A question about a Class E Tesla Coil circuit

Question

The link to the original website and the schematics :

http://uzzors2k.4hv.org/index.php?page=4MHzclassE1 http://uzzors2k.4hv.org/projectfiles/4MHzclassE1/proper_4MHz_classE_SSTC.gif

I'm trying to build a 2n3904/2n3906 mosfet driver based on the same design, but at 13.5MHz, but am having some problems.

Here are my questions :

1. I understand the purpose of the 9:4 transformer to be that of impedance matching, however I cannot find any reference on how to calculate the output impedance of the 2n390x push-pull output stage. Judging from the circuit it looks like it's about 200Ohms, but how is this value calculated?

2. The bottom part of the schematic uses 2 2n3904 BJTs. I'm guessing it's a cascode configuration, but that's not the usual arrangement found in the cascode circuits. What exactly is it?

3. Is there any hope I can make this work on my 13.5MHz driver driving IRF740 with C_iss=1400pF?

Answer 1

Operating a class E at ~13MHz with a IRF740? Hmm. Since you asked questions, let's go through those. We'll get back to the IRF740.

1. Gate drive transformer: It's not really about matching the impedance of the gate as it is resonating with the gate circuit. The transformer is needed to float or offset the drive around the FETs threshold voltage, so that the FET without drive is just on the verge of conducting. To get the rise and fall times out of the drive the transformer has to have a fair amount of leakage inductance. You could also say that the transformer is poorly coupled when compared to most gate drive transformers. The leakage inductance is necessary to add Q to the drive so that it is a resonant gate drive. Most likely if the primary is driven with 12Vpp the secondary will have a near sine output with about 20Vpp to 24Vpp. So, not matching, but resonant.

2. The stackup of 2 2N3904s is not a Cascode. Top 2N3904 is just a biased saturating inverter to buffer the crystal oscillator output. The bottom 2N3904 is just a switch to turn on or off the pre-driver. An enable switch for the circuit.

3. Using the IRF740 at 13MHz. Datasheet of the IRF740 doesn't talk about \$R_g\$, the gate resistance inside the package of the part. That's too bad here since it's a really important parameter for an application like this. But, the datasheet does give a dV/dt value of 4V/nSec, that can be used to estimate \$R_g\$. The dV/dt rating gives the rate of rise that can be applied to \$V_{\text{ds}}\$ that will cause 3V or 4V to appear as \$V_{\text{gs}}\$ internal to the FET when the gate is shorted to the source. An equation (reference 60427) to make this calculation is:

   \$V_{\text{gs}}\$ = \$C_{\text{gd}} V_{\text{dsSlp}} R_g \left(1-e^{-\frac{t}{R_g \left(C_{\text{gd}}+C_{\text{gs}}\right)}}\right)\$

   Iterate values of \$R_g\$ to find \$V_{\text{gs}}\$ ~4V, and you'll find that \$R_g\$ ~ 8 Ohms for the IRF740. That amount of \$R_g\$ is a problem since it means a dissipative gate drive will have a time constant of at least 12nSec, while a resonant drive will be hard to get the desired Q to operate. At 13MHz the period is about 75nSec. Class E amplifiers usually operate at ~50% duty cycle, so FET on time will be ~37nSec. It becomes hard to get enough amplitude on the gate to turn the FET on and off in time. This and gate power loss is why a resonant gate drive becomes attractive at higher frequencies. Q needs to be at least 4 or 5, and the frequency for the sine shaped drive needs to be at or near 13MHz. In this case with the IRF740 resonant gate drive needs an inductance of ~ 100nH for 13MHz operation, but that is only a Q of ~1. Qs approaching 5 have a resonant frequency closer to 500kHz. Finally, even if the gate worked out, the dV/dt would limit the peak \$V_{\text{ds}}\$ at 13MHz to ~ 100V. Not sure what your needs are here, but that could be a problem.