Simple Shunt Regulator

Rick Martin

Joined Jun 14, 2009
31
Hi Rick
Sounds good, yes I need to do some more grunt work first, almost there. Problem is I don't have a tutor, so here I am.

Jony
In the Q diagrams the current is coming out of the collector on the second pic. So, umm, if E is the sum of B and C how is the current being "emitted" from C:confused:
No worries mate and to your question to Jony, yes it is and it may help if you flip the diagram up side down so the emitter is on the bottom to help understand the negative relationship of the PNP (2nd pic) compared to the PNP (1st pic), but Jony will expand on that I would say.

Good luck with it all mate.
 

Jony130

Joined Feb 17, 2009
5,335
Jony
In the Q diagrams the current is coming out of the collector on the second pic. So, umm, if E is the sum of B and C how is the current being "emitted" from C:confused:
This diagram show

conventional current flow from "+" to "-".
But in reality in NPN we have "electrons" current from "-" (emitter) to " +" (collector)
And for PNP "hole current" from "+" (emitter) to "-"(collector)
So in NPN emitter emits electrons and in PNP emitter emits holes.


Jony
"What is the voltage drop on R4 and R5?" Huh? I understand the diagrams (thanks!) but the question doesnt make sense. Which R4 and R5 for which diagram? And no voltages are shown so I assume you mean from a KVL perspective.:confused::confused:
never mind, for example for the left circuit


VR4=V3+V2 or VR4=VB-V1
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
never mind, for example for the left circuit
Jony if your point is that resistors in parallel have the same voltage across them your point it taken. Thanks.

Your posts are really valuable and I really like that circuit that shows the current flow with NPN and PNP Q's! That circuit is great! Those tiny little flow arrows make all the difference! Wish they would make that a standard for drawing circuits!

It helps me understand PNP and NPN flow, finally! I still don't get why the PNP and NPN have the collector and emitter on opposite sides so that one always appears to be emitting from the collector while the other emits from the emitter (depending on conventional current view or electron flow view). Wierd.:(

It should not matter whether a Q emits holes or electrons, the emitter should always be the output, otherwise why call it the emitter?
 

Jony130

Joined Feb 17, 2009
5,335
It helps me understand PNP and NPN flow, finally! I still don't get why the PNP and NPN have the collector and emitter on opposite sides so that one always appears to be emitting from the collector while the other emits from the emitter (depending on conventional current view or electron flow view). Wierd.:(
It should not matter whether a Q emits holes or electrons, the emitter should always be the output, otherwise why call it the emitter?
Hmm, don't bother about this, note that arrow on a BJT (on a emitter) symbol show how current will be flow through BJT. Of course from "+" to "-".

To open NPN transistor you need to forward bias the base-emitter junction. Voltage on the base must be larger then voltage on emitter for about 0.6V.
If NPN transistor is full on (saturated) then he act just like a short (switch).
Short collector to GND. Or in general shot collector to emitter.




We get similarly situation in PNP BJT. But this time to open PNP transistor the voltage on a base must be lower (0.6V) than emitter voltage. Or emitter voltage must be higher (0.6V) then base voltage.



And again when PNP is saturate (full on) then he act like switch and shorts emitter with collector. But this time PNP short collector to the Vcc




And maybe you try to analysis this "simple" circuit. When you can treat BJT as a controlled switch.



And tell me:
1. switch S1 -off ; which BJT are open and which are close and which LEDs light.
2. Switch S1 on, which BJT are open and which are close and which LEDs light.
 
Last edited:

Ron H

Joined Apr 14, 2005
7,014
I have a suggestion, Jony130. It may be confusing to some people when you refer to a saturated transistor as "open", since an open switch does not conduct. I know that your implied analogy was to a valve. Others might not.
I believe "on" and Off" are unambiguous.
As I said, just a suggestion.:)
 

Jony130

Joined Feb 17, 2009
5,335
I have a suggestion, Jony130. It may be confusing to some people when you refer to a saturated transistor as "open", since an open switch does not conduct. I know that your implied analogy was to a valve. Others might not.
I believe "on" and Off" are unambiguous.
As I said, just a suggestion.:)
hehe, maybe you're right.
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
Hello Jony

Thanks for the test. Somehow I am sure I have this wrong. I could have built the circuit but that would be cheating. So I did think about it.

But my analysis told me the LED's will be on in both cases so I think I am wrong.

Here is my analysis:

For S1 open:
T3 PNP is in the active region because the emitter is higher than the base because it get extra current from T2. So D2 is on.

T2 PNP is in the active region because the emitter is higher than the base because the base has a 10k resistor. So D1 is on.

T1 is inactive because S1 is open and there is no current to base.

For S1 closed:
As above except T1 is in the active region because S1 is closed and there is current to base which is higher than the current to the emitter because there is 20k R on emitter compared to 10k R on the base.

I am sure my analysis is flawed but I am really looking forward to your explaination of my faulty analysis as I am keen to learn why I can't read this circuit properly but I sense I am close to understanding it with your help. Thanks again Jony.
 

Jony130

Joined Feb 17, 2009
5,335
Oki, If we start analysis circuit we analysed them from left to left right.

S1 - is open:
There is no path (from +) for base current for T1. So T1 is cutoff.
Now T2, since T1 is cutoff there is no path for T2 base current to the gnd. So T2 is cutoff to.
T3 is "on" because there is a path for T3 base current to gnd through.
Vcc--->emiter-base--->R5--->D1-- minus Vcc (GND).
The flow of a base current will "on" the T3 and Ic current will start to flow.
Vcc--->Emitter--->Collector--->R6--->D2--->GND
So we have this situation


S1 closed:
For sure T1 is "on" because there is a path for T1 base current from Vcc through:
Vcc-->R1-->S1-->base-emiter--->gnd.
of course there is a flow of a current through R2, but this current is very small (0.7V/R2).
So T1 is "on" and this mean that collector is short to the GND (T1 is saturated)
So now there is a path for T2 base current to GND through:
Vcc-->Emitter--base T2--->R4--->collector-emitter T1--->gnd
and again there is a flow of a current through R3 that adds to the T1 collector. Ic1=Ib2+IR3
So T2 is "on" to and this mean that collector of a T2 is short to the Vcc.
And there is of course non chance to T3 base current to flow.
T3 is cutoff. And Ic2 current flow:
Vcc-->emitter-collector T2--->R5-->D1--->GND



And maybe you download this simple simulation program.
http://www.mikrokontrolery.net/pawel/crocodile%20clips.rar
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
Hey Jony,
Thanks so much for that sim program, its terrific! I can't believe it has meters and even an occilloscope too! This will save me alot of time and help me learn faster. Its really good.

And there is of course non chance to T3 base current to flow.
T3 is cutoff. And Ic2 current flow:
Vcc-->emitter-collector T2--->R5-->D1--->GND
You lost me here. I will try that software and see if it helps. If current flows through T2 then the base of T3 has a current so T3 should conduct.

Also:

S1 - is open:
There is no path (from +) for base current for T1. So T1 is cutoff.
Now T2, since T1 is cutoff there is no path for T2 base current to the gnd...
Why can't T2 base current go to ground via the collector :confused:

Your analysis is exellent Jony, thanks again. I will spend more time until I undertsand it completely...I will be back soon.
 

Jony130

Joined Feb 17, 2009
5,335
And there is of course non chance to T3 base current to flow.
T3 is cutoff. And Ic2 current flow:
Vcc-->emitter-collector T2--->R5-->D1--->GND
You lost me here. I will try that software and see if it helps. If current flows through T2 then the base of T3 has a current so T3 should conduct.
With this beginners always have problems.
But you have to remember that when T2 is saturate the voltage on a collector is almost equal Vcc (8.8V).
Voltage on the base is too high to open the T3. And that why T3 is cutoff

S1 - is open:
There is no path (from +) for base current for T1. So T1 is cutoff.
Now T2, since T1 is cutoff there is no path for T2 base current to the gnd...
Why can't T2 base current go to ground via the collector :confused:
Now I am confused.
By which collector you see path to the GND?
T1 is cutoff, so there is no any other way to GND, at least I do not see it

And I think I must remind you that in PNP the base current is flow out from the base and "search" the path to gnd.


 

Ron H

Joined Apr 14, 2005
7,014
Why can't T2 base current go to ground via the collector :confused:
Current cannot flow from the collector to the base in an NPN due to the fact that the junction is reverse biased. The only time current flows through the C-B junction is when the transistor is saturated (full on), and then the current will be flowing from the base to the collector, as when S1 is on.
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
And I think I must remind you that in PNP the base current is flow out from the base and "search" the path to gnd.
Yes Jony, I need alot of reminding. This part of your answer explains it to me to some extent. I forgot current on a PNP flows out of the base and not into to it, thats why it searches for ground from the base.

So does that also mean there is no current from T2 E to C because T1 is open and so T2 base is also open. In other words, T2 does not conduct at all beacuse the base of T2 cannot conduct. Is this correct?

If so, all make sense to me for this question.

But you have to remember that when T2 is saturate the voltage on a collector is almost equal Vcc (8.8V).
Voltage on the base is too high to open the T3. And that why T3 is cutoff
What is the limit for the voltage on T3? Is it 0.6v? So the high "8.8v" voltage from T2 to the base of T3 is causing T3 to stay open (not conduct). Is that right? And if so what is the voltage range for the base of T3 i.e. what voltage will make T3 conduct?
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
Hi Ron
Current cannot flow from the collector to the base in an NPN due to the fact that the junction is reverse biased. The only time current flows through the C-B junction is when the transistor is saturated (full on), and then the current will be flowing from the base to the collector, as when S1 is on.
Yes that clears it up for me except I cannot remember what voltage cause a Q to be saturated?

With PNP and NPN are they both reversed biased in normal operation?

Sorry, I have forgotten how Q's work when it comes to being saturated and reversed biased...
 

Jony130

Joined Feb 17, 2009
5,335
So does that also mean there is no current from T2 E to C because T1 is open and so T2 base is also open. In other words, T2 does not conduct at all beacuse the base of T2 cannot conduct. Is this correct?
Yes, correct.

What is the limit for the voltage on T3? Is it 0.6v? So the high "8.8v" voltage from T2 to the base of T3 is causing T3 to stay open (not conduct). Is that right? And if so what is the voltage range for the base of T3 i.e. what voltage will make T3 conduct ?
To force T3 (PNP) to conduct the voltage on a base must be lower then 0.6V voltage at the emitter. In our case emitter voltage is equal 9V.
So to "on" T3 voltage on the base must be equal 8.4V.
I cannot remember what voltage cause a Q to be saturated?
Saturation depends entirely on the current gain β and resistors value Rb and Rc.

For example for this circuit:



β =100
Base current is equal:
Ib=(10V-0.623V)/Rb=10uA
And collector current:
Ic=β*Ib=100*10uA=1mA
And the last emitter current
Ie=Ib+Ic=1.01mA .
Collector current flow through the Rc and causes voltage drop on a resistor.
VRc=Ic*Rc=1V So collector voltage will be equal:
Vce=Vcc-Vrc=9V
If now we change Rb to Rb=187KΩ know we get:
Ib=(10V-0.65V)/187kΩ=50uA---->Ic=100*50uA=5mA
VRc=5mA*1KΩ=5V
Vce=10V-5V=5V
Now it can be seen that if we increase Ib (Vbe increas) ->Ic is increase to and collector (Vce) voltage decreases. And that why we say that CE amplifier gives 180 degree phase shift.
Now we can wonder, what is the max Ic that can flow in this circuit.
We knows Ohms law, so Ic_max=Vcc/Rc=10mA and that give as Ib_max=Ic_max/β=100uA
Out of curiosity we check what is going on if Ib=200uA (Rb=46.5KΩ)
collector current Ic=100*200uA=20mA
VRc=1KΩ*20mA=20V
and collector voltage
Vce=10-20V=-10V
Of course Ohms law and kirchhoff law must hold.
So Ic can not be greater Ic_max(Vcc-Vce(sat))≈10mA and Vce=Vce(sat)=0.1V; Ie=Ib+Ic=200uA+10mA=10.2mA and this "state" is called saturation.
So transistor is in saturation when (Rc/Rb)<β

 

Ron H

Joined Apr 14, 2005
7,014
Saturation in a bipolar junction transistor (BJT) occurs when the base-emitter junction and the base-collector junction are both forward-biased. Vce will be low, usually less than 200mV.
Regarding Jony130's comments, β (beta) is low when Vce is low, so the spec value for β will not work for saturation calculations. Most BJTs saturation specs are defined when Ic/Ib (called forced beta) = 10.
 

Jony130

Joined Feb 17, 2009
5,335
Well I assume that β=100=const.
And I know that in a saturation Ic=Ib*β dont hold anymore
And I built the test circuit, and check my BJT Vce(sat) for:
Rb=200KΩ (2*100K); Rc=1KΩ; Vcc=10V And all my small-signal BJT was saturated. Vce<0.2V for 23xNPN (mostly BC548B;BC337-25) Vce<0.17V for 20 x PNP (BC559B; BC327-25).
And I was surprised that PNP has a smaller Vce(sat) then NPN.
And in my country the most popular BJT is a BC546...9/BC556...9 (SMD version BC84x/BC85x); BC337/BC327 from current gain group B or C.
With minimum gain of 200.
I also check 4xBD139-16; BD140-16.
And they were also saturated for Rb=100K; Rc=1K; Vcc=10V
1xBD244 was saturated to, in this circuit (Rb=100K)
2XBC558 Vce≈2.4V For RB=200K and 0.2V for Rb=100K
So my rule of thumb for saturate is Rb<10..50*Rc in most cases.
 

Thread Starter

howartthou

Joined Apr 18, 2009
111
Jony your answer to my saturation question has lost me for the moment.

You two guru's (Jony and Ron) have left me behind. I am only a grasshopper here. I am just getting to grips with that test Jony gave me with the 3 Q circuit and how it behaves with the switch on and off.

I am going to rewind this thread a little (for my own sanity) so I am off to model that circuit using the software Jony gave me, and see if I really understand the behaviour. I will be back and let you know. I think I have it because you have answered all my questions really well but I need to test the circuit to confirm my understanding.

I will also read up more on saturation so I can also revisit Jony's valuable answer and Ron's clever reply. But I want to model Jony's circuit first just to confirm my undertsanding of the current flow and general behavior (then I will have another go at saturation).

I wonder how long you guys have been at this because your knowledge is amazing.

Thank you so much for your help so far. I shall return soon.
 

Jony130

Joined Feb 17, 2009
5,335
I'm not a EE, and I don't have any electrical education. I'm a hobbyist with
eleven year of experience, and still learning.
 
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