How to forward the body diode of the mosfet without triggering the parasitic bjt?

Question

For some applications the body diode of the mosfet is desired but how to make it forwarded without turning the parasitic bjt on ? The bjt's base is the body and to turn the diode on we must give its anode a voltage higher than its cathode and the diode's anode is the body so we must give the body a voltage to turn the diode on but that will turn the bjt on too and the mosfet will not operate correctly then how in some applications the body diode turns on ?? Edited : This is the same structure TemeV used and here is my question : if the substrate for some reason wasn't shorted to source because some mosfets have a substrate terminal non shorted now that means the anode of the diode and the base of the transistor will have the same voltage on them and if the diode turned on wouldn't the bjt turn on too ? If it does then turning the diode means that no current will pass from d to s and it will all pass through the diode

Answer 1

To inject current into the body, but avoid run-away currents, the 3_D (3 dimensional) structure must be designed to provide a bipolar current gain less than ONE. That can be done.

In the inherited layout/design I hardened against high-slew-rate transients, the cure was to greatly expand the connections to the bulk, whether the wells or the substrate.

Insights came after many hours of staring at thermal-emissions (recombination photon emisisons) microscope images.

Answer 2

FET's body is the base of the internal BJT, source is the emitter and drain is the collector. Typically FET's body terminal and source terminal are connected. This connetion is drawn with red in the picture below. It means that internal BJT's base is short circuited to the emitter. Thus V_BE voltage is zero and BJT is not turned on.

With source and body short circuited the remaining PN junction is between body and drain. Because body and source terminals are connected, there is a diode between drain and source. You don't have to care about the internal BJT.

You can use the body diode just like any other diode. Just apply negative G_DS voltage (the source is at higher voltage than the drain) and the diode will conduct. Just like any diode, there will be roughly 0.7 V voltage drop when there is current flowing through the diode.

The picture below depicts how the internal BJT and internal diode form from the P and N substrates of the MOSFET. The PN junction of the diode is of course the same PN junction as BJT's base-collector junction.

Edit:

if the substrate for some reason wasn't shorted to source because some mosfets have a substrate terminal non shorted now that means the anode of the diode and the base of the transistor will have the same voltage on them and if the diode turned on wouldn't the bjt turn on too ? If it does then turning the diode means that no current will pass from d to s and it will all pass through the diode

Oh, you are talking about the rare case of four terminal MOSFETs. Sorry for the misunderstanding.

Yes, if you drive current through that diode, the BJT will turn on. Though, now the emitter is the drain and collector is the source.

The current will flow from source to drain if you:

1. Apply greater voltage to source than to drain (The BJT is unidirectional component and requires positive V_CE to conduct). Then the BJT will conduct as the BJTs do. Though the current gain is likely very low.
2. Apply sufficient gate-body voltage. Then a channel forms between drain and source, and the MOSFET conducts like the MOSFETs do.

With the option 2, the current will also flow from drain to source, because the MOSFET conducts in both directions.

The PN junctions from body to drain and from body source will work like diodes. Just like BJTs base to emitter and base to collector junctions.