Hot answers tagged

63

Buffers are used whenever you need... well... a buffer. As in the literal meaning of the word. They're used when you need to buffer the input from the output. There are countless ways to use a buffer. There are digital logic gate buffers, which are passthroughs logicwise, and there are analog buffers, which act as passthroughs but for an analog voltage. The ...


27

This is a buffer. Two gates means twice the output current. But why use AND gates instead of buffers, you might ask? I was originally going to say that they probably used one or two AND gates elsewhere in the circuit and just popped down a single quad AND gate chip, using two of the AND gates as buffers instead of calling for an actual buffer to save ...


20

You should always tie unused inputs to a valid logic level. That could be tied to GND or to the VDD voltage rail. Never leave unused inputs floating in that it can cause excessive power dissipation in that IC package and introduce extra noise into the voltage / GND rails. It is common practice to use a pullup or pulldown on the unused inputs on unused ...


20

Chip suppliers are keen that their users avoid common design errors, shown by application examples in their data sheets. This one is addressed by Linear Technology in their data sheet for LTC6241. It also applies to many other opamps: The good noise performance of these op amps can be attributed to large input devices in the differential pair. Above ...


19

No, there is no way to make an inverting buffer with just an op-amp that does not depend on the resistor values. You can get resistors with very fine accuracy and stability (at an equally impressive price) or you can get networks with matched (in value and in temperature coefficient) where the absolute accuracy may not be so impressive but the ratio is ...


18

You are right. In most cases this is silly, adds offset voltage, and uses another part. Most likely this is just someone's knee jerk reaction, or blindly following a rule of "always buffer the signal" without thinking about it too hard. Not all schematics out there are the result of good design. There are some subtle advantages to the second buffer-only ...


16

If the datasheet says it is unity gain stable, then yes. Unity gain stable means precisely that the op-amp will be stable when used as you describe. Do be sure the feedback path is short. If you make it long, then its inductance is increased, and maybe weird things will happen. No need to be extremely paranoid on this point; just don't go routing it 10 ...


16

An output pin (GPIO) with both source and sink capability is one that can be used for driving a load up (towards Vcc) or low (towards ground). In other words, it is effectively a push-pull type output driver. The "high capability" means it can support a relatively low impedance load, providing "high amount" (40 mA per pin for ATmega328) of current from ...


14

To make an Audio Voltage source, you want the crossover voltage distortion to be null which requires some quiescent DC current > 1% of the max current. This modest distortion and output impedance is reduced further by negative feedback or excess open loop gain. The active diode bias DC voltage can be predicted in mV for the differential Darlington output ...


14

MOSFETS used to be more common in power amplifiers, but they were often the lateral type power MOSFETS. Most modern MOSFETs (Vertical MOSFETs / HEXFETs) are highly optimized for switching and require very careful design in a linear amp design. For example these modern switching types have a large nonlinear gate capacitance that is difficult to drive. In ...


13

10 kΩ is a typical value for a pullup/pulldown, and the 500 µA loss at 5 V is usually not a problem; the optocoupler's LED will usually need a multiple of that. For low power applications you can increase the value, and the upper limit is determined by the 74HC244's input leakage current. The datasheet says that is maximum 1 µA, then a 1 M&...


13

You aren't allowed to put more than VCC+0.5V on the output pin, even if it's in high-Z state. If you do, you could damage the chip. What's happening in your circuit is there's a diode connected from the output pin to VCC, oriented so that it will be reverse biased in normal operation (cathode connected to VCC). This diode is there to shunt current during ...


12

Your op amp is oscillating because your open-loop gain is larger than 1 at a frequency at which the phase shift is 180°. The op amp in your circuit is driving an almost entirely capacitive load - the MOSFET's gate. There are many possible ways to correct this using just a well-placed resistor or a capacitor. It might be best to use a series resistor or a ...


12

To increase drive capability. A microcontroller I/O pin might only be able to source or sink 8 mA (for example). A buffer chip might be able to source and sink 64 mA. If we want to drive a long wire, or have large fanout to many receiving chips, the we have a large capacitive load and we might need to use a buffer to maintain fast rise and fall times. For ...


12

You have many options. If you need to connect very few optocouplers, you can connect them directly to the GPIO of your microcontroller (through a resistor), provided that: You do not exceed the GPIO output current. You do not exceed the total port current. You do not exceed the total gnd/vdd current. If you need to connect more optocouplers, you can try ...


12

Second Breakdown (Many) Audio amplifiers operate the output stage in their linear region. Modern power MOSFETs are not designed to operate in the linear region. Many of them (HEXFETS) are composed of a grid of hundreds of thousands of smaller FET elements to increase power density and switching speed. Other switching-optimized MOSFET families have similar ...


11

This is really simple - use an N channel FET and have it as a source follower. You can even use a BJT. The one below has gain due to the 3k3 feedback and the 1k to ground from -Vin. If you don't want gain connect the output directly to -Vin and omit the 1k. A unity gain buffer on the output of an op-amp is either an emitter follower or a source follower. ...


11

It doesn't have much effect on the performance, except to make it somewhat slower because there are two poles in the transfer function. Chances are the designer only needed the one op-amp in the dual and chose to do something benign with the remaining amplifier to keep it out of trouble. This is a common situation with LM324 quad and LM358 dual amplifiers. ...


11

Very few do. If your chip's datasheet does not say it has 50-ohm termination, then it almost certainly does not. Traditional CMOS and TTL logic does not provide matching termination, though a few specialized types (line drivers?) might. Typically drivers are low impedance and receivers are high impedance (with some capacitance). Traditional ECL (emitter-...


11

A simple BJT like MMBT3904 or any switching BJT will do the job. You can get a reel of 100 for two bucks.


11

Let's start at the given output driver topology: simulate this circuit – Schematic created using CircuitLab On the left is the basic idea, but it is worth some discussion already. \$C_1\$ exists, in part, because a speaker is a complicated device and doesn't entirely enjoy an average DC current flowing through it and because you aren't using a ...


11

The designers have probably chosen the AND gates because of availability or some other such convenience. Maybe they had the AND gates on the bill of materials for that board already. The gates are used as a buffer for driving the cable. The two gates can output more current than the part which generates the signal initially. There are ICs that do just ...


11

The resistor adds additional hysteresis beyond what the chip provides. With the nature of what the internet is these days quite possibly you may have found a circuit idea that was designed before the time that the a Schmidt Trigger type IC was used in the circuit. It is quite common for photo detectors to operate with very long rise and fall times. When a ...


10

I'm afraid the answer is going to be a microcontroller or other programmable device. My reasoning is as follows: The sensor is a slave device, not a master. Slave devices need an external clock to clock the data out. But only 24 clock pulses. Simple logic devices aren't generally set up to make their own clock. Therefore, the simplest solution will be a ...


10

No, the 50-Ohms is not a convention for PCB tracks to carry signals. The 50 Ohms is a standard for coaxial cables and corresponding interconnects - dozens and dozens various SMA/SMB, BNC, type-N, etc. connectors. In fact, typical (thin) PCB traces have 65 - 80 - 100 Ohm characteristic impedance on a typical stack-up (7 mils or 12 mils of FR4 between ground ...


10

With any CMOS logic IC, you MUST connect unused logic inputs to a known logic level. You may connect unused inputs to either High or Low, whicever is convenient (or whichever is necessary to make the part work as desired). An unconnected CMOS input may take on any level - if it sits at a "maybe" state, the input circuit will draw excessive current, and ...


9

Typically, the pull down resistors for CMOS logic are between 10kΩ and 100kΩ, up to 1MΩ for battery powered devices with low power consumption. At the same time, the LED in the optocoupler is a current-driven device, as opposed to a voltage-driven device. If an LED is connected to high impedance, it will not produce light. Effectively, an LED works as a ...


9

ECL outputs are referenced to the most positive supply rail. This means that any noise appearing on the most positive supply rail will be directly coupled onto the output signal. For example, if the power supply is 5V and GND, then all outputs would be referenced to 5V, and any noise on the 5V supply would also be seen at the ECL outputs. Therefore, the ...


9

This is a typical powered headphone splitter/amplifier. It is a single gain amp to multiple non-amplifying buffers. The 5.1kΩ Resistors represent the load. As it will be in parallel with your actual load, they won't affect much. Notice, no pull-down resistors needed between the Gain Amp and Buffer Amps. This is as generic as it gets.


9

The "buffer" is just there to, as the name implies, "buffer" the output. Since OA1 is part of a feedback network, some of it's output is used already (lost through R2 and R1.) Which means OA1 has less drive capability. So if you were to connect OA1 to some other part of a circuit, unintended things could happen. OA2 simply "follows" or "buffers" the output ...


Only top voted, non community-wiki answers of a minimum length are eligible