What is the use of the PNP transistor in this circuit?

Question

The schematic below describes the structure of a single channel from the 8-Channel Darlington Current Driver TD62783A.

_{Image source: 8-Channel Darlington Current Driver TD62783A}

I understand most of the circuit's workings but I fail to understand why the PNP transistor exists.

I was thinking that it exists to push current from its base to the collector of the transistor Q1 and that the whole PNP and Darlington section of the circuit acts as an active load for the constant current source centered on Q1. But the presence of the high-value resistor R3 confuses me as I don't see any clear purpose for its presence, perhaps to help with biasing the PNP?

How would I go about analyzing the functioning of the PNP transistor and can I determine the values of the currents passing through it?

EDIT:

As it stands, my current understanding of the circuit is:

• The purpose of the circuit is to ensure as big as possible of an output current, not a constant current.

• The constant current source is used to "absorb" the variations in supply voltage.

• The Darlington pair formed by Q3 and Q4 must be driven high-side, hence the use of Q2.

• R3, R4 and R5 are high-valued resistors that discharge the base-emitter capacitance and insure a quick turn-off.

• The gain of the "high-current triple" formed by Q2, Q3 and Q4 is gigantic and as such any leakage current from Q1 (EM interference, stored charge or otherwise) might turn on the LEDs unwantedly. We aim to use R3 to solve this issue.

It's still a bit unclear to me if Q2 is supposed to be saturated (and act as a switch) or in the active region (and boost the current gain of the Darlington pair).

SECOND EDIT:

I've come across this answer on stack exchange and I now can better understand sphero's and AnalogKid's answers.

^{simulate this circuit – Schematic created using CircuitLab}

^{simulate this circuit}

Q2 is indeed an inverted high-side switch. When we turn on Q1, VB2 will be VCC - 0.7V and the PNP inverter will have it's output high. When Q1 is off R3 pulls the base of Q2 to VCC and because VB2 > VCC - 0.7V the inverter outputs a very low voltage.

So in my mind, R3 now has two roles:

• to speed-up the turnoff
• to decrease the voltage at the base of Q2 so that Q2 isn't always off

I also think that the number of LEDs connected to the output is also important, if we have too few connected, Q2 will be in the active region because VEC2 will be much larger than VEB2.

When no load is connected I think we rely on the diode nearest to R5 to keep VOUT high and keep Q2 saturated. During normal operations the diode will only divert an extremely small reverse leakage current from the main flow.

^{simulate this circuit}

Answer 1

The Q2 is nothing more than transferring the signal from Q1 (low side) to the output Darlington pair.
It is because Q3 and Q4 need a high-side driving which is available with PNP (Q2) only.

More precise:
The Q1 current sink sinks R3 and Q2_base current. Than Q2_base current sets the Q2 collector current according Ic_Q2 = Hfe * Ib_Q2
And finally Q2 collector current sets the output Darlington pair current.

Answer 2

It is a digital circuit designed to be able to cope with a wide range of voltages for VCC but with input high and low voltages essentially independent of VCC. The data sheet gives a maximum value of 50V with no minimum specified but will probably function correctly down to 5V or so. The lower limit will be dependent upon the base-emitter voltages of Q3 and Q4.

As you have correctly deduced Q1 forms a constant current circuit that sinks current from the base of Q2 with little dependence of the value of VCC. The current will be about 250uA determined by the voltage from the two diodes at the base of Q1 and the 2.6k ohm value of R2.

The collector current from Q2 then passes into the base of Q3 and Q4 to be amplified to drive the output.

The resistors R3, R4 and R5 are to ensure that the small leakage currents of the transistor do not cause the transistor to be partially turned on when the input voltage is low. The amount of current diverted into the resistors is small relative to the current driving the base but large compared to the nano amp leakage currents even at high temperature.

The resistors also assist in discharging the device capacitance and charges stored in the junction when the input transitions from on to off.

Answer 3

That is the complete schematic for each stage of the part.

Q1, the two diodes, and R2 form a constant current sink at the Q1 collector. This is why there is no resistor in series with the Q2 base-emitter junction to limit the Q2 base current. The diodes do not need to be precisely matched with anything, but they in fact are because they are formed on the same die as all of the other semiconductors.

You need a constant-current drive for Q2 because of the huge operating voltage range of the part. A simple current limiting resistor would mean a change in the Q2 base current of over 10-to-1.

The third diode (to GND) increases the transition voltage of the circuit, the input voltage at which the output stage changes state.

Q2 is an inverting saturated switch. Its job is to yank the output darlington pair up to Vcc as fast as possible. R3 turns off Q2, and R4 and R5 assure a rapid turn-off of Q3 and Q4.

Here is the CC circuit, and a link to a thread on it. One diode and the value of R2 set the sink current, while the other diode is there for temperature compensation to help keep the current "more constant" as the Q1 base-emitter junction heats up during operation.

Cheap, low-quality constant-current sink over wide voltage range

Answer 4

In your analysis you should not consider the PNP to be much like a discrete PNP transistor. It's likely a lateral PNP and as such has very low hFE, more like 10 than 200 or 300.

Q1 forms a switched constant-current sink with approximate 'on' current of 0.7V/R2.

Q2 pulls its collector up to close to Vcc when the base current is 'on'.

The output is a Darlington emitter-follower that does not need much base current to fully switch on (it's degraded a bit by R4/R5 to reduce the effects of leakage when 'off' and at high temperature), so Q2 (when 'on') will be in saturation for rated output currents and the output voltage will be roughly a volt less than Vcc.

Answer 5

When the input voltage rises, Q1 turns on and pulls current from the base of Q2. Q2 delivers current to the base of Q3, and the Q3/Q4 pair turns on.

R3 is there to insure that Q2 positively turns off, even if Q1 has some leakage current (which may be the case at high temperatures, over manufacturing variance), or just from local EMI that might otherwise "blip" the base of Q2 low enough that it might turn on.

Answer 6

Transistors Q2 and Q3 actually forma a Sziklai pair as shown In Figure 1. Q4 modifies the configuration by forming a darlington pair with Q3. The three transistors still function like the Sziklai pair but with higher current gain.

Combining the three transistors into a single element, Qequiv, reduces the circuit to the one shown in figure 2. The configuration might be mistaken for an emitter follower, but at actually acts like a high-gain, high-current open collector transistor.

The Sziklai-Darlington triple has such a high current gain that leakage through Q1 could provide output current. R3 is used to shunt the leakage around Q2's base emitter diode. It must be small enough so that when the input is zero, the output current is also zero.

The circuit around Q1 is a current source controlled by a ground referenced voltage to provide level translation to Q2. R3 also provides a VCC referenced voltage to control the output current through Q2. Q2 directs the current to the output as a source.

So Q2 converts the sinking current through Q1 to a sourcing current to the output.

^{simulate this circuit – Schematic created using CircuitLab}