stryker1080 said:But what is preventing all the current from being sucked through the NPN when it is turned on and as a result not providing enought voltage to the FET thereby shutting off the LEDs completely?
Because the NPN will NOT turn ON until the current flowing in the R2/FET/LED reaches a value that creates a voltage drop across R1 equal to +0.7V,
When the NPN is turned on, can you not consider the collector-emmiter line to be like a straight wire with no resistance?
NO, It has a an 'apparent' resistance
Do you think you could possibly go though calculations for that circuit to determine what R1, R2 and FET/NPN transistors i need to draw 1A of current to the leds?
If you want a FET/LED current of about 1amp, then the R2 value would be 0.7ohms.. [ Volts= amps * resistance] V= 1 * 0.7R = 0.7V, which turns on the transistor.
That way i would be able to go through the design process and how you are supposed to calculate these. My problem is that i am not sure what depends on what. Like the FET will give a certain current based on base voltage regardless of the emmiter-collectro voltage drop? Im also confused on how if the e-c voltage drop is less than 2 (or close to the base voltage) then the current is controled by the voltage of the e-c, and if the e-c voltage is more than the gate voltage controls it?
But what is preventing all the current from being sucked through the NPN when it is turned on and as a result not providing enought voltage to the FET thereby shutting off the LEDs completely?
Because the NPN will NOT turn ON until the current flowing in the R2/FET/LED reaches a value that creates a voltage drop across R1 equal to +0.7V,
hi,
The FET is a gate voltage controlled device, within the limits of the FET design, for the FET shown in your drawing, the higher the gate voltage the more current it passes. Until the R2 voltage drop switches on the NPN and the gate voltage is pulled down by the NPN thus reducing the FET current.
If you want to check out a circuit without blowing an LED, then in place of the LED connect a current meter able to indicate say 3 amps and the adjust the value of R2 until 1Amp flows in the meter.
The value of R1 would be around 10K
Is this clear.?
The value of R1 would be around 10K
The current will settle at a value of about 0.7V across R2 since that is the voltage needed to turn on the NPN transistor. Thus the current is ~0.7V/R2. If the current varies from this the NPN will either conduct more or less which adjusts the current through the FET to bring everything back into balance. It's the miracle of negative feedback.stryker1080 said:Ok this makes sense, but how can you tell what the actual final current in that branch will be? Sure the NPN can turn on and redirect the current, but what if it doesnt redirect enough and the current ends up stabilizing at 1.5 A? All that NPN does is ensure the switch turns on at 1A, but whats preventing from the current from overshooting.
I'm not sure what this source is talking about. To reduce the collector current you simply reduce the base current (Which slightly reduces the base-emitter voltage. The base emitter junction looks like a diode.)stryker1080 said:There is another question i had, from other Electronic design sources, i read that in order to reduce the current that runs through the collector-emmiter you need to apply a -ve voltage. How can you possibly get a -ve voltage at the base with the circuit that you posted?
hi,stryker1080 said:Ok, i have some updates:
I went back to the site which had the circuit on how to construct this current limiter and they were using a Fairchild FQP50N06L FET transistor. I looked up the specs of it and it is a MOSFET N-channel Enhanchement mode. I looked up what this means, and this particular type of FET is ALWAYS OFF unless a small POSITIVE voltage to the base is applied. Now this is starting to make sense how the NPN will actually limit.
Ok so i think i have a decent understanding of how the negative feedback works usiing these transistors. The only thing i still want to know is what is the purpose of R1? My guess would be that it is there to limit the current that will be going into the FET so to not "blow" it up? How would you go about calculating the value of resistor you need?
Consider that the NPN is NOT conducting, then the full supply voltage would be applied to the FET gate via the R1 resistor
The value of R1 is chosen to suit the gain required by the NPN, I have chosen 10K as a starting point.
Well the battery voltage is going to be variable. I wanted this circuit to be able to drive 1, 2 or 3 or however many LEDs i want. The brief instructions on how to set it up mentioned that it can work with as many LEDs as you want as long as the voltage source is higher than LEDs draw + ~.5 V for dropout voltage (guessing this is the NPN). The author wrote that if you select a voltage source that is higher than required by the LEDs, then the FET will just disipate this extra energy as heat and with a proper heatsink it would be ok.Whats the battery voltage going to be.??
stryker1080 said:Ok, this is starting to make more sense but has also raised a few more questions (bear with me...).
If the NPN is not conducting, how would the full supply voltage be applied to the FET gate? If there is almost no current going through the FET through the base, how would you do a voltage drop loop in this circuit?
When the NPN is NOT conducting its open circuit, so no current flows thru collector resistor, therfore there is NO voltage drop across the R1 resistor, so the full battery voltage is applied to the FET gate. The FET switches very quickly and it starts tp pass current in the LED circuit and R2. The voltage drop across R2 [0.6V] starts the NPN conducting and that causes current to flow in the collector resistor pulling the gate voltage down.
Remember the FET gate draws NO current.
What do you mean chosen to suit the gain?
I have chosen a NPN current of 1mA, which according to the datasheet gives a gain of about 50.
Well the battery voltage is going to be variable. I wanted this circuit to be able to drive 1, 2 or 3 or however many LEDs i want. The brief instructions on how to set it up mentioned that it can work with as many LEDs as you want as long as the voltage source is higher than LEDs draw + ~.5 V for dropout voltage (guessing this is the NPN). The author wrote that if you select a voltage source that is higher than required by the LEDs, then the FET will just disipate this extra energy as heat and with a proper heatsink it would be ok.
If the NPN is not conducting, how would the full supply voltage be applied to the FET gate? If there is almost no current going through the FET through the base, how would you do a voltage drop loop in this circuit?
When the NPN is NOT conducting its open circuit, so no current flows thru collector resistor, therfore there is NO voltage drop across the R1 resistor, so the full battery voltage is applied to the FET gate. The FET switches very quickly and it starts tp pass current in the LED circuit and R2. The voltage drop across R2 [0.6V] starts the NPN conducting and that causes current to flow in the collector resistor pulling the gate voltage down.
What do you mean chosen to suit the gain?
I have chosen a NPN current of 1mA, which according to the datasheet gives a gain of about 50.
crutschow said:I don't mean to confuse the issue but if you're interested in maximum battery life than a switching regulator configured to generate a constant current can be used. This has higher efficiency than a series regulator, particularly if the battery voltage is significantly higher than the sum of the LED voltages. If you're interested in such an approach an example of this is discussed at **broken link removed**.
So just to clarify, you have selected to send 1ma to the BASE which turns on. This NPN has a gain of 50 so the current going through the collector-emmiter will be 50ma. Is that correct? If it is, i have several questions stemming from this:
Its not the base current I have set at 1mA, its the collector current
Also, one other thing that just popped in my head. I looked at the specs for the MOSFET, and if a 6+ volt is applied , the emmiter-collector current is 100 A?! If i had 3 leds and i need about 12V to supply them, and i need 1A (needs a 2V gate), where would the leftover 10V go??
The FET current can never get higher than 1Amp with the values shown.
Lookup negative feedback, from the NPN, the current in the FET is a Constant Current of 1Amp.
stryker1080 said:So you selected the 1ma based on the BASE EMMITER ON voltage VS collector chart on the NPN specs sheet? Basically you can get the collector current based off the on voltage to the base... On the NPN i am looking at i would have about 10mA of current with the 0.7V base voltage.
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