WimRFP
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Hello,
The circuit below you cannot simulate in AC analysis as without sufficient drive, there will be no current through the transistors.
I assume that you drive this circuit with about 12Vpp (so you have almost full output swing). The power you can extract from this circuit depends on the current handling of the transistors and dissipation.
Maximum current rating is 200mA, so your load should draw less then that. When your drive this transistors into saturation (square wave drive signal of about 13Vpp), dissipation will be low. This is because when there is current through the transistors, there is almost no voltage across C and E.
When you use a small drive, you will have larger product of Ic*(Vce). Depending on the package of the transistor, it may run too hot. When you have the T092 or SOT223 version the risk over too high die temperature will not be so high.
When you use a 50 ohm load with about 13Vpp drive, the output current will be about 5.6/50 = 110 mA. Given a square wave, output power will be 5.6*110m = 610mW. With a sine wave drive (13Vpp) output will be half (so about 300mW). When you want to drive your transmitter coil, you things become more complicated.
Note that at 100mA, the HFE of the transistors drop significantly so you source should be able to supply at least 5mA. When using a low impedance source and applying too much voltage, you may bias the BC diodes too much. Some resistance between source and input will reduce the change on this.
tip: when you flip Q1 vertically, you can change your drawing so it is easier to read.
The circuit below you cannot simulate in AC analysis as without sufficient drive, there will be no current through the transistors.
I assume that you drive this circuit with about 12Vpp (so you have almost full output swing). The power you can extract from this circuit depends on the current handling of the transistors and dissipation.
Maximum current rating is 200mA, so your load should draw less then that. When your drive this transistors into saturation (square wave drive signal of about 13Vpp), dissipation will be low. This is because when there is current through the transistors, there is almost no voltage across C and E.
When you use a small drive, you will have larger product of Ic*(Vce). Depending on the package of the transistor, it may run too hot. When you have the T092 or SOT223 version the risk over too high die temperature will not be so high.
When you use a 50 ohm load with about 13Vpp drive, the output current will be about 5.6/50 = 110 mA. Given a square wave, output power will be 5.6*110m = 610mW. With a sine wave drive (13Vpp) output will be half (so about 300mW). When you want to drive your transmitter coil, you things become more complicated.
Note that at 100mA, the HFE of the transistors drop significantly so you source should be able to supply at least 5mA. When using a low impedance source and applying too much voltage, you may bias the BC diodes too much. Some resistance between source and input will reduce the change on this.
tip: when you flip Q1 vertically, you can change your drawing so it is easier to read.