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Solid States Devices => solid state devices => Topic started by: Bizzy on March 01, 2011, 02:25:07 PM
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Good morning
Can anyone tell me what is the lowest voltage needed in the base of a transistor for it to switch on and allow electricty to flow from the collector to the emitter? Particularly a TIP 120
Thanks
Bizzy
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Good morning
Can anyone tell me what is the lowest voltage needed in the base of a transistor for it to switch on and allow electricty to flow from the collector to the emitter? Particularly a TIP 120
Thanks
Bizzy
Principalle
Germanium Transistors need about 0,25 volt to switch on
Silicon Transistor 0,7 volts. (up to 3 time for power transistors and fully collector currents!)
so TIP120 and some BDX.. as 33, 53 enz) are
DARLINGTON Transistors.
2 Times Base-Emitter Follower are to "fill".
So Minimum Voltage to start is 1,4 Volts.
(for fully current even more)
Seaching in google
TIP120 datasheet , without ""
http://www.datasheetcatalog.com/datasheets_pdf/T/I/P/1/TIP120.shtml
find out the Vbe Caracteristic at the needed IC (Colector Current)
ALS give MANY attenteion ti Vce(sat)
that is the satuartion voltages.
The Transistor will not switch to ZERO volt!!
Evenn an Darlingtin have higher Satuartion values as
normal Transistors.
For Low saturation use GermaniumTransistors or FETs.
PESE
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Adding to what pese has said.
In non-mosfet ie bipolar transistors the base drive works based
on current so you need to look at both the base inherent
device resistance and external circuit resistance to get adequate
current at low voltages for things to begin conducting.
Germanium semiconductors, while having very desirable
characteristics, the germanium semiconductor dopants are not
pinned and can migrate - changing their characteristics over time,
although the temperature and base voltage both play a role.
A silicon Schottky semiconductor can have a transfer characteristic
similar to germanium but at slightly more expensive. For high-efficiency
required circuits, I would try to use Schottky semi's to minimise losses
where available.
:S:MarkCoffman
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Right,
the gaim at darlington transitors is about 10000 times
so it need base current lower than 1 mA to drive many amps.
Als the vltage level that is need on the Darlington TIP120 is also
about less than 1 Volt to switch the fully needet Amp in Collector load.
More details you find in the datasheets
Gustav Pese
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Right,
the gaim at darlington transitors is about 10000 times
so it need base current lower than 1 mA to drive many amps.
Als the vltage level that is need on the Darlington TIP120 is also
about less than 1 Volt to switch the fully needet Amp in Collector load.
More details you find in the datasheets
Gustav Pese
Hi Pese
If What I am understanding then. a TIP 120 would require less than 1 Volt coming into the base to swicth. is that correct?
thanks
Bizzy
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@Bizzy .The actual voltage on the base depends on how "switched on " you want the transistor to be . A transistor is not like a light switch which is either on or off . Varying the base voltage regulates the current flow between collector and emitter .That is how the transistor works as an amplifier .Hope this helps .
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@Bizzy .The actual voltage on the base depends on how "switched on " you want the transistor to be . A transistor is not like a light switch which is either on or off . Varying the base voltage regulates the current flow between collector and emitter .That is how the transistor works as an amplifier .Hope this helps .
Hi Neptune
So if I put more voltage through the base more voltage will be allowed through the collector to the emitter. IS that correct?
Thanks
and thanks for your patience with my questions.
Bizzy
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It depends on whether it's NPN or PNP transistor, which determines how you should hook it up... what you are trying to drive?
Here is an example of a transistor circuit which can drive a relay (energize 9V or 12V relay coils based on a small input signal like a 555 timer)...
http://www.jaycar.com.au/images_uploaded/relaydrv.pdf
I have built this circuit myself with parts from Radio Shack. Hope this helps.
P.S. R1 going into the base of the transistor should be in the ballpark of 4.7k , depending on your current requirements.
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There are guys on here who know much more than I and Feynman is a good example , but having said that one needs to start at the beginning . You stated in your last post , " the more voltage on the base , the more voltage between collector and emitter . " Not quite true . The higher the voltage on the base , the higher the CURRENT between collector and emitter . Think of the collector and emitter as the leads of a variable resistor . the value of this resistor is controlled by the voltage on the base , within certain limits . The name transistor is an abbreviation for TRANSFER RESISTOR . If you are using the transistor to drive a relay , you are really using it as a switch . To switch it on , your base voltage needs to be high enough to pass enough current through the relay to operate it . Build a circuit with a cheap transistor , instead of a relay , use a low voltage bulb with a resistor in series with it to limit current . Apply different low voltages on the base , and watch the results . On a diagram , a NPN transistor has the emitter arrow pointing out ]Not Pointing iN] That's how I remember it .
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Maybe I can sum it up in a view word....
Transistor:
Amplifies current.
A transistor has a significant parameter which gives the current amplification.
This parameter is B/Beta B as plain DC amplification and Beta for AC compounds on top of DC bias.
B is in the range of 150 to > 500 for small signal tranistors and <150 for power transistors - can be even <=10
Additional - the B-E thing is a diode.
For a silicon transistor you have a voltage drop of 350-800mV on the B-E path - depending on the current you feed into Base to Emitter.
Lets take a power transitor with B=50.
If you feed a current into the Base with 100mA -> you can drive a C-E current of 5Amps.
If your load - is in away that its not possible to drive more than 5Amps - even a current of 200mA will not cause more C-E current. In that case you speak about "saturation". Transistors are not switches - even if you put a multiple of the necessary current into the Base - there will still be a voltage drop between C-E. This is the saturation voltage VceSat (another important parameter).
Darlington transistors are 2 transistors within one case which helps to get high current amplifications -even for power transistors. The driver transistor has high current amplification - and the power transistor lower one.
The sum current amplification is B1*B2 - results in a sum B of >500 or even > 1000. Because of the nature of a darlington - The B-E path consists of 2 diodes in series - so the minimume base voltage is >700mV
If it comes to switching - mosfets are the better choice.
They are kind of voltage controlled resistors/current sources.
You need no input current - just an input voltage is enough -
and the Drain-Source voltage can be almost zero - depending on the Rds "on" (thats the resistance between drain and source) and your load current- if switched on.
Depending on technology - the voltage needed between gate and source for switched "on" can be 6V down to 1.5Volts.
The gate can be somewhat seen as tiny capacitor. If its loaded the mosfet conducts, if empty the circuit is open.
Power mosfets have a big capacity between Gate and source >10nF, smaller ones as low as 100pf.
To keep switching losses at minimum - the driver has to fill this cap quite fast - otherwise it may take some time until Rds goes down from open to almost zero - which will cause heating.
So you need energy to keep a transistor conducting -
A mosfet needs only control energy on switching on - or off.
If in on state - a mosfet needs no further input current.
To optimize switching with transistors and mosfets - you use driver stages with assure that the proper current/voltage is fed into base/gate to minimize switching losses. Additional there are effects that the load current influences your base or gate impedance - keeping parasitary effects in mind.
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http://books.google.at/books?id=bkOMDgwFA28C&printsec=frontcover&dq=the+art+of+electronics&source=bl&ots=F2hoQCa4Yo&sig=h1FliBkfus0oM16iDBw9GCGE_ds&hl=de&ei=esRuTffRC4bGtAbwgt3dAQ&sa=X&oi=book_result&ct=result&resnum=8&ved=0CGEQ6AEwBw#v=onepage&q&f=false
In this book you can find driver stages, and lots of other information which may help.
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This is very infomrative and helpfull.
Basically I want to use a transistor to control a pulse motor. Turning on and off at certain intervals. So it sounds like a Mosfet would suit my needs better.
Bizzy
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In actuality, NPN & PNP Xistors can be turned on by MicroVolts.
For proof, look at a Transistor Radio. The voltage on the antenna that feeds the First Stage, is typically Microvolts.
.
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@ Fatbird . What you say is only partly true , and is likely to confuse a novice .
Microvolts from the antenna are fed into a high Q tuned circuit , to magnify the voltage . Also base bias resistors are used to "partly turn on " the transistor already , so that a small signal voltage will turn on the transistor a further amount . That is how signal amplification works .
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@ Fatbird . What you say is only partly true , and is likely to confuse a novice .Microvolts from the antenna are fed into a high Q tuned circuit , to magnify the voltage . Also base bias resistors are used to "partly turn on " the transistor already , so that a small signal voltage will turn on the transistor a further amount . That is how signal amplification works .
Yes I am confused enough...
So is that the reason why you put a resistor at the base of a transistor in a joule theif?
Bizzy
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@Bizzy . I can not comment on the joule thief as I am not familiar with the circuit . So I can only speak in general terms . In a circuit where a transistor is used as an amplifier , the base of the transistor often has two resistors connected to the base . The other ends of these resisters are connected to the positive and negative supply rails of the circuit . So you therefore have two resistors connected in series across the battery , with the base connected to the point where they join . This arrangement is called a potential divider , or voltage divider . The voltage on the base , called the base bias voltage is dependent on the value of the two resisters . this voltage is measured relative to the negative supply rail in a circuit which uses a NPN transistor . Suppose the battery is 6volts . and the resistor between the base and the positive supply rail is 5K [5000 ohms] and the resistor between the base and the negative supply rail is 1K[ 1000ohms . The base bias voltage will then be 1volt .Why not get a book on basic transistor stuff or study on line ? My knowledge is basic , but I will help if I can .
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Yes I am confused enough...
So is that the reason why you put a resistor at the base of a transistor in a joule theif?
Bizzy
In configurations where the emitter is tied to ground (NPN-tr) or to positive supply (PNP-tr) its quite essential to have a resistor in series to the base.
As already mentioned the transistor is a current amplifier - and the resistor transforms an applied voltage into a base current I=U/R. For your calculation you have to keep in mind that the B-E path has a voltage drop of 0.5V - so you have to subtract that 0.5V from your voltage.
To optimise switching you can put a capacitor (some nf - up to hundreds of nf) in parallel to the resistor. This helps with switching.
The circuit with a voltage divider (two resistors from power to ground with base connected to the middle is used for small signal amplification - and not applicable for switching).
In a scenario where Collector is connected to supply(NPN) or ground(pnp) you have a circuit called "voltage follower". In such a case the emitter voltage follows the base voltage (with 0.5V drop). In that case you don´t need a base resistor. The emitter voltage follows the base voltage and you can drive more current there. But you don´t gain voltage amplification - if you feed a 3V pulse into the base - you will have the same pulse at the emitter - but can draw more curren.
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voltage followers are typical used in driver circuits.(to driver power transistors and power mosfets)
The final stage of a transistor power amplifier is formed of 2 voltage followers back2back (one npn - collector on positive supply - one pnp - collector on negative supply - both bases connected, both emitters connected)
While the earlier described switches only drive current in one direction - such a circuit (also known as push-pull stage) can drive a load from the connected emitters to ground in both directions.
If you have for example a weak 5V output of a controller or ´555 - you can use such push-pull stage to get a strong signal to drive a hefty power mosfet gate. Other simple ways to drive a mosfet gate are to take a CMOS4049 or CMOS4050 ic. This consists of 6 buffers(with push pull output) - inverting or non-inverting. These 6 buffers can be operated in parallel - and are an oldschool way to drive a mosfet gate;-))
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@fritz . like I said my knowledge is limited . Thanks for giving information about switching circuits .
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I noticed you guys mentioned switching circuits such as Joule Thief... we just got in preliminary results today courtesy of Physics Prof, and it appears COP>1 (1.05 - 1.4 or so), based on power waveform integration with lab-grade equipment.
http://www.overunityresearch.com/index.php?topic=717.msg11590;topicseen#msg11590
Results are pending further confirmation. Just thought I'd drop the info for anyone investigating with transistors.
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Hi Pese
If What I am understanding then. a TIP 120 would require less than 1 Volt coming into the base to swicth. is that correct?
thanks
Bizzy
The TIP120 , also repacement circuits made with 2 transistors in darlington configuation.
have:
1. current gain amplification of 10.000 and more
Vce must be about 1,2 to 2,5Volt
(see datasheet for the related collector-currents.
Vce(sat) about 1Volt and more
(see datascheets.
With an serie resistor to drive the first bas , you can enlarge the driving voltage and
limiting the input currents
to adapt to driving circuits
Pese
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With an serie resistor to drive the first bas , you can enlarge the driving voltage and
limiting the input currents
to adapt to driving circuits
Pese
Hi Pese
Could you post a schematic of a series resistor for the base?
Thanks
Bizzy