-1
\$\begingroup\$

circuit

I'm building a MOSFET driver like this. The signal source is 7V peak-to-peak. Vcc is 12V. The push-pull stage is 2n3904/2n3906.

I've tested the push-pull output before adding the transformer, and using a 50 ohm load, the output voltage is reduced by half, so I concluded that the output resistance of the push-pull stage is 50 ohm.

Then according to the MOSFET spec (IRF740), C_iss = 1400pF, at 13.5MHz I calculated the impedance to be 8 ohm. So I used a 2:1 transformer (therefore 4:1 impedance transformation) to match the load at the gate.

But now I get a flat signal at the Q3 gate, AC is almost non-existent. If I disconnect the MOSFET Q3, I get a good sine-wave. Can anybody help me find out what's the problem with this circuit?

C2 is a 104 capacitor which is practically a short at RF. The 2n3904 has a spec of peak 200mA current capacity, which I translate to be 2.5V for 1/4 period, which should get doubled by the transformer and is supposed to be enough to charge the gate to a observable voltage.

\$\endgroup\$
  • \$\begingroup\$ For completeness you might want to mention what C1 is and the frequency of your sine signal. \$\endgroup\$ – PlasmaHH Dec 16 '14 at 13:59
  • \$\begingroup\$ drain and source connected to ground. \$\endgroup\$ – JIm Dearden Dec 16 '14 at 13:59
  • 2
    \$\begingroup\$ What happens to the drive signal at the bases when you connect the MOSFET gate? The current gain of those transistors won't be much at 13MHz, maybe 20, so you're loading the oscillator output with < 1K. \$\endgroup\$ – Spehro Pefhany Dec 16 '14 at 14:11
  • 1
    \$\begingroup\$ For digital (square wave drive), a gate driver chip such as Micrel MIC4420, for a video amplifier such as MAX9650. Both are cheap. An Apex power amplifier would be better again, but $$$. \$\endgroup\$ – Spehro Pefhany Dec 16 '14 at 14:37
  • 1
    \$\begingroup\$ Start simple. Leave out T1 and Q3. Load the output at C2 with the equivalent RC of Q3 gate. That way you don't worry about T1 and Q3 now. Expect to need a class AB bias for Q1/Q2. With the bases tied together there is a dead spot in the emitter drive will be killer at 13 MHz. Don't do any of this on a protoboard. Physical layout of all the parts and routes will be very important ... keep all loop areas as small as possible. \$\endgroup\$ – gsills Dec 18 '14 at 4:11
1
\$\begingroup\$

To fully turn on an IRF740 it needs about 7V on the Gate. You have 7Vpp available, but using AC coupling cuts the usable voltage down to only 3.5V. Since the IRF740 doesn't even start to turn on until about 3.7V, you won't get anything out of it.

If you want to drive the FET with AC then you need to step the voltage up (not down) by about 2:1. Alternatively you could use a 1:1 transformer and apply a bias of about 3.5V to the 'cold' side of its secondary winding, raising the DC Gate voltage voltage above ground to make use of the full 7Vpp. Or you might be able to simply DC couple your push-pull driver to the FET.

I get a flat signal at the Q3 gate, AC is almost non-existent. If I disconnect the MOSFET Q3, I get a good sine-wave.

Have you measured the output impedance of your oscillator? My guess is that it is quite high, so the voltage collapses when a load is applied. You may need another stage of amplification to provide a low impedance rf source. Most rf transmitters rated for 5~10W have at least two stages of power amplification between the oscillator and final output.

The 2n3904 has a spec of peak 200mA current capacity, which I translate to be 2.5V for 1/4 period,

It might be able to take such abuse, but above 100mA its gain collapses so it won't provide the impedance reduction you need. You should treat the 200mA rating as an absolute maximum, not normal operating current. You need transistors that can deliver 800mA or more while still providing a reasonable current gain.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.