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Author Topic: a question about transistors  (Read 14033 times)

Offline neptune

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Re: a question about transistors
« Reply #15 on: March 03, 2011, 09:10:12 PM »
@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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Re: a question about transistors
« Reply #15 on: March 03, 2011, 09:10:12 PM »

Offline fritz

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Re: a question about transistors
« Reply #16 on: March 03, 2011, 09:46:22 PM »
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.

Offline fritz

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Re: a question about transistors
« Reply #17 on: March 03, 2011, 10:20:22 PM »
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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Re: a question about transistors
« Reply #17 on: March 03, 2011, 10:20:22 PM »
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Offline neptune

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Re: a question about transistors
« Reply #18 on: March 03, 2011, 10:57:06 PM »
@fritz . like I said my knowledge is limited . Thanks for giving information about switching circuits .

Offline Feynman

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Re: a question about transistors
« Reply #19 on: March 04, 2011, 06:18:41 AM »
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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Re: a question about transistors
« Reply #19 on: March 04, 2011, 06:18:41 AM »
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Offline pese

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Re: a question about transistors
« Reply #20 on: March 04, 2011, 08:54:51 AM »
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

Offline Bizzy

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Re: a question about transistors
« Reply #21 on: March 04, 2011, 01:35:38 PM »

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

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Re: a question about transistors
« Reply #21 on: March 04, 2011, 01:35:38 PM »
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