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11th November 2021

Universal Transistor (UniT) Driver

What is a transistor driver

Transistors are so called "active" components, meaning that they need external power source to work as they should. Transistors are 3 terminal components (usually Collector, Base, Emitter). Path between the Collector and Emitter (C-E) is the high current path that is being switched with low power Base-Emitter (B-E) signal. With ordinary bipolar transistors, the current flow through C-E is determined by current flow B-E. Current flow through C-E is limited to current flow through B-E times some gain constant. This constant is called either Beta or hFE. BJTs (Bipolar Junction Transistors) have Beta in order of 100s, meaning that current flow through C-E is limited to 100 times the current flow through B-E. In other words; Ic = 100*Ib. For low power applications, base current doesn't have to be any high to switch the load. For example: switching 100mA power-LED only requires 1mA of base current (assuming Beta = 100). This current can be easily provided by direct output of a microprocessor. But that is a rare case of very low power load. If on the other hand the load consumes 100A, we would need 1A of base current. This cannnot be provided by a microprocessor and we need an amplifier stage between the microprocessor and the transistor itself. This amplifier stage is usually called a transistor driver.


In modern power applications we need to switch very high currents at high frequencies. While BJT transistors are useful for a lot of applications, switching high power and high frequency loads is not one of them. BJT transistors are quite slow. They may be able to switch very quickly (>1MHz) but very low powers. And vice-versa, they can only switch low frequencies when it comes to high power loads. Thankfully Metal Oxide Semiconductor Field Effect Transistors (MOSFETs) solve this issue. Unlike BJTs, MOSFETs can switch high frequencies at quite high currents. We can switch a lot of Amperes at few MHz frequencies when driven correctly. MOSFET's terminals are called (Drain, Gate, Source). While our appliances and electronics get more and more demanding, another type of transistor was invented; Insulated Gate Bipolar Junction Transistor (IGBT). IGBTs are a compromise between switching frequency and switching current. IGBTs are usually used for medium fast applications as they can only handle lower frequencies than MOSFETs. On the other hand, IGBTs can switch very high currents. They are even made to withstand short circuit conditions for several microseconds. They are perfect for pulse applications. 

In summary: BJTs are obsolete in high power applications, they are slow and inefficient. MOSFETs are made for high frequency, medium currents and IGBTs are made for medium frequencies and very high currents. 

When it comes to power loss, we can model BJT and MOSFET transistors the following way:


Equivalent simplified BJT schematics

Equivalent simplified MOSFET schematics

Both these transistor types have their own advantages. Notice how MOSFET Gate is simply a capacitor, this means we don't need any current to flow once we charge it. MOSFET will stay conductive as long as its gate is charged. This of course means less power loss. Yet in high current circuits BJT is way more efficient. 

BJT C-E path can be modeled as a simple diode with more or less constant voltage drop. On the other hand MOSFETs act as resistors between D-S. Voltage drop of a resistor increases with flowing current. 

Therefore to calculate power loss on an BJT we would use following formula:

P = U * I

Where U is the constant voltage drop across C-E diode. The only variable in this equation is the current I. While for MOSFETs both voltage and current would change. For MOSFETs we can rewrite the formula the following way:

P = R * I^2

Where R is the constant on-state resistance of the MOSFET. Notice now that the power loss is not linearly proportional to the flowing current. It is now proportional as a square of current. This means that for a BJT, raising the flowing current by a factor of 10, means 10 times the power loss, but for a MOSFET it would mean 100x the power loss! Modern MOSFETs usually have very low on-state resistance (we also call it Rds(on) ) and therefore are usually more efficient than BJTs even in higher power applications. But there will always be a current for which BJT will be eventually more efficient. Wouldn't it be nice, if we had some transistor that would act as a capacitor on the input and as a diode on the output? 


Equivalent simplified IGBT schematics

And that is exactly what IGBT is. It is quite literally a combination of a MOSFET switching a BJT therefore it acts as a MOSFET on the input and as BJT on the output. Notice the names of the terminals: C, G ,E (Collector, Gate, Emitter). It's a mishmash of MOSFET and BJT terminals as it is a mix of these two.

IGBTs combine both advantages of BJTs and MOSFETs; no current is needed for the transistor to stay conductive after gate is charged and the power loss is linearly proportional to flowing C-E current.


On the first glance it may seem like MOSFETs and IGBTs need no amplifiers to switch huge currents, but unfortunately that's not the case.

First of all, to properly switch MOSFET transistor, the gate needs to be charged to at least ~12V, for IGBT it's even more. Microcontrollers usually use 5V or 3.3V logic. And the second of all, the gate capacitance is actually not that low. Power transistor' gate capacitance can be in order of 10s of nano-Farads. Doesn't sound like much, but if you need your transistors to switch fast; let's say 200ns from switched off to fully switched on, means that you need to charge 10nF to ~12V-20V in 200ns! Let's calculate the average current flow during this gate charging:

I = C*dU/dt = 10nF * 20V/200 ns = 1A

1 Ampere of current. So Initially the current flow can be quite big, some Gate drivers are built to provide up to 10s of Amperes current to switch very big transistors with huge gate capacitances (>100nF, usually the bigger the transistor is, the higher is its gate capacitance).


High side switching

Another problem comes in if we switch transistors with "floating" Source pin. Floating means that it is not connected to the ground of the circuit. This unconnected node can be at any voltage potential, it can even be "floating" on high voltage and/or high frequency signals. Any components that can be floating on high voltage signals are called "high side" components.

On the following schematics we can see a full bridge of transistors switching a load and voltage waveform on one of the high side transistors:



Full bridge of MOSFET transistors

Voltage waveforms on high side transistor' Source pin


4th February 2022

It is clearly visible that for high side transistors we cannot simply use any Gate driver. We need a gate driver with floating ground reference as the Gate voltage always has to be in reference to the Source of the transistor. There are some IC gate drivers on the market which achieve floating grounds thanks to charge pumps or capacitive insulation, but they are usually not very robust and do not offer many features and protections. I want to design a robust driver with protection against overcurrent, desaturation, under-voltage and cross-conduction. But before the work on the driver can start, I have to design insulated power supply so I can connect the gate driver to floating Source of high-side transistors.

Insulated 4 channel power supply

The transistor topology I work the most with is the full bridge, it is also very often used in commercial high power power supplies. Therefore I need an insulated power supply with 4 channels to switch 4 transistors. 

So without much thinking I designed this:


Low voltage half-bridge driven by IR2153


PCB design of 4channel insulated power supply

I haven't even received the PCB and I already know this is not the final design. I only want to try if 12V input voltage is enough to provide enough power. Half-bridge design generates half the source voltage on the output, this way the primary of the insulating transformer will only get 6Vrms, which is quite low considering I would like this power supply to provide at least 50W. Also the secondary circuits are quite inefficient, they use linear voltage regulators which aren't just very lossy but also quite fragile. I the next revision I will probably use boost converters with voltage feedback for better efficiency and stability of the output voltage.


30th May 2022

So unfortunately I have made a few mistakes while designing the 4channel PSU. I have redesigned the PCB and schematics very slightly.




Modular PSU channels with linear stabilization (unsoldered negative rail)

Now the PCBs are modular and output voltages are adjustable with LM317 linear regulator IC. Each channel was tested to provide over 10W at 24V. Since the voltage regulation is done linearly the efficiency increases as output voltage reaches higher. The reason I decided for linear voltage regulation was simplicity. Also cascading multiple switching power supplies results in too much noise and shorter life expectancy. 

The big issue with this supply is that it's quite big. There is no way of fitting all of this inside a small laboratory power supply let alone a mobile phone charger. But this is a problem I will deal with in the future, right now I only need a reliable 4 channel power supply.

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