Current will flow though R1 to the base of Q2, so there will be a considerable voltage drop in R1, resulting in the base of Q2 being far less than 5 V, so reverse biasing the base-emitter junction will only need about 3 V. Pull up or pull down are not push-pull as they do not have two active devices.Ģ) Yes, it is very common to have a diode like that to protect the devices from negative voltages.ģ) No. Perhaps, the author is just trying clarify a point without making an exact calculation.ġ) I have understood "push-pull" to imply that there are active devices driving the circuit high and low, This can be two devices of the same polarity driving though a centre-tapped transformer. How are these voltages, 2.1V, 1.4V, and 0.7V calculated? They add up to 4.2V and not to 5V. It says Vcc=5V but wouldn't it then require HIGH to be greater than 5V if the base emitter junction of Q1 transistor is to be reverse biased. when the voltage goes below the ground level temporarily, the current flows from ground terminal through D1 to the input? This way current wouldn't flow through Q1 transistor. You can see here four circuits and all of them look different from each other considering the layout of components but function-wise they are pretty much the same in the context of given circuit.Ĭircuit #1 and #2 were taken from this webpage.ĭoes this mean that when there is a negative spike at the input, i.e. The totem-pole circuit also does the same thing, in my opinion, making the output low or high but its use is mostly restricted to turning on/off of FET transistors. Such a combination is used to pull the output voltage toward ground or toward the supply voltage in other words low and high. It is my understanding that when the word pull-up/pull-down is used, it refers to combination of a resistor and switch such as this one.
I believe that the words pull-up/pull-up network and push-pull network are synonymous in the context of circuits.
Could you please help me with the queries below? Thank you.
0 kommentar(er)
