# Reducing transistor switching time

I am currently required to study different ways of reducing the transistor switching time. From what i understand, the two most used methods are speed-up capacitors and Using a negative, non-linear voltage reaction to avoid saturation.

Unfortunately, i don't really understant how either of them work and I haven't been able to find any useful resource online.

Could anybody explain them to me or point me to some useful material ?

• when saturated, a transistor's reaction is much slower. There was a recent post with a link to a 'how transistors work' and it talks about insulation and voltages etc. Maybe it's useful to you. Commented Oct 26, 2014 at 11:00

When transistors saturate there is stored charge in the base that must be removed before it will turn off. One way of doing that is to use a speed-up capacitor, which helps to suck the charge out of the base by increasing the drive current during switching. More information in this on-line source (source of the below image).

It's usually better to prevent saturation rather than trying to deal with it after it occurs, and a popular method is the Schottky transistor - which is an integrated structure combining a BJT and a Schottky diode from collector to base. The Schottky diode prevents saturation by diverting base current to the collector when the transistor gets close to saturation. It's a modification of the Baker Clamp, which performs a similar function without requiring Schottky junctions.

See the Wikipedia article I lifted those images out of for more information.

As you can see from the above diagram, the structure of the Schottky transistor is extraordinarily simple and compact (not much different from a regular transistor) so it was very popular in the era of bipolar logic- the 74LS family (and 74S where high speed was required) was very popular in its time.

• Another method is to use the transistor in its linear region, google 'Emitter-coupled logic'. Commented Oct 26, 2014 at 11:42

This is great. Taking nothing away from the question or the excellent answers.

Shottky Baker clamp and Speedup Capacitor were really important in the 1970/80s.

I had a similar challenge recently. Trying to drive NeoPixels (WS2812B) at 7.5V from a 5V MCU, needed a level shift up to 7.5V to make the first NeoPixel in the chain read the logic high correctly. So I needed a fast switch for the level shift.

I was running the NeoPixels at 1Mhz, but the short pulses (which mean logic zero), are only half as long , so the effective frequency is more like 2 or 3Mhz. So even with both, a Shottky Baker clamp and a "Speed up Capacitor" , I was still struggling to level shift the NeoPixel waveform with a BJT without distorting it badly.

However, perhaps the "most standard small signal MOSFET" out there, the 2N7000, costs pence these days, comes in all sorts of packages including a through-hole TO92, and eats this speed for breakfast. I ordered ten for the box, plugged one in and the problem was solved. The turn on/off times of the MOSFET are faster than what the MCU was able to put out, so the waveform was perfectly preserved.

Same Circuit (minus Shottky and any SpeedupCapacitor).

You can drive the gate of the 2N7000 directly from the MCU, or use a ~470Ohm resistor to "protect the MCU". Choose a pullup resistor on the Drain of the MOSFET which suits the capacitance of your load.

The moral of the story is: BJTs are not very fast at on/off switching, because they have excess charge carriers in the base region when you drive them to saturation. Baker clamp and Speedup capacitor were tricks used in the past to address this issue. They help, but a MOSFET is just much faster, because it just doesn't have the charge carrier build up problem.

MOSFETs have other challenges, but for this kind of small signal high speed on/off switching they are trivial to use, cheap and readily available. So one answer to the OPs question of "How to reduce transistor switching time" is "Don't. Use a MOSFET".

There is another clever technique to quickly switch off a transistor by reverse biasing its base emitter junction. This scheme is demonstrated in this high voltage switching circuit whereby the upper transistor is forced off by having the load current pass through a diode when the lower transistor turns on. The load current through the diode applies a reverse bias across the upper transistor B-E junction causing it to turn off quickly.

The advantage of this circuit is that it is not necessary to have a high voltage control signal for the upper transistor which operates with a self applied BIAS through R3. The main disadvantage is that the output does not switch completely to GND due to the forward voltage drop of the diode.