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Bootstrapping with transistor

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Indiann

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Can anyone explain, how to use the concept of BOOTSTRAP with transistor in order to raise the output.
For example: If we are using a transistor, we can able to amplify the signal by using the supply voltage.
i.e vin=5 volt and supply volt=12 volt in a transistor, we can able to get a output of 12 Volt
Here my question is how to increase the output as greater than supply volt. I am learning and doing a basic research in electronics, hope it will help me if anyone reply. And sorry, if my question asked is wrong.Advance thanks
 

Hi,

i.e vin=5 volt and supply volt=12 volt in a transistor, we can able to get a output of 12 Volt
Click to expand...
I can´t see your problem. If there is 12V supply..what is the problem to output this 12V?

Klaus
 

A "boost converter" uses an inductor that produces a high voltage spike each time its driving transistor turns off. If you want a high voltage DC then simply use a peak detector at the spikes.
You can also use a "voltage multiplier" circuit made with diodes and capacitors to boost an aC voltage to a higher DC voltage.

A bootstrap diode and capacitor is used to feed a voltage higher than the supply voltage to the gate of a common-source Mosfet.
 
Here's the LTspice simulation of an example bootstrap circuit driving a source-follower (common-drain) N-MOSFET high-side switch.
Notice that the bootstrap generates a gate voltage sufficiently higher than the drain (and source ON) voltage to fully turn on the MOSFET.

Capture.PNG
 
Last edited:
@post #4: Vin =2 V, then Ib~1.3/1k=1.3 mA.
Ic~14.3/5k=2.86 mA
Ic/Ib=2.2 = forced beta.

Why oversaturating the transistor ? I would have thought of Ic/Ib in the range from 5 to 20, not that low.
 
Last edited:

CataM said:
@post #4: Vin =2 V, then Ib~1.3/1k=1.3 mA.
Ic~14.3/5k=2.86 mA
Ic/Ib=2.2 = forced beta.

Why oversaturating the transistor ? I would have thought of Ic/Ib in the range from 5 to 20, not that low.
Click to expand...
No particular reason.
It's just a demo circuit so I didn't pay close attention to the overdrive value.
Typically a value of 10 is used for Ic/Ib.
 

My problem is how to increase the output voltage which should be greater than the supply.
For example , my supply is 12 volt and my output should be greater than 12v.
 

Why not asking a complete question?

Signal type (AC or DC, frequency range, waveform), intended output voltage range, load impedance?
 

Hi,

yes. confusing.

Befor you wanted to output the same voltage than the supply voltage: post#1: Supply voltage = 12V..Output voltage = 12V

But now you say "greater" than 12V.
What do you mean by greater? 12.001V or 120V?

A draft, a schematic, an almost complete explanation, values, units....is needed to avoid guessing. I don´t like to guess here.

Klaus
 

Thanks for your response. I will now clearly explain my problem.
My question is , how could we amplify the circuit to get an output which should be greater than supply voltage.
Example: I have a battery of 12 volt which i want to make use of it, in a circuit to yield an output which should be greater than 12 volt.
 

Unfortunately you are only repeating the vague specification you gave before.

"Bootstrap" is specifically describing a circuit concept where an AC voltage swings above the supply rail, it's e.g. not applicable for DC output voltage. Presently we don't even know if the intended function falls in the bootstrap circuit category.
 
A "joule thief" circuit (look in Google) takes the 1V from a battery cell that is almost dead and creates an oscillator that applies pulses of current to an inductor that makes a higher voltage swing that is rectified and sometimes is filtered into DC to drive an LED blinking circuit that uses 4.5V. My solar garden lights do that.
 
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