# Double Trapezoidal Wave Transmitting System with Controllable Turn-Off Edge
## 3 Double Trapezoidal Wave Transmitting System
### Turn-Off Edge Control Circuit  
Many studies have been carried out to realize the rapid drop of emission current by high voltage clamps, and have been widely used in practical measurement [16; 17]. Whereas clamp technology requires the clamping voltage to be higher than the battery voltage of the transmitting bridge, otherwise the clamp circuit will be broken down. This limits the turn-off time of the emission current. We designed the turn-off edge control circuit, which broke through the battery voltage limit. The turn-off edge control circuit is located at both ends of the coil, which can be divided into a high voltage TVS diode and a low voltage TVS diode circuit. The high voltage TVS diode clamp voltage is much higher than the battery voltage of the transmitting bridge. The transmitting bridge routes a high-power diode (D1) and four MOSFETs (S1, S2, S3, and S4). When the emission current is turned off, the transmitting coil produces a voltage overshoot with high amplitude in the same direction as the current because of the existence of inductance. Due to the diode D1, the voltage overshoot cannot  
Figure 3: The overall structure of the double trapezoidal wave transmitting system.  
be released through the continuous-current diode of the MOSFET, which constitutes the transmitting bridge circuit. When the voltage at both ends of the transmitting coil is higher than the clamping voltage of the high voltage TVS diode, the high voltage TVS1 will be broken, and TVS1 will turn on instantaneously and keep the voltage at both ends of the transmitting coil at the clamping voltage \(V_{1}\) of TVS1 until the current disappears completely. The coil is discharged by TVS1 in the process. Due to the high clamp voltage of \(V_{1}\), the emission current decreases rapidly. The influence of stray capacitance is ignored in the analysis and the discharge circuit of the fast turn-off process is shown in Figure 4.  
The clamping voltage \(V_{1}\) of TVS1 is introduced into the formula to obtain the current fast turn-off time \(t_{f}\):  
\[t_{f}=-\frac{L_{\text{C}}}{R_{\text{C}}}ln\frac{V_{1}}{V_{1}+U} \tag{6}\]  
The low voltage TVS diode circuit is composed of NOR gates, optocouplers, MOSFETs (S5, S6, S7, and S8), and low voltage TVS2. The low-voltage TVS diode clamp voltage \(V_{2}\) is lower than the battery voltage of the transmitting bridge. In the current fast turn-off mode, S5/S6 are off and the low-voltage diode TVS2 does not affect the fast turn-off process.  
In the slow turn-off mode, S5/S6 are on. The control signals of the transmitting bridge are the input signals of the NOR gates. When the transmitting bridge works, the outputs of the NOR gates are low and S7/S8 are off. TVS2 is isolated from the coil. However, when the current is turned off, the control signals of the transmitting bridge reach a low level and then S7/S8 are on. The NOR gates control the TVS2 to connect to both ends of the coil. Next, the coil is discharged by TVS2. The voltage at both ends of the coil is clamped on a very low voltage to realize the slow turn-off of emission current. The discharge circuit of the slow turn-off process is shown in Figure 5.  
Figure 4: The discharge circuit of fast turn-off process.  
Figure 5: The discharge circuit of slow turn-off process.  
In the process of slow turn-off of emission current, the discharge current circuit is as follows: the coil\(\rightarrow\)S8\(\rightarrow\)S6\(\rightarrow\)low-voltage TVS diode\(\rightarrow\)S5\(\rightarrow\)S7\(\rightarrow\)the coil or the coil\(\rightarrow\)S7\(\rightarrow\)S5\(\rightarrow\)low-voltage TVS diode\(\rightarrow\)S6\(\rightarrow\)S8\(\rightarrow\)the coil. The slow turn-off time \(t_{s}\) of the emission current is:  
\[t_{s}=-\frac{L_{\mathrm{C}}}{R_{\mathrm{C}}+4R_{\mathrm{ON}}}ln\frac{V_{2}}{V_ {2}+U} \tag{7}\]  
where \(R_{ON}\) is the on-resistance of MOSFET (S5, S6, S7, and S8).