Because the circuit diagram drawn by this structure is a bit like an Indian totem pole, it is called a totem pole output (also called totem output). The output electrode uses an upper resistor connected to the collector of an NPN transistor, and the emitter of this tube is connected to the collector of the lower tube for simultaneous output; the emitter of the lower tube is grounded. The bases of the two tubes are respectively connected to the control of the previous stage. It is the upper and lower output tubes, which are connected in series from a DC perspective, and the connection point of the two tubes is the output end. The upper tube is turned on and the lower tube is turned off, and the output is high level. The lower tube is turned on and the upper tube is turned off, and the output is low level. If the circuit logic allows both the upper and lower tubes to be turned off, the output is high impedance. In switching power supplies, similar circuits are often called "half-bridges". An interesting explanation: Totems are mostly derived from the worship of reproductive organs and their abilities in tribes. Because human life span was very short and survival was difficult in ancient times, they attached great importance to reproductive power that could increase survival ability. When it comes to men, it means that this person has a strong ability, and people in the tribe will admire him very much. The totem pole drive also has the same ability in the circuit: the upward and downward push and pull force is very strong, the speed is very fast, and as long as there is electricity, it will not tire. The role and principle of totem pole drive The role of totem pole drive: The role of totem pole drive circuit is to improve the current driving ability and quickly complete the charging or discharging process of gate charge. When is totem pole driver used? Some tubes may require a relatively large drive current or sink current, in which case a totem pole circuit is needed. Analyze the principle of totem pole boost drive
Description of device function: Qn: N BJT Qp: P BJT Qmos: NMOS to be driven Rb: base resistor Cb: acceleration capacitor Rc: collector resistor Rg: driving resistor Principle analysis: On the left, an input drive signal Drv_b (very weak driving ability) passes through a totem pole output circuit and comes out from the common end of the emitter of the transistor to obtain a signal Drv_g with greatly enhanced driving ability (carrying capacity); from the energy point of view, the weak energy signal Drv_b takes power (obtains energy) from Vcc through the action of Qn and Qp, thus becoming a Drv_g signal carrying high energy; in this energy transfer process, Qn and Qp work alternately in the cut-off and saturation states respectively; the specific working process (logical analysis) is as follows: here, taking a square wave as an example, 1 represents a high level, 0 represents a zero level, and -1 represents a negative level; Vb represents the common base voltage of Qn and Qp, Vqn_c represents the collector voltage of the Qn tube, Vqn_be represents the base-emitter voltage of the Qn tube, and Vqp_be represents the base-emitter voltage of the Qp tube. When the input drive signal Drv_b=1, Vb=1, Vqn_be=1, because: there is a Vcc voltage at both ends of Qn, that is, Vqn_ce=1, so the Qn tube is saturated and turned on, and the Qn tube current mainly flows from the collector to the emitter, Drv_g=1, and the MOS tube junction capacitance is rapidly charged; (Qn tube is saturated and turned on, and the energy provided by Vcc greatly enhances the driving capability) When the input voltage is low level Drv_b=0, Vb=0, Vqp_be=-1, because there is a voltage on the junction capacitance of the MOS tube, that is, Vqp_ec=1, so the Qp tube is saturated and turned on, and the Qp tube current mainly flows from the emitter to the collector, Drv_g=0; at this time, the MOS tube junction capacitance is rapidly discharged; (Qp tube is saturated and turned on, and the MOS tube discharge speed is accelerated)
Circuit Description The totem pole amplifier circuit is composed of two transistors Q2 and Q3, the upper tube is an NPN transistor, and the lower tube is a PNP transistor; the collector of the NPN transistor is connected to the power supply output end of the transformer auxiliary winding, connected to R7, and shares the same VCC with the chip. The power supply voltage is 20V. From a DC perspective, the circuit is connected in series, and the common emitter connection of the two pairs of tubes is the output end. The circuit structure is similar to that of a Class B push-pull power amplifier OCL. Theoretical Analysis The positive and negative half cycles (high-low level) of the square wave signal output by GATE are respectively amplified by the two "arms" of the push-pull output stage Q2 and Q3 in turn. The conduction time of each "arm" is half a cycle of the pulse. The operating frequency of the square wave pulse here is 25-50KHz (the frequency varies according to the load). The logical process of the circuit operation is that when the input is high level, the upper tube is turned on and the lower tube is turned off, and the output is high level; when the input is low level, the lower tube is turned on and the upper tube is turned off, and the output is low level; when the upper and lower tubes of the circuit logic are both turned off, the output is high impedance. In the switching power supply circuit, similar circuits are often called "half-bridge". The simplified and equivalent circuit diagram of the totem pole is as follows
The theoretical calculation is as follows: A. Working state analysis Static: Vi=Vo→→Q2, Q3 are not working, Vo=0V Dynamic: Vi=H (high level)→→Q2 is turned on, Q3 is turned off; Vi=L (low level) Q3 is turned on, Q2 is turned off; the two transistors work in half a cycle respectively, and the working principle of this circuit is similar to that of a Class B push-pull amplifier. From the equivalent circuit, it can be seen that: the driving current Io=C×(Vgs÷Dt)=(Vcc-Vgs)÷R, from which the following relationship is derived: Vcc=Vgs*(1+RC/Dt) ∵て=RC<