremented again when I add the wave. Therefore,i` becomes channel_count * 2 + 1. This was the mistake because we have channel_count * 2 + 2 ...


1

Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab The input voltage is a rectified mains voltage, which can be mathematically written as follows: $$\text{v}_\text{in}\left(t\right)=\left|\hat{\text{V}}_\text{i}\sin\left(\omega t\right)\right|\tag1$$ Using Laplace transform, we can write: $$\text{V}_\...


1

If all UART comms protocol originates at the Port of the MCU then this may be a workable solution to MUX to the various target devices. On the other hand if UART comms can asynchronously originate from any of the target devices at any time then this multiplexing scheme is a non-starter.


1

I'm not sure how big each UART is (and the necessary multiplexers). Some time ago I had the intention to use several UARTs (like 4-6) for MIDI. My idea was to use STM32F103C8T6, which have 3 UARTs (each). Using such MCU, you can easily gather all UART info and send it (e.g. via SPI) to your LPC MCU. This also prevents the problem related to buffer/timing ...


1

simulate this circuit – Schematic created using CircuitLab the bounce time must be less than T, otherwise increase T Rev 1 Added buffer hinted by Elliot. (chose Schmitt Nand) corrected polarity for negative logic resets. added Power on Reset (POR), low leakage Si diode. added Watch Dog Timer reset input (WDT) (useful) For more wisdom, read what ...


1

The LM324 is powered from 5 volts and is being driven by a voltage source centred at 2.5 volts with a positive peak taking it up to 5 volts and this is your basic problem. The LM324 does not have an input voltage range that is close to the positive rail. Additionally, the LM324 cannot produce output voltages that are sufficient to drive the MOSFET at the ...


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