# Transformer-coupled MOSFET gate drive circuits

I provided two pictures below regarding the transformer coupled gate driver circuit.

I know that capacitor blocks the dc when its fully charged and allows AC to pass.

However, how does the capacitor reduces the outpus pulse by Vc?

In the second picture on the secondary side of the transformer there is another capacitor which it's polarity is reversed, why? Why there is a second capacitor?

How is the pulse reduced by Vd?

• Well, draw out the waveforms; what would it be? Commented Jul 15 at 19:52
• The linked app note from TI is very thorough and explains these capacitors. What are you not understanding? Commented Jul 15 at 19:59

However, how does the capacitor reduces the outpus pulse by Vc ?

You are feeding a single-ended pulse train (0-VDRV). You can think of this as the sum of a DC offset and an AC pulse train:

Since you are feeding this signal to a transformer you need to remove the DC component otherwise the transformer core will saturate. So the coupling cap removes the DC component which equals the mean (average) value of the pulse train:

$$V_C=V_{avg}=D \cdot V_{DRV}$$

where $$\D\$$ is the duty cycle of the pulse train, or in other words, the ratio of the pulse (on) time and the period.

So the pulses reaching the transformer primary will have negative and positive peaks because the DC component has been removed by the coupling cap. So, effectively, 0V before the capacitor became -VC after the coupling cap, and the positive peak (VDRV) before the coupling cap became VDRV-VC after the coupling cap. The peak-to-peak is still VDRV.

In the second picture on the secondary side of the transformer there is another capacitor which it's polarity is reversed, why ? why there is a second capacitor ?

At the transformer secondary we have a pulse train having negative and positive peaks. An NMOS cannot be turned on with a negative gate-to-source voltage so the negative peaks are simply unusable. The DC component has been removed before feeding the signal to the primary and therefore effective positive peak voltage is reduced by some amount, VC. The resultant positive peak may not be enough to turn the NMOS on, so the lost DC component must be regained. This is called "DC reconstruction".

In the 2nd image, the capacitor-diode pair does the DC reconstruction. Let's first assume that the diode is ideal. What they do is, in the simplest way, store the negative peak (VC) and then push the whole secondary signal up by this amount (You can work out how this happens by following the dot and no-dot voltages for both positive and negative half cycles). With the help of the diode and the stored VC by the capacitor, the secondary signal will be pushed up by this amount and ultimately the negative peak will be -VC + VC = 0 and the positive peak will be VDRV-VC + VC = VDRV.

How is the pulse reduced by Vd ?

We assumed the diode to be ideal, but in reality, it's not. Because of the presence of the non-ideal diode, there's going to be a drop of VD, hence the negative peak of -VD and the reduction in the positive peak by -VD.

First schematic:

After few periods the Vc becomes stable at some value Vc. If duty is 50% the primary has equal positive and negative voltage levels. If the duty is different the Vc gets steady at different voltage, so primary has different positive and negative voltage levels.

At some duty like 10% the secondary voltage levels are not sufficient to drive the gate (positive part too high). The second schematic avoids it because it makes the gate voltage levels always the same.

The figure is for single ended drive. This means you will have 0-10V (or something) and thus you will have a DC voltage across your transformer. This will saturate the transformer, at any duty cycle except 0. The capacitor removes this DC effect. Even on differential drive you could have this problem as small offset voltages would eventually saturate the transformer, but this depends on the offset, saturation current of transformer, and resistances of driver and transformer.

See figure 38,39,40 in your app note -- they don't use a coupling capacitor because they use a differential drive.