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I saw this schematic at the link below and I would like to know if there is any use in particular for Q2. Since the transistor Q2 is just, somehow, operating as a simple diode that is following the voltage from the collector of Q1. Is it worth to add the transistor Q2?

schematic

Image source:
Post: Electrical Engineering Stack Exchange question: Driving MOSFET from MCU with BJT
Author: Microsoft Linux TM License: CC BY-SA 4.0

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  • \$\begingroup\$ Simulate the output impedance (change in voltage that's required to deliver a change in current) with, and without, Q2. Then you'll see whether it's useful or not, and its limitations. \$\endgroup\$
    – Neil_UK
    Commented Nov 19, 2023 at 19:29
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    \$\begingroup\$ Nedou - Hi, FYI part of the relevant site rule is that the author of copied material should be named (giving attribution is part of the CC BY-SA license for material copied from Stack Exchange). Therefore I improved the citation to the source of that schematic. \$\endgroup\$
    – SamGibson
    Commented Nov 19, 2023 at 19:36
  • \$\begingroup\$ "... a simple diode that is following the voltage from the collector of Q1"? I hadn't heard of a follower made with a diode... \$\endgroup\$ Commented Nov 19, 2023 at 19:46
  • \$\begingroup\$ I said that because the Base-emitter of Q2 behaves just like a simple Diode so the output VO=(10-15)-0.7V \$\endgroup\$
    – Nedou
    Commented Nov 19, 2023 at 19:53

7 Answers 7

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Q2 is not simply a B-E diode, it is an emitter follower.

Without Q2, R1 pulls up with about 10mA, driving the load rather modestly. This is okay for small loads, like logic-level inputs, or even small MOSFET gates, but anywhere more current is required, the resistor value would get prohibitively small.

Note that R1's current flows continuously while Q1 is on, thus affecting overall efficiency.

With Q2, emitter current can be hFE+1 times the base current, which means R1 can be about as many times larger in value for the same output current. Thus, output current is increased, the load on Q1 reduced leaving more current available for pull-down, and efficiency is improved.

The one downside is the VBE offset, which limits output voltage to pull up to a diode drop below the supply. For a MOSFET gate driver, or line driver, this is fine, or easily compensated by raising the supply voltage a little bit. It's not so acceptable where a more precision voltage is required, or for high efficiency (where you would usually use a proper half-bridge circuit for a high-power load).

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    \$\begingroup\$ Indeed, I made two simulations both for a 100 ohms load: The first one without Q2 and D1 and the second jus alike the schematic above and I observed what you have just said. In the first case I am limited at 10 mA and for the second case I can go up to 100mA. \$\endgroup\$
    – Nedou
    Commented Nov 19, 2023 at 20:34
  • \$\begingroup\$ @Nedou I actually prefer Tim's explanation to Spehro's here. It starts immediately, on line one, where it needs to start. And goes from there. Very simple. (+1 to Tim. I would have muddied things more, I am sure.) \$\endgroup\$ Commented Nov 20, 2023 at 7:14
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Compare the performance with and without Q2 (100kHz with 0.6Ω resistive load):

enter image description here

Without Q2 (specifically with Q2 connected as a diode) there is only the 1k resistor to charge the gate capacitance. So the switching the MOSFET 'on' is very slow.

With Q2 present the transistor current gain greatly increases the current available when driving the gate 'high'.

Aside from poor waveforms, in this particular example the power dissipation of the MOSFET is predicted to be (typically) 580mW. Without Q2 the power dissipation is more than 12 watts, so the MOSFET would fail in seconds.

Expanding a mite, when the MOSFET is to be turned off, the lower 2N4401 pulls the charge out of the gate to ground and we have the advantage of the current gain multiplying the base current. With a 5V drive, base current is about 4.3mA so with a gain of 100+, there is maybe 500mA available.

When the gate is turned off, the base current on the upper emitter follower 2N4401 is reduced to zero (when the base goes to a negative voltage with respect to the emitter as the diode begins to conduct). Then the lower 2N4401 draws gate charge out through the diode. It won't quickly pull it quite to zero, because of the series diode, but this circuit is typically used with MOSFETs that have a much higher Vgs(th) threshold voltage than 0.6V.

Here is the gate current waveform with the first circuit (Q2 functioning as a transistor):

enter image description here

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  • \$\begingroup\$ The case when the MOSFET turns off should also be considered, and the role of D1 should be clarified. \$\endgroup\$ Commented Nov 19, 2023 at 21:27
  • \$\begingroup\$ Nice diode trick... In TTL it is implemented by the input multiple-emitter transistor. Obviously the problem is that the simple NPN emitter follower is an open-emitter stage that can only source current. \$\endgroup\$ Commented Nov 20, 2023 at 8:01
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What is the purpose of Q2 on the included schematic?

Just consider the circuit without the diode (just for convenience): -

enter image description here

Q2 can deliver significant power a load (when the input voltage is zero). If you want to deactivate the load, you have to raise the input voltage to activate Q1. This cuts the base drive to Q2 and, Q2 is capable of supplying hundreds of milliamps to the load (compared to R1).

However, there are two things that can be troublesome with this circuit configuration: -

  • If the load is a capacitor that has been charged up when Q2 was previously activated, when Q2's base is dragged to ground/0 volts (Q1 activating), the reverse bias base-emitter voltage seen by Q2 could be damaging.
  • The load cannot be readily discharged so if you want a kind of push-pull output you won't get it without the diode.

So, with the diode fitted, you run no risk of applying too much reverse bias base-emitter voltage and, Q1 can discharge the load via the diode.

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  • \$\begingroup\$ I don't think they're asking about the diode. \$\endgroup\$
    – Hearth
    Commented Nov 21, 2023 at 5:58
  • \$\begingroup\$ @Hearth I haven't done a very good job of explaining this have I!! I shall make it clearer. \$\endgroup\$
    – Andy aka
    Commented Nov 21, 2023 at 11:20
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The answer to the question you cite almost tells you what you want to know.
When Vi is high Q1 is driven into saturation and Vo is dragged to ground. Q2 is off as its base is -0.7V negative WRT its emitter thanks to D1.
When Vi is low, Q1 is turned off so the base of Q2 is effectively connected to Vcc via R1 and it is consequently driven into saturation. This significantly increases the current that can be drawn from the output over what is possible with only R1 in circuit and/or improves the switching speed (rising edge).

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    \$\begingroup\$ Q2 cannot be saturated; it always works in active mode (of course, if the load is not too low). \$\endgroup\$ Commented Nov 19, 2023 at 21:07
  • \$\begingroup\$ I also think that this is not the role of the diode, but rather to discharge the (capacitive) load, and less likely, to protect the base-emitter junction from reverse voltage. \$\endgroup\$ Commented Nov 19, 2023 at 21:21
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    \$\begingroup\$ @Circuitfantasist Surely, without the diode, any capacitance on the load, when Vi is high, would be discharged slightly faster via Q1 and the Vo would be only Vce sat of Q1 rather than an additional 0.7V. Consider Q2 when Q1 is saturated. The load current flows through D1 and Q1 and establishes Q2's Vbe at -0.7V thus ensuring Q2 is firmly off. I agree D1 may have a secondary function of ensuring Q2's Vbe never gets more then 0.7V negative. \$\endgroup\$ Commented Nov 20, 2023 at 11:35
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Without Q2 and depending on Vi, the output is driven low by the low impedance transistor Q1 and driven high by the high impedance resistor R1. This is an asymmetrical output impedance.

Q2 is there to provide a low impedance output for the output high state, too.

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Q2 makes it easier to switch off the output. When Vi is high, Q1 is active and steals the base current from Q2. This base current is much smaller than the output current; without Q2, all the output current would need to flow through Q1 (and would be wasted).

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    \$\begingroup\$ "switch off the output". Don't you mean switch on the output? Or, if you don't want to get into a fight about "high voltage is on, low voltage is off", perhaps "switch the output high". \$\endgroup\$
    – TimWescott
    Commented Nov 19, 2023 at 20:16
  • \$\begingroup\$ "Active" is not the right word here for the Q1 stage because it works as a switch with two end states off and on. By "active" is meant the intermediate semiconductor state exploited by analog amplifiers. So Q1 is saturated. \$\endgroup\$ Commented Nov 20, 2023 at 20:48
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I think a systematic exposition supported by experiments would be a good addition to the other questions. The idea can be revealed in the sequence below.

Building the circuit

Unloaded CE stage

Let's first remove the second stage (Q2). So we consider a classic common-emitter stage that works as a transistor switch. The interesting state of this stage is off; so let's apply zero input voltage Vin.

schematic

simulate this circuit – Schematic created using CircuitLab

Because the voltmeter is "ideal", no currents flow through the circuit, and the output voltage is equal to the supply voltage.

Loaded CE stage...

Let's now load the circuit with a significant load (100 Ω). Since we are interested in the voltage across it, and do not want to complicate the circuit diagram, we can apply a small "2 in 1" trick by reducing the voltmeter resistance to 100 Ω (CircuitLab allows this frivolity). Thus the load is "voltage visualized".

schematic

simulate this circuit

The low-resistance load draws significant current that creates significant voltage drop. Most of the voltage is lost across R1 and little is left for the load.

... diode inserted...

We move closer to the OP's circuit by inserting a diode D between the output and the load. Almost nothing changes because the diode forward-voltage drop is negligible (0.7 V) compared to the supply voltage (10 V).

schematic

simulate this circuit

... base-emitter junction inserted

We move even closer to the OP's circuit by replacing the diode with the base-emitter junction of the transistor Q2 (or simply disconnecting its collector if the circuit is already configured).

schematic

simulate this circuit

As we expected, nothing changes because the base-emitter junction of a transistor is just a diode.

Buffered CE stage

Now comes the most interesting moment - we connect the collector to the power supply to see what this "magic" (aka negative feedback) is here. The "diode" becomes a transistor, and immediately begins to restore the disturbed equilibrium. Now it (not the resistor R1) begins to pass a much larger collector current through the load, and thus increases its voltage until it approaches the input (base) voltage. There is a redistribution of the current (current steering) from the base-emitter junction to the collector-emitter section.

schematic

simulate this circuit

The transistor settles only when it achieves the equilibrium ratio Ic/Ib = β or Ic/Vbe = gm (transconductance). So the Q1 high-resistance (1 kΩ) output is buffered with the Q2 emitter follower that provides the significant load current.

Discharging diode added

High output voltage: But another problem arises if the load is mostly capacitive. When Q1 is off the load capacitance C is charged to the output voltage of the Q2 emitter follower. Apparently the D diode (for now backward-biased) is put in place to prevent something unwanted from happening the next moment...

schematic

simulate this circuit

Low output voltage: ... when Q1 turns on. Aha, clear... The charged capacitance (represented by the voltage source Vch in the schematic below) is discharged through the diode. Re limits the current.

schematic

simulate this circuit

Conclusion

The simple NPN emitter follower is an open-emitter stage that can only source current. So, when the load is mostly capacitive, a path must be provided for its discharge current. A possible way to do this is through a backward-biased discharging diode.

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