Darlington pair as a switch

[Rusttree]

I'm experimenting with driving a larger voltage with a uC. I thought I understood how to do this using a darlington pair, but the circuit is not behaving as expected.

I'll spell out my line of thinking and I'd very much appreciate if someone could explain where I'm going wrong. In the schematic, pin 28 of the uC goes high and allows current to flow into the base of T1 (very low current). A higher current flows through the emitter of T1 and directly into the base of T2. This increases the current gain even more. I believe the collector-emitter voltage drop on a 2n3904 is about 0.3V, so I expected a voltage of 4.2V across Rload.

Instead, I'm getting a collector-emitter voltage drop of 3V across T2, which leaves only 1.5V for the load. Reviewing the datasheet for the 3904s, I don't see that I'm exceeding any capabilities of the transistors. I suspect there's something fundamental I'm not understanding. Thanks in advance for your help.

-Dan

 

[MikeMl]

What is the supply voltage for the PIC?

Even if it were 5V, the PIC pin will not pull all the way to 5V when sourcing more than a few mA. The Vbe of the first transistor is about 0.7V, while at higher current, the Vbe of the second is likely to be ~0.9V, so 4.5-0.6-0.6 is about 3V at the emitter of the second transistor. Remember that you are using these as two emitter followers in series, so the Vbe drops are additive.

You really should be using a high-side PFET as a switch. Doing it that way, you will get almost 4.5V across the load.

 

[Rusttree]

A 3.3V linear regulator.

 

[MikeMl]

A 3.3V linear regulator.

Oh, that's even worse. The pic pin will only pull to ~3V, so 3V -0.7 -0.9 = 1.4V.

 

[Rusttree]

Ok, so conceptually I have it right. But the 3904 is a bad transistor to use considering the voltages I'm working with. Would you agree with that?

 

[MikeMl]

,,,But the 3904 is a bad transistor to use considering the voltages I'm working with.

I would say that using an NPN transistor to drive the load is a bad choice for your application.

 

[crutschow]

Ok, so conceptually I have it right. But the 3904 is a bad transistor to use considering the voltages I'm working with. Would you agree with that?

Conceptually that's not right for your requirements. NPN devices, when driving a grounded load from a positive voltage, are acting as emitter followers, thus the output voltage is at least one diode drop (or two diode drops for a Darlington) below the 3.3V base drive voltage. Thus you need to use a high side driver (PNP bipolar or PMOSFET) transistor to drive a grounded load with minimum voltage drop, as MikeML showed.

 

[Rusttree]

Understood. Thanks for the explanation. And thank you MikeMI for taking the time to draw that circuit diagram.

-Dan

 

[MikeMl]

Understood. Thanks for the explanation. And thank you MikeMI for taking the time to draw that circuit diagram.

-Dan

That's not just a drawing; it is a running simulation of the drawn circuit. The green trace is the simulated voltage across the simulated load resistor. Note that it is just one Vce-sat (~0.1V) less than the 4.5V supply voltage.

 

[audioguru]

A darlington pair of transistors is not needed and emitter-followers are not a switch.

Since the original circuit used 2N3904 transistors that work poorly above 100mA then the load current is probably about 50mA.
If the load is at the collector of a 2N3904 that has a base current of 5mA (the PIC can provide up to 25mA) then the collector-emitter saturation voltage loss is typically only 0.08V and the load gets (4.5V - 0.08V=) 4.42V.

 

[Rusttree]

Alright, I've been going back over this in my head and I think there is something more fundamental I'm misunderstanding about BJTs. I realize I have two distinct notions about BJTs that are in direct conflict with each other (I'm visualizing the attached image as I type this out):

Notion 1: A transistor can act as a switch if the base is saturated. As the current through the base approaches the saturation point, Ve will approach Vc. Ve will never actually reach Vc because of the Vce voltage drop.

Notion 2: Ve is equal to Vb minus the Vbe voltage drop.

So, in notion 1, Ve is a function of both Vb and Vc. In notion 2, Ve is only a function of Vb. Which one is true? Or are they both flawed?

 

[audioguru]

Your transistor is an emitter follower that does not saturate. The emitter voltage is about 0.7V less than the base voltage and the collector voltage does not matter (if it is high enough so that the transistor does something). If the collector voltage is 10V and the base voltage is 5V then the emitter voltage is about 4.3V.

A common-emitter transistor circuit saturates the transistor. The base voltage must be about 0.7V and then current can flow in the base current-limiting resistor. Usually the base current is 1/10th the collector current. Then the collector voltage goes down almost to 0V. So almost the entire supply voltage will be across the load like if a switch was used.

 

[Rusttree]

Ok, thank you. That helps me understand the theory a little better.

It would seem, then, that the common emitter switch is the way to go in terms of simplicity. Does the simulated circuit MikeMI posted above have any advantages over the common emitter switch?

 

[audioguru]

Does the simulated circuit MikeMI posted above have any advantages over the common emitter switch?

Mike didn't answer yet so I will guess.
Maybe he thinks that your load must be connected to ground like in a car.
Then he uses a common-emitter NPN transistor to drive a common-emitter PNP transistor that drives the load.

 

[MikeMl]

Since the PIC pin can only drive from 0V to ~3V+, the NPN must be there to reference the base drive of the PNP from 4.5V (Off) to ~3.8V (On).

 

[Roff]

If the load must be grounded, then Mike's circuit is required. If the load can return to +4.5V, then Audioguru's grounded-emitter NPN can be used.

 

[audioguru]

Why does the PIC output go up to only 3V? It is not the same as old TTL. If its supply is 4.5V then its output voltage is at about +4V with a load of 10mA.
Its outputs are the same as high-speed Cmos (74HCxxx) so they can supply 25mA with a fairly low voltage loss.

 

[Roff]

Why does the PIC output go up to only 3V? It is not the same as old TTL. If its supply is 4.5V then its output voltage is at about +4V with a load of 10mA.
Its outputs are the same as high-speed Cmos (74HCxxx) so they can supply 25mA with a fairly low voltage loss.

The OP said the PIC is powered by 3.3V. The output should go almost to the rail if the load is light.