Please explain 'no' current to PNP transistor

[windmill john]

Hi All,

Please have a read of the attached picture.
I can see why the NPN transistor turns on the top half of the LED. Not totally sure why the PNP doesn't turn on the bottom half at the same time.
To clarify my lack of understanding, if that's not a contradiction in term! I guess this is an OHM's law question.
There is a positive feed to the base of the PNP via the 2.2K resistor. So is the reason it does not light the bottom half of the LED purely because the resistance 2.2K ohm is a lot higher than the 1K resistance to the base of the NPN?

The comment about the diode confuses me because surely current flow is left to right through it in the diagram?

Thanks.
Beginner section, so be gentle 

[langwadt]

flip the picture upside down then it is much easier to understand

[shapirus]

When the switch is open, current flows (remember that current direction is positive->negative, and the arrows on the transistor symbols follow this convention) like this:

Code: [Select]

battery_pos ->  1K resistor -> NPN base -> NPN emitter -> LED anodes -> LED cathode (COM) -> battery_neg
             `->     NPN collector    -^

It can't flow via the PNP base, because its possible path is blocked with the diode.

When the switch is closed, NPN transistor can't turn on, because current is diverted back through the diode to battery negative, being the path of lowest resistance, instead of going into its base. But now current can flow via:

Code: [Select]

battery_pos -> PNP emitter -> PNP base -> 4.7K -> switch -> battery_neg
                           `-> PNP collector -> LED anodes -> LED cathode -> battery_neg

The 2.2K resistor is optional, as I see it. Possibly added to pull the PNP base high to prevent reverse current flow through the germanium diode (I don't remember off the top of my head what their typical values were).

It's easy to understand cicruits like this when you consider the possible paths and directions for current to flow.

Using animated simulators like https://everycircuit.com/app helps a lot in understanding the basics.

[bdunham7]

There is a positive feed to the base of the PNP via the 2.2K resistor.

And?  What is needed to turn on a PNP transistor according to the last sentence of the first paragraph?

The operation of the diode and the need for it to be germanium is a bit more complex, but yes current will flow through it from left to right when the switch is closed.

[shapirus]

The operation of the diode and the need for it to be germanium is a bit more complex, but yes current will flow through it from left to right when the switch is closed.

This is a good point btw. Considering that germanium diodes are now mostly phased out, this circuit isn't quite an ideal educational example, and even if they were still used, it would need an explanation. A good homework task will be to redesign this circuit with a requirement that the germanium diode must be replaced by a silicon one (no cheating! P-N diodes!), and still have it working properly

up: in fact, I forgot about those LEDs and their junction drop. This circuit, unless I'm wrong, will work as expected with a silicon diode too, provided that the NPN transistor is also a silicon one.

[floobydust]

The NPN is normally on all the time, due to base current flowing in 46, supplied by the 1k resistor.
The PNP is normally off all the time, there is no path for base current to flow out of 40 to ground. The diode blocks current from flowing that way.

When the key is pressed, the NPN is turned off because the key shunts the base current from the 1k through the diode to ground, so it does not reach the transistor enough to light the LED's.
When the key is pressed, the PNP is turned on because the key provides a path for its base current, to ground. Current flows in the opposite direction in a PNP compared to NPN.

Not sure if the transistors are germanium like in the old 100-in-one kits from the 1970's. They could be silicon here but I remember Radio Shack kits calling them the "2SA" and "2SC".
The difference is in the emitter-base voltage needed to turn them on, i.e. 0.3V verses 0.7V

[Zero999]

It's a good educational example of how not to draw a circuit.

Upside down.
No 0V/earth/ground point.
No +V supply voltage reference.

I've redrawn it. Hopefully this will make it easier to understand.

[MrAl]

[ Specified attachment is not available ] [ Specified attachment is not available ]

Hi All,

Please have a read of the attached picture.
I can see why the NPN transistor turns on the top half of the LED. Not totally sure why the PNP doesn't turn on the bottom half at the same time.
To clarify my lack of understanding, if that's not a contradiction in term! I guess this is an OHM's law question.
There is a positive feed to the base of the PNP via the 2.2K resistor. So is the reason it does not light the bottom half of the LED purely because the resistance 2.2K ohm is a lot higher than the 1K resistance to the base of the NPN?

The comment about the diode confuses me because surely current flow is left to right through it in the diagram?

Thanks.
Beginner section, so be gentle 

Hi,

I was off the computer for a while and didnt see that Zero999 had the same idea i had.
I already drew it up though so here is a better rendition of the circuit maybe this will help.
I also removed the random node numbering for clarity.
You have to click on the thumbnail to view the diagram it would not let me post it full size.

I have included five different color LED approximate voltages that would appear at the emitter as shown with the switch off.  The base is just about 0.7v higher than the LED voltage if a regular silicon transistor or a bit less if germanium so you can take it from there.

[Doctorandus_P]

I'd say trow away that book and get another one.

If you have a book trying to explain the basic workings of tranistors, and they start by turning the thing upside down and obfuscaiting the power supply, then that book has already gathered so much negativity that it would be a one way ticket to the garbage bin for me.

Also, transistors are not "turned on by voltage", but they amplify current. That is another reason to throw that book in the garbage.

About the simplest (and incomplete) way to describe a transistor is that te BE junction is a diode, and the CE current is equal to the BE current, but amplified by a constant factor (called Beta, or Hfe).
And of course, if the "programmed CE current can not flow, then it is limited by external circuitry, this is called "saturation", and in that case the voltage over CE can be quite low (can be as low as 100mV).

[windmill john]

Thanks all. Being very new I still find it a little baffling, especially as I thought and you say the current can flow left to right through the diode..

I think I’ll wait until tomorrow, print out a copy of the circuits, so I can read all your posts again at the same time as studying the original and the modded ones; bedtime now!

[shapirus]

I'd say trow away that book and get another one.

In addition, it is either very old (germanium diode, what, really?) or uses circuits taken from old sources.

Not that it is necessarily a bad thing, but for education it is generally better to use more recent literature which illustrates circuits as they are normally drawn today, as well as uses components commonly used and available today.

[james_s]

I'd say trow away that book and get another one.

In addition, it is either very old (germanium diode, what, really?) or uses circuits taken from old sources.

Not that it is necessarily a bad thing, but for education it is generally better to use more recent literature which illustrates circuits as they are normally drawn today, as well as uses components commonly used and available today.

It's obviously the book from one of those 150 (or whatever) in one electronics kits with the little spring terminals, hence the numbers at each connection. If you throw away the book you won't be able to build the circuits by following the instructions. These old kits are mostly collectibles at this point, usually they did have at least one germanium diode which was also used as the detector in the radio receiver circuits.

[shapirus]

usually they did have at least one germanium diode which was also used as the detector in the radio receiver circuits.

which, by the way, could be used to listen to AM radio without batteries .

[ledtester]

This Falstad simulation might be helpful:

(Falstad simulation)

When you press the switch note that the current through the 1K resistor flows through the diode instead of the NPN base -- this is the reason why the NPN turns off and the top LED goes dark.

Why does no current flow through the PNP when the switch is open?

In order for the PNP to conduct current would have to flow out of the base (from 42 to 40). But from node 40 the only path to ground is through the diode and the diode blocks current going from 125 to 126. So no current flows out of the PNP base and thus the PNP does not conduct current from 42 to 41.

When the switch is closed, current out of the PNP base can get back to ground via the switch.

[floobydust]

The ground point is wrong, it's at the mid point of the 9V battery and I don't see how you select that. edit: Inputs and Sources -> Add Ground... nuthin' happening

[MrAl]

This Falstad simulation might be helpful:

(Falstad simulation)

When you press the switch note that the current through the 1K resistor flows through the diode instead of the NPN base -- this is the reason why the NPN turns off and the top LED goes dark.

Why does no current flow through the PNP when the switch is open?

In order for the PNP to conduct current would have to flow out of the base (from 42 to 40). But from node 40 the only path to ground is through the diode and the diode blocks current going from 125 to 126. So no current flows out of the PNP base and thus the PNP does not conduct current from 42 to 41.

When the switch is closed, current out of the PNP base can get back to ground via the switch.

Hi,

That's cute but can you get the node voltages to show up too.  That is also important to understand why the currents flow as they do.

[shapirus]

The ground point is wrong, it's at the mid point of the 9V battery

"Ground" is always a virtual, relative thing (unless we're talking real earth ground) and it can be wherever you decide to assign it. There can be no "ground" at all (however many simulators will refuse to run until a ground point is defined). Where is ground in a circuit consisting only of a LED, a dropper resistor and a battery?

[windmill john]

This Falstad simulation might be helpful:

When you press the switch note that the current through the 1K resistor flows through the diode instead of the NPN base -- this is the reason why the NPN turns off and the top LED goes dark.

Thanks again all, just hanging in there......

ledtester, just need clarification as to why the current doesn't flow through the NPN base as well as the diode. Cannot see the reason for that, even in zero999's picture. Sorry for my (thickness) misunderstanding.

[shapirus]

Thanks again all, just hanging in there......

ledtester, just need clarification as to why the current doesn't flow through the NPN base as well as the diode. Cannot see the reason for that, even in zero999's picture. Sorry for my (thickness) misunderstanding.

because in order to do that, it has to overcome the Vbe drop of the transistor (~0.3V for Ge, ~0.6V for Si) plus the junction drop of the LED (~1.8V? not sure exactly) whereas to flow through the diode, it has to only overcome the ~0.3V drop of the Ge diode junction, and that is the way it "chooses": the least resistance path. There is no proportional division of current taking place here (like you see in resistors), since we're talking about potential barriers (correct me if I'm wrong), and voltage at point 46 doesn't have a chance to rise above the point where it would be able to cause current flowing into the NPN base.

Imagine a dam on a river divided into two halves: one is a couple meters higher than the other. All the water, unless there is too much of it, flows through the lower section, never having a reason to flow through the upper one. Hydraulic analogies often work very well to help understand electricity, so they do in this case: water height = voltage, volume = charge, flow rate (volume/time) = current, pipe cross section = 1/resistance, p-n junction voltage drop = dam height.

[Zero999]

I'd say trow away that book and get another one.

If you have a book trying to explain the basic workings of tranistors, and they start by turning the thing upside down and obfuscaiting the power supply, then that book has already gathered so much negativity that it would be a one way ticket to the garbage bin for me.

Agreed.

Also, transistors are not "turned on by voltage", but they amplify current. That is another reason to throw that book in the garbage.

That's not strictly true.

About the simplest (and incomplete) way to describe a transistor is that te BE junction is a diode, and the CE current is equal to the BE current, but amplified by a constant factor (called Beta, or Hfe).
And of course, if the "programmed CE current can not flow, then it is limited by external circuitry, this is called "saturation", and in that case the voltage over CE can be quite low (can be as low as 100mV).

That's only valid when considering a transistor as a switch. In a linear application, it's easier to analyses a BJT as a voltage controlled current source, with the base current being necessary to bias the transistor.

The ground point is wrong, it's at the mid point of the 9V battery

"Ground" is always a virtual, relative thing (unless we're talking real earth ground) and it can be wherever you decide to assign it. There can be no "ground" at all (however many simulators will refuse to run until a ground point is defined). Where is ground in a circuit consisting only of a LED, a dropper resistor and a battery?

You're right. It is arbitrary, but in this case, the circuit is easiest to understand with 0V being on the negative of the two 4.5V batteries.

A 0V reference is required to perform any calculations. Some simulators will just put it on a node which they think is convenient such as the negative side of a voltage source, in this case. It might be putting the 0V between the two batteries, rather than one end, because that configuration is commonly used to get a bipolar supply for an op-amp circuit, or it could be the negative of first voltage source it sees.

This Falstad simulation might be helpful:

When you press the switch note that the current through the 1K resistor flows through the diode instead of the NPN base -- this is the reason why the NPN turns off and the top LED goes dark.

Thanks again all, just hanging in there......

ledtester, just need clarification as to why the current doesn't flow through the NPN base as well as the diode. Cannot see the reason for that, even in zero999's picture. Sorry for my (thickness) misunderstanding.

The diode is germanium, which has a forward voltage of 0.3V, whist the transistor is silicon and has a base-emitter forward voltage of 0.6V. The transistor's emitter is also connected in series with an LED, which has a forward voltage of around 2V.

[windmill john]

Path of least resistance…..

Thanks both again. It does look like Ohm’s law rules the roost. I do appreciate everyone’s posts, the dam analogy works at my level 

[ledtester]

That's cute but can you get the node voltages to show up too.  That is also important to understand why the currents flow as they do.

Simulation with voltage probe across the NPN B-E junction added:

(Falstad simulation)

You can add more probes with Draw -> Outputs and Labels -> Add voltmeter/Scope Probe.

I also added a ground terminal -- kinda surprised that Falstad doesn't require it. It probably just picks an arbitrary node if you don't have a ground symbol in your circuit.

[floobydust]

Thank you for the Falstad simulations you post. I think it makes it much easier for a beginner to see current flow animated and mouse-over for voltages as well. With the oddball ground reference point the sim had +/-4.5V rails which is not straightforward when learning how this works.
Funny thing is shorting the diode does nothing, I had to laugh at these circuits they are sometimes what was in the designer's head.
The old 100-in-one from the 1970's didn't waste a single part or wire.

[james_s]

The original circuit doesn't have a split supply. the kit just has two banks of AA cells which are connected in series for this particular circuit. I'm actually pretty sure I had this same kit when I was a kid, I vaguely remember this specific circuit, although I suspect the same circuits were used across many of the kits. It's actually pretty impressive they managed to make so many different circuits with such a limited collection of parts.

[windmill john]

How dare you all criticise my 130 in 1 Maxitronix kit!   

I’ll order some breadboard and bits. The kit isn’t visual as components are so spread out. If I stick them on a breadboard and lay then out in a… better manner, it may help me understand better…