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Saturation questions.

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Hello there,

I learned a little bit about electronics when I was studying my O-Levels, and I'm very interested on this part very much.

I've been reading a lot about transistors lately, but I couldn't seem to understand what's a Saturation.

Okay, I know it's something like the transistor is fully turned 'ON', and base supply voltage is around 0.6V. But I heard that when the Transistor is saturated the transistor will stop working and will turn 'OFF' instead.

I'm very confused about these - please enlighten me, because now I'm a super newbie about the world of electronics. :) :confused:
 

Marks256

New Member
Saturation is when the transistor is fully turned on. :)
 

philba

New Member
Horowitz and hill define it as the point where the collector can get no closer to gnd. typically .05 to .2V above gnd. increasing Ib will not result in increased Ic.
 

hjames

New Member
There's another usage of "saturation" when dealing with MOSFET type transistors. In that case, saturation is something you want to avoid when switching large currents.
 
I'm building a simple blinking LED system here, so I need some more info about the transistors there.

I can just stick a huge IC 555 inside the circuit and just make do with the system, but I wanna know how can I build this without ever using IC (just plain two-transistors, one PNP and one NPN).

I found out that the blinker (or the astable vibrator system) depends on the transistor's saturation point to make the switch disconnected for a while... is that true? :confused:
 

lord loh.

Member
In an astable system, one transistor saturates. The base of the second is connected to the collector of the first. So the base of the second is grounded and the second transistor switches off (cut off) after some time (depending on the Resistor Capacitor time constant), the second transistor receives the base current and turns on switching off the 1st transistor...

The Allen Mottorshed book has all the working vewry well exlplained with waveforms of all different points (nodes) in the circuit.
 
Okay - I got it - the transistor saturates when it's fully turned on.

When searching for single LED blinker circuit (the most basic oscillator example) and found different different versions of these itself.

Check one of these:
http://www.talkingelectronics.com/te_interactive_index.html

and this also:
http://ourworld.compuserve.com/homepages/Bill_Bowden/page5.htm#flash2.gif

Is there a simple explaination about these? I know the capacitor in the system is used to hold the charge for a while, and then release it to the base of one of the transistors until it runs out of current and diminishes.

But the arrangement for the Capacitor seems to be strange - sometimes another version of these saying that the capacitor is connected to the negative side of the battery.

Also it needs more than 0.6 volts to saturate the transistor - what will happen when the charge from the capacitor is completely released? I'm sure it'll be back to around 0.6 volts to the base? :eek:

I'm really a super newbie in these system, especially oscilliator circuits, because they are very complex. :confused:
 

Hero999

Banned
hjames said:
There's another usage of "saturation" when dealing with MOSFET type transistors. In that case, saturation is something you want to avoid when switching large currents.
I think you're confusing it with core saturation which only applies to inductors with iron cores.

FOr more information on transistors check out this:
http://www.williamson-labs.com/480_xtor.htm
 
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hjames

New Member
Hero999 said:
I think you're confusing it with core saturation which only applies to inductors with iron cores.

FOr more information on transistors check out this:
http://www.williamson-labs.com/480_xtor.htm
Not to go too far off topic, but saturation I meant in mosfets refers to "current saturation", i.e. constant current mode. It's the part on the Id vs Vds curves where Id is constant for a given Vgs, and the mosfet is operating in it's linear region. Blah Blah Blah (insert more tech mumbo jumbo). Since the transistor type wasn't specified, it was a possibility.

Hey. that link doesn't even deal with mosfets...

James
 

Hero999

Banned
I understand, the link wasn't aimed at answering you, it was to provide littletransistor with more information on transistors.
 

hjames

New Member
Hero999 said:
I understand, the link wasn't aimed at answering you, it was to provide littletransistor with more information on transistors.
Ah, well, the dangers of toe-stepping...

NPN transistors will turn on at Vbase-Vcollector = ~.6V, and saturate somewhere above that. Saturation is described in terms of Ib - the current going into the base of the transistor. When a transistor is in it's linear region, Icollector = Ibase * Hfe. At some point the transistor goes non-linear and as philba says, increasing the base current does nothing.

For someone starting out, this theory junk is something to read and remember - you'll understand the meaning as you work through more experience. For these oscillator circuits, it's usually enough to think of the transistor as a switch - if current flows through the base to the emitter, a larger current is going to flow through the collector to the emitter. Nearly all the oscillator circuits are going to vary in how they're hooked up, and most of them will work just fine - you'll need to mentally trace through what happens step by step.

For oscillator circuits, just keep an eye on what happens to the voltage across the capacitor - that'll tell you everything. And if in doubt, grab a simulation program or just wire it up and watch it work.
 
Okay, I think I need to mentally trace every step which is happening in the circuit...

I'll keep an eye out on the capacitor's voltage. But I'm still confused why the transistor can suddenly turned off eventhough it has some voltages on the Base from the supply. aah...I need to think more... :confused:
 

Marks256

New Member
That is because the power is going lower than what it needs to be turned on.
 
The explainations in the talkingelectronics website about the Single LED flasher is almost detailed, but there are certain points which I couldn't understand...

http://www.talkingelectronics.com/Projects/FlasherCircuits/Page83FlasherCircuitsP1.html
Taken from the site:

Here is the technical description of the operation of the circuit:
When the supply is connected, both transistors are off and the electrolytic charges via the 330k resistor and 22R. When the voltage on the base of Q1 rises to about .6v, the transistor begins to turn on and the resistance between its collector-emitter terminals is reduced. This allows current to flow in the collector-emitter circuit and Q2 is turned on via the 1k resistor. The 10n reduces the effect (the resistance) of the 1k resistor. Q2 conducts and the LED is illuminated. The current through the LED is limited by the 22R resistor and at this point in the cycle a voltage is developed across the 22R. The negative end of the electrolytic is `jacked up' by this voltage and the positive end pushes the charge on the electrolytic into the base of Q1 to turn it on even harder. In a very short time all the energy in the electrolytic has been delivered to Q1 and it cannot hold Q1 ON any longer. The transistor turns off slightly and this has the effect of turning off Q2 a small amount. The LED begins to turn off and the voltage across the 22R reduces. The negative lead of the electro drops a small amount and so does the positive lead. This action continues around the circuit until Q1 is fully turned off. This turns off Q2 and the LED is extinguished. The cycle starts again by the 10u charging. The charge-time is considerably longer than the discharge time and this gives the LED a very brief flash.
Eh... what's a 'jacked up' and negative feedback? The bolded sentence is a little bit difficult to be understood.

The thing is the power is still around 0.6 volts when the capacitor is charging... and when the capacitor is discharging, uh... the charge is sent back into the base and make it to be saturated. However, when the cap is run out of charge, there's still little electricity from the source right to the base? :confused:
 

hjames

New Member
Eh... what's a 'jacked up' and negative feedback? The bolded sentence is a little bit difficult to be understood.
It's just an idiom - for example car jacks - something that raises objects above ground level. The current through the resistor sets the voltage across the resistor which adds to the voltage already across the capacitor. Since a capacitor can't instantaneously change it's charge/voltage, it pulls it ends up increasing the voltage, and thereby current going into the transistor...

Well, when the cap runs out of charge, say it goes to 0V, and the LED turns off, what happens to the voltage across the 22R? What's the voltage at the base of the transistor at that point, and how much current is going into it?
 

freeskier89

New Member
My analog knowledge has never been really good and I have a question that pertains to this thread's subject. On this page under the flip flop schematic: http://www.talkingelectronics.com/Projects/FlasherCircuits/Page83FlasherCircuitsP1.html (the page littletransitor posted), They have the polarized cap configured so that the positive lead is connected to the collector of the transistor and I have seen it this way everywhere else too. When the transistor saturates, the positive lead of the cap would be tied to ground and there would be a positive charge on the negative lead of the capacitor from the 10K resistor. Something in my methodology must be entirely wrong, because normally, I do not think that applying a positive charge on the - terminal of the cap and negative charge on the + terminal would be good at all for a polarized capacitor. If someone could clear this up for me, I would be very greatful :)
 

audioguru

Well-Known Member
Most Helpful Member
Hi Freeskier,
You are very confusing because you don't attach the schematic that you are talking about.
You linked to an LED Flasher circuit but you seem to be talking about the flip flop circuit at the end of the page. Just copy then upload the circuit that you are talking about.

The charge on the capacitor stays for a while. When the transistor saturates and grounds the + end of the capacitor, then the - end of the capacitor is driven to a negative voltage which cuts-off the other transistor. Then the capacitor is slowly discharged by the 10k resistor then begins charging a little in reverse.
 

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Roff

Well-Known Member
AG, I was also confused until I scrolled down the page and found the "flip-flop" (actually, I've always seen this circuit called an astable multivibrator).
Coincidentally, though, I believe the timing cap in the other circuit you posted (below, modified) is backward. The cap will briefly be reverse biased by about 3 volts when the LED flashes. If it is swapped around (as shown), it will be reverse biased by a maximum of only 0.7V when the LED is off.
 

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freeskier89

New Member
Sorry about the confusion guys. I was talking about the astable monovibrator or "flip flop" at the bottom of the page. Thanks audioguru for your information, that clarifies it a bit for me. I still do not seem to fully understand how a negative charge relative to ground is developed on the negative lead of the capacitor, but that I will look into that very soon. Its about time I get rid of my ignorance towards analog stuff. :) Thanks again!
 

audioguru

Well-Known Member
Most Helpful Member
Like I said, the charge on the capacitor stays on it for a while. The only thing to discharge it is the 10k resistor, and it will discharge to 37% of its voltage in 100uF x 10k= 1.0 second.
So with the capacitor charged to a little more than 1V, when its + end is grounded then its - end goes below ground to a little more than -1.0V.
 
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