16
\$\begingroup\$

I commonly see weak pull down resistors at the base of NPN transistors. Many electronic sites even recommend doing such things, usually specifying the value as something like 10x the base current limiting resistor.

Bipolar transistors are current driven, so if the base is left floating, I see no need to pull it to ground.

Also, I commonly see gate current limiting resistors on FETs.

They are voltage driven and there is no need to limit current feeding the gate.

Are these two situations examples of people confusing the rules between transistors (which need base limiting resistors) and FETs (which need pull down resistors) or combining the rules or something...

or am I missing something here?

\$\endgroup\$
  • 2
    \$\begingroup\$ Bipolar transistors really aren't current driven. They're fundamentally voltage controlled devices as illustrated by the Ebers-Moll model; the base current is just an imperfection! \$\endgroup\$ – Bitrex Jun 30 '11 at 0:39
  • 3
    \$\begingroup\$ @Bitrex: Bipolar transistors conduct due to minority carrier injection into the base region. These are "used up" in the process of allowing collector-emitter conduction, so more need to be injected to keep conduction going. Constant injection of carriers is current. You can look at phenomena all kinds of different ways, but to me thinking of bipolar transistors operating on current seems closest to the physics and also the most useful for designing circuits with these transistors. \$\endgroup\$ – Olin Lathrop Aug 1 '12 at 13:03
16
\$\begingroup\$

The reasons become clear when you are considering not only the ideal behaviour of the transistors but also their parasitic elements.

The pull-down resistor at a npn-type BJT's base helps keeping the base "low" whenever the driving element for the base resistor should be unconnected or in a tristate mode. Without this resistor, charge entering the base via the capacitance between collector and base ("Miller capacitance") could remain there and turn on the transistor.

There are two common reasons for a series gate resistor in a MOSFET circuit. One is that the resistor limits the driving current and allows for some control of the gate charge current (think of the gate as a capacitor that needs to be dis/charged in order to turn the MOSFET off or on). With a carefully chosen resistor, you are able to get some control over the turn-on or turn-off transition times of the MOSFET. Sometimes, you even use a resistor paralleled by a diode and another resistor to have different charge and discharge currents, i.e. a chance to influence the turn-on-time in a different way than the turn-off time. The second reason for a base resistor is that the trace inductances around the MOSFET form a resonant LC tank with the MOSFET's parasitic capacitances. When all you want is a clean transition of the gate voltage (rectangular waveform), you may get a lot of ringing in reality. The ringing may be so severe that the MOSFET turns on and off a couple of times before settling and finally obeys to what the driver requests. A resistor inside of the LC resonant circuit around the gate driver is able to damp this resonance and the path between driver and gate is the easiest spot to put the resistor. For small-signal circuits, these resistors may not be necessary, but when driving power MOSFETs, you absolutely need them.

\$\endgroup\$
15
\$\begingroup\$

A series resistor in the gate line of a MOSFET will protect the driver (microcontroller) from ringing effects caused by parasitic inductances.

The optimum value for Rg is very application dependant. You want the MOSFET to switch as quickly as possible to minimise switching losses, but not so fast that parasitic inductances and capacitances associated with the pcb layout and any wiring to a load, will cause high di/dt voltage spiking or ringing.If you find that an optimised value of Rg controls switch on OK but slows the turn off too much, then a fix is to place a diode across Rg with its cathode towards the gate drive circuit. This will bypass Rg during turn off thereby speeding up the turn off. Placing a resistor in series with the diode will enable you to control turn off time independantly of turn on. Further reading (for all aspects of mosfet switching).

For switching small loads (like 100mA), or when a real MOSFET driver chip is used, the gate resistor is probably not needed.

(Note: these links were on the first G results page for "mosfet gate resistor")

\$\endgroup\$
  • 2
    \$\begingroup\$ The gate resistor is known as an "anti-snivet resistor" according to page 88 of Pease, Troubleshooting Analog Circuits. \$\endgroup\$ – markrages Jun 29 '11 at 6:54
  • 2
    \$\begingroup\$ @markrages - Ah, good old Bob "RAP" Pease. I've no idea what a "snivet" is though... \$\endgroup\$ – stevenvh Jun 29 '11 at 10:23
  • 3
    \$\begingroup\$ RIP, RAP. And RIP, Jim Williams, too. How sad to see them go. \$\endgroup\$ – zebonaut Jun 29 '11 at 19:35
11
\$\begingroup\$

The series resistor to the MOSFET gate is sometimes required to reduce a current peak when switching due to the gate's capacitance. Logic circuit, esp. microcontrollers allow only for a very low capacitive load. It can also be used to reduce slew-rate (the switching speed).
A pull-down on the gate is used to prevent the gate from floating if the controlling I/O is configured as input. In this case the resistor's value can be chosen pretty large (1~10M\$\Omega\$).

The base resistor on the BJT is often combined with a pull-up, and this combination is used to set a stable quiescent point. [our teacher in college, not very good at English and apparently only having seen the word in print pronounced it as "keskent". It took us a while to understand what he meant :-)]

\$\endgroup\$
4
\$\begingroup\$

Most transistors have a small amount of collector-base leakage; if there isn't any pull-down, this current will be amplified by the transistor's gain. In situations where leakage isn't a concern, the resistor may be omitted, but if leakage current is a concern, adding the resistor can reduce it.

\$\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.