Pull-up because the floating wire in space now is connected to a voltage (or pulled up) via a resistor. In order for open drain to communicate a ONE, they need an external component that connects a TTL level. When not driving a ZERO the output is a floating point out there like a piece of wire without any voltage. Some special outputs, based on the design of the chip, can ONLY output/drive a ZERO (0v). For TTL this voltage level is anywhere between 2v and 5v, that is considered a ONE. To represent a ONE the chip is allowed to output a voltage level. A digital chip (not analog) has the ability to output a ZERO or a ONE. The nmos output type is probably the same as open-drain, but I am not 100% sure. The 5v output is considered weak because any significant current draw will cause the voltage to drop a bit. When the chip wants to output 5v, it turns the low-side switch on and the output goes to 5v. Since the resistance of the switch is essentially zero, the resistor doesnt matter and the output is 0v. In order to output 0v, the chip turns on the low-side switch. This is called a pull-up resistor and is usually around 10k ohms. Outputs from the chip must have a resistor connected to 5v. The chip turns on the low-side switch to drive the output is 0v, and turns on the high side switch to drive the output is 5v (it never turns on both at once).Ĭhips with an output type of open-drain have a low-side switch but no high side switch. One is connected to 0v (low-side switch) and the other is connected to 5v (high-side switch). They use two switches (BJTs or MOSFETs) to do this. Push-pull, totem-pole, complimentary, cmos, and ttlĪre capable of outputting either 0 or 5 volts directly. A digital chip's output should output either 0 volts or 5 volts (assuming 5 volt logic).
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