BJT Load Position

Thread Starter

k1ng 1337

Joined Sep 11, 2020
85
Hi I am interested in the fundamental differences in a circuit with these conditions:

Vin = 5v
Base resistor = 1k
Load resistor = 1k

With a large base resistor in relation to the load I assume the BJT is saturated

NPN (2n3904) + PNP (2n3906)

1) load between + and collecter, emitter to ground

2) + to collecter, load between emitter and ground

3) load between ground and collector, emitter to +

4) ground to collector, load between emitter and +

Thanks :)
 

Ian0

Joined Aug 7, 2020
2,227
The minimum base current required is equal to load current divided by Hfe. With a base current larger than that, the device will be saturated - A larger base current will require a smaller base resistor.
1) is the usual connection for a NPN transistor switching a load - known as "common emitter"
3) is the same thing for a PNP transistor
2) and 4) are "common collector" or "emitter follower" circuits for NPN and PNP respectively.
The same base current is required in all four circumstances.
With an NPN transistor the base voltage will need to be 0.6V ABOVE the emitter voltage.
With an PNP transistor the base voltage will need to be 0.6V BELOW the emitter voltage.

So far everything looks similar. The differences appear when the load is connected to the emitter.
WIth the load connected to the collector the base voltage only has to change by a fraction of a volt to switch the load from OFF to ON. Changing the base voltage from 0.4V to 0.7V can switch the voltage across the load from 0V to 5V
With the load connected to the emitter, the base voltage has to change by the same fraction of a volt PLUS the voltage change across the load. So the base has to go from 0V to 5.6V to switch 5V to the load.
 

DickCappels

Joined Aug 21, 2008
7,441
Since running with Ib = Ic/hfe is not an easy operating point to hit in saturated switches and it moves around readily and hardly predictable It is common for transistor datasheets to specify saturation conditions when Ib = Ic/10. That way the transistors tend to meet the spec.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
85
The minimum base current required is equal to load current divided by Hfe. With a base current larger than that, the device will be saturated - A larger base current will require a smaller base resistor.
1) is the usual connection for a NPN transistor switching a load - known as "common emitter"
3) is the same thing for a PNP transistor
2) and 4) are "common collector" or "emitter follower" circuits for NPN and PNP respectively.
The same base current is required in all four circumstances.
With an NPN transistor the base voltage will need to be 0.6V ABOVE the emitter voltage.
With an PNP transistor the base voltage will need to be 0.6V BELOW the emitter voltage.

So far everything looks similar. The differences appear when the load is connected to the emitter.
WIth the load connected to the collector the base voltage only has to change by a fraction of a volt to switch the load from OFF to ON. Changing the base voltage from 0.4V to 0.7V can switch the voltage across the load from 0V to 5V
With the load connected to the emitter, the base voltage has to change by the same fraction of a volt PLUS the voltage change across the load. So the base has to go from 0V to 5.6V to switch 5V to the load.
Thanks, but you provided me with an explanation for only half the conditions (8 total 4 for each type of BJT), NPN - 1 and 2 / PNP 3 and 4

Clearly these are what are used to make most of the transistors but I am curious to the function in ALL configurations, not just what are logical and commonplace

What happens when a PNP is under 1 and 2? And when an NPN is under 3 and 4?
 
Last edited:

Ian0

Joined Aug 7, 2020
2,227
Generally, I don't concern myself with exactly what happens when I connect up a transistor the wrong way round. I try to avoid doing it.
Maybe it will just sit there and do nothing, maybe it will blow up. Sometimes it can work, but very badly - this happens in some transistor if you swap Collector and Emitter.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
85
Generally, I don't concern myself with exactly what happens when I connect up a transistor the wrong way round. I try to avoid doing it.
Maybe it will just sit there and do nothing, maybe it will blow up. Sometimes it can work, but very badly - this happens in some transistor if you swap Collector and Emitter.
The general idea I've read is simply don't do it but I can't find explicit resources that say why much less any algebra

I've burnt a few transistors with these configurations, my most recent was a half bridge using two NPN, obviously I created an internal short circuit

I suppose a solution is to draw out all the junction configurations in every state -16 total - 4 NPN ON, 4 NPN OFF, 4 PNP ON and 4 PNP OFF then measure voltage across every pin, luckily I have a ton of BJTs to light on fire :)

Thanks for your insight
 

Ian0

Joined Aug 7, 2020
2,227
Nothing wrong with a half bridge with two NPN transistors if driven properly:
https://sound-au.com/jll_hood.htm

Generally with bipolar transistors, the arrow should point from positive supply to negative. If it doesn't it's probably backwards.

(Just to confuse, in MOSFETs the arrow points the other way)
 
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Thread Starter

k1ng 1337

Joined Sep 11, 2020
85
Nothing wrong with a half bridge with two NPN transistors if driven properly:
https://sound-au.com/jll_hood.htm

Generally with bipolar transistors, the arrow should point from positive supply to negative. If it doesn't it's probably backwards.
(Just to confuse, in MOSFETs it points the other way)
I was experimenting but I did get it to work with the help of very well known circuits but I can't help but wonder where I went wrong and if what I did "wrong" actually has a beneficial use elsewhere so I think I'll be drawing up a matrix / table that details all 16 configurations, at the very least I'll identify which ones produce invalid / shoot thru conditions
 

DickCappels

Joined Aug 21, 2008
7,441
Personally I do not consider emitter followers as being capable of saturation because they are active unless the base is close to or below the negative power supply.
 

Ian0

Joined Aug 7, 2020
2,227
Personally I do not consider emitter followers as being capable of saturation because they are active unless the base is close to or below the negative power supply.
I suppose, that if the base voltage were increased above the positive supply voltage, there would be a point at which no further collector current could flow, and a similar situation to saturation would be reached. It becomes debatable as to whether it is still operating "common collector".
 

DickCappels

Joined Aug 21, 2008
7,441
Talking NPN at the moment, if you took the base more positive than the collector then you would be driving the forward biased B-C junction. Let's not forget that you can swap the collector and emitter and still have a working transistor, except the B-E breakdown voltage is lower than the C-B voltage so that voltage would have to be adjusted. In that case it would be saturated but the current flowing in revers of what it would normally be. This "trick" is sometimes used in muting and blanking circuits.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
85
Talking NPN at the moment, if you took the base more positive than the collector then you would be driving the forward biased B-C junction. Let's not forget that you can swap the collector and emitter and still have a working transistor, except the B-E breakdown voltage is lower than the C-B voltage so that voltage would have to be adjusted. In that case it would be saturated but the current flowing in revers of what it would normally be. This "trick" is sometimes used in muting and blanking circuits.
So the basic principle between emitter and collecter is some function of the doped materials area (volume?), how heavily doped and it's position relative to the base region?

What other factors are involved?

I'm trying to get an idea of what's physically happening in the transistor and then relate the algebra.. I suspect that mankind will never really be able to imagine the fundamental forces but the reality is that it is happening and it blows my mind how I have a device switching 100 000 times per second.. almost seems impossible and makes our "real world" problems seem very small
 

DickCappels

Joined Aug 21, 2008
7,441
As far as I can remember the collector and the emitter differ mainly in doping level and geometry. If we are lucky somebody more familiar with transistor design can offer more information.
 

neonstrobe

Joined May 15, 2009
153
I'm trying to get an idea of what's physically happening in the transistor and then relate the algebra.
That's a challenge. You need to understand a diode first.
But essentially the emitter is heavily doped (lots of phosphorus or arsenic in the silicon, for an NPN) and the base is less heavily doped (but P type for an NPN) and the collector lighter still. Forward biasing the emitter and base causes a lot more electrons to flow into the base than holes into the emitter (in a PNP this is the other way around). The electrons are moving in the direction of the collector and end up in the collector to base depletion region, where the electric field sucks them into the collector. The doping differences give rise to the gain so that only about 1/100 of the collector current flows in the base and emitter.
 
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