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How can a transistor connected in emitter follower configuration be driven into saturation?

I know that for saturation, the base voltage should be greater than the emitter voltage, but also than collector voltage.

I am asking because some H bridges use only NPN transistors with the top transistors connected in emitter follower configuration. I don't understand how can they be driven into saturation.

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5 Answers 5

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Emitter Followers do not saturate. There is a reason for it. Since the load current is pulled by the emitter it can never rise above the base voltage for in-order to get Vce<Vbe. It would require a Vb>Vcc to saturate Vce.

The popular old L298N is rated for low voltage and 2A but not together. If you used a single supply Vcc =5V with a 2A load the voltage across the load worst case is almost zero, meaning almost a short circuit is needed to get 2A. But for low currents a few volts across the load is possible.

It is an emitter follower high side and common emitter low side in an H bridge while both are used to drive a differential load. There is a large drop allowance on the external low side current sense resistors which must be minimized preferably 50mV max then amplified for feedback.

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BJT emitter followers have higher linear operation and better hFE without the saturation effects but also with a higher Vce drop than Vce(sat).

BJT saturation depends on the CB junction no longer being reverse polarized and the Ic current rise and voltage drop of Vce less than Vbe.

This apparent forward conduction of collector-base also reduces the maximum linear hFE current gain into this non-linear mode, as the collector is no longer a high resistance current source but with lower current gain and becomes a low resistance voltage source or switch. Whereas emitter followers have the emitter drawing a base current Ib=Ic/hFE ( and Ic~Ie)

You won’t see these with complementary shared collector outputs because the risk of shoot-thru across Vcc+/- or gnd will short the supply with inductive loads unlike CMOS which share common drains. CMOS or complementary power FETs are easier to control the required “dead-band” or dead-time where both drivers are turned off or partially conducting yet must be protected from flyback voltage.

Generally, Vce=Vce(sat) is often defined by Ic/Ib=10 at some nominal rated currents. When you see transistors rated for Ic/Ib = 20 or 50 it is only because their linear maximum hFE is > 10 greater than these ratios. There exists hFE’s > 1000 rated at 50:1= Ic/Ib but this comes at the expense of other parameters like Rce= delta Vce/delta Ic in saturation and also BW reduces for possible GBW products so they are not common for switching transistors. Diodes In has many patents on on transistor design to improve this and rate their best devices with low Rce in mOhms vs the typical 1 Ohm for a PN2222A but this comes at a great cost,so you won’t likely find it in motor driver IC that need fast dV/dt.

I recall the audio transistor 2N5088 has these high hFE’s with gold doping but also limited to audio Bw applications.

Since Vce is created by the emitter load current pulling down the base current, and the base voltage always being slightly lower than Vcc, Vce can never go lower than Vbe or in other words the CB junction can never see a forward voltage and thus Vce never reach Vce(sat). Thus these “old school motor drivers are only useful where the Vcc is much higher than Vbe to avoid the dropout voltage that is also common the BJT Op Amps. With emitter follower Darlington drivers, it’s even worse with more voltage drop and motor drive power-loss with two BE diode drops.

Thus unless cost is an overwhelming bias to choice, FET bridges are preferred for Vcc <= 12V for lower heat rise and greater efficiency.

Of course you can always sink that heat and raise Vcc to generate a voltage to match the motor limits but this is a big cost-performance tradeoff for low voltage motors.

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  • \$\begingroup\$ If I made any errors in writing, pls correct. \$\endgroup\$ Commented Jul 12, 2021 at 15:44
  • \$\begingroup\$ You said: these “old school motor drivers are only useful where the Vcc is much higher than Vbe to avoid the dropout voltage. What is the point of raising the Vcc in such a circuit? \$\endgroup\$
    – Anna
    Commented Jul 12, 2021 at 16:45
  • \$\begingroup\$ Although 12V L298N motor drivers are cheap and common , worst case 4.9V drop @ 2A means ~ 7.1V/ 4ohm coils or an 8V motor with 4 ohm coils at reduced max RPM is the limit and you need a heatsink for that. So sizing the supply requires more voltage than the motor rating with poor efficiency. \$\endgroup\$ Commented Jul 12, 2021 at 17:13
  • \$\begingroup\$ FET H bridges offer much better RdsOn than this apparent Rce of 4.9V/2 / 2A of 1.25 Ohms per driver average. \$\endgroup\$ Commented Jul 12, 2021 at 17:18
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You need to provide a base voltage higher than the collector voltage. This is the problem with N-channel/NPN high-side switching transistors: you need a voltage higher than the supply rail to turn them fully on.

This isn't as much of a problem as it may seem, though; certainly not enough to outweigh the advantages of using N-channel transistors over P-channel ones.

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Saturation occurs when the collector current is limited externally at the collector, rather than being under control of the base current. They can only saturate when driven such that Vbase > Vcc + 0.6 V - Vcesat.

You're correct that an H-bridge using NPNs won't reach saturation, although if driven at +12 V, the transistor power dissipation isn't too bad, since it's dropping 0.6 V across the collector-emitter path, rather than the 0.2 or so volts in saturation. Your example here does not do this; an integrated H bridge that uses NPNs (like the L293d) will include level shifting.

Something that operates over-the-top, like a charge pump, can also work to achieve true saturation.

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How can a transistor connected in emitter follower configuration cbe driven into saturation?

I know that for saturation, the base voltage should be greater than the emitter voltage, but also than collector voltage.

There's your answer -- the base needs to be brought above the collector supply.

I am asking because some H bridges use only NPN transistors with the top transistors connected in emitter follower configuration. I don't understand how can they be driven into saturation.

If you're thinking of the L293 -- that's an ancient design, from the 1970's or perhaps 1980's. It has huge voltage drops from supply to driven pins because when it first came out it was absolutely awesome for it's time. Now it's what technical folks call "obsolete".

If you look at H-bridge driver designs from the last 20 or 30 years, they're almost always based on FET final stages, and they either use P-channel FETs on the high side, or they use voltage boost circuits to drive the gate of the high-side N-channel FET.

You could design a pretty efficient H-bridge driver using all NPN transistors, if you used "new" (from the 1990's) "superbeta" transistors and if you drove the bases properly. But driving the bases properly means you need at least as elaborate circuit for those bases as you do the gates of FETs. There was a company (since absorbed into Diodes, Inc) that tried popularizing that, based around their BJT technology. But the design decision basically boiled down to "we can make a circuit almost as good as a FET-based design, with a lot more work, and will these BJT's be able to keep up in the long run?"

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I am asking because some H bridges use only NPN transistors with the top transistors connected in emitter follower configuration. I don't understand how can they be driven into saturation.

When NPN BJTs or N-Channel MOSFETs are used as a high side switch in H-bridges, they are driven using an isolated base/gate driver circuit, kinda like the one below. BJTs in an H-bridge are never operated in the emitter follower mode and are driven into saturation.

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Using a gate/base drive transformer (usually 1:1 with a ferrite core) makes the PWM or square wave signal appear between the base and emitter (or gate and source) of the transistor and turns the transistor fully on.

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