No, I'm not allowed to buy components from Mouser Electronics until I show I understand the circuit or/and math involved.
I have several power sources, a 6000 counts DMM, LEDs, MOSFETS, Tempature Sensors, Arduino Uno, some resistors, and a SSR.
How do you turn on an NPN BJT Transistor?
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I have several power sources, a 6000 counts DMM, LEDs, MOSFETS, Tempature Sensors, Arduino Uno, some resistors, and a SSR.
Ok, I looked closer and you have a better start than your voltage labels show...
You calculated Base Current correctly. 200uA
You have 5V supply feeding a 21500 resistor that is connected to the transistor base.
We then assume, to a first approximation, that the voltage drop from base to emitter is 0.7 volts.
That means the VOLTAGE DROP ACROSS the resistor (Rb) will be 4.3 V.
4.3/21500 = base current.
My confusion was that you put 4.3V at the base of the transistor. The 4.3V dropped across the resistor. So the base pin of the transistor should be 0.7V (relative to ground).
Is that clear?
Sorry for the bad photo - I'm in the dark ages here.
The parenthesis show the voltage drop from the top to bottom of the parenthesis. To find the VoLTE vs ground, add up the stack of parenthesis bottom to your point of interest.
Also, the voltage drop from emitter to collector of the transistor is assumed to be 0.1 V if the transistor is saturated. We will get to some details on that later as well.
Right, I just didn't like that you wrote 4.3 V by the base. 4.3 is the voltage difference between the two ends of the resistor. The base is at about 0.7V.
I screwed up. It's not 200uA, and it's not 4.3v supply voltage. Sorry about that.
5v comes off the power supply, goes into the new pre calculated 12,500Ω Resistor, and supplies the base of the NPN Transistor with 0.0004A or 400uA.
What happens after that, I don't know. Whether the base-emitter, or just the emitter has the diode that drops 0.7v, is the transistors problem. I fed the transistor with 400uA, with the new 12,500Ω resistor.
5v / 12,500Ω = 0.0004 A
0.0004 A / 50 Hfe = 0.02 A (20mA) Ic.
HFE = Ic / Ib, 0.02A / 0.0004A = 50 HFE. Perfect, fully on.
Calculating base current, according to the HFE.
0.02A Ic / 50 HFE = 0.0004A Ib
Calculating the base resistor, based on Supply voltage, and base calculated base current.
5 Vs / 0.0004A Ib = 12,500Ω
Calculating resistor power dissipation.
0.0004A² x 12,500Ω = 0.002 Watts or (2mW).
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The error is in "calculate base current, assuming 0.7 V". Build the circuit. Measure base voltage.
So then this circuit is correct.. Right? Or does the Diode not effect the supply voltage, and it's actually 5v going into the resistor at the base?
Ok, you are understanding and all of your numbers look in the right range.
Now, we are ready to talk about a next step in understanding.
Saturation.
What would happen if you use 5000 ohms for the base resistor? And that is the only change. How would the current on the LED change (or not change) and why?
Now, we are ready to talk about a next step in understanding.
Saturation.
What would happen if you use 5000 ohms for the base resistor? And that is the only change. How would the current on the LED change (or not change) and why?
I'll take a crack at it now, and see if I get it.
5Vs / 5000Ω Rb = 0.001A Ib
HFE = Ic / Ib, 0.02A Ic / 0.001A Ib = 20 HFE
20 HFE.. 20HFE / 50HFE = 0.4 x 100 = 40% Brightness.
The LED will only light @ 40% brightness.
0.4 x 0.02A = 0.008A, the LED, regardless of the resistor for the LED, will only draw 8mA.
Because the NPN Transistor is only being forward biased 40%.
But I'm still confused about the Diode on the base-emitter junction. Every video I've seen shows that the supply voltage get's dropped by 0.6 or 0.7v. So the actual supply voltage going to the base resistor is 4.3 - 4.4v. So the brightness changes according to that factor, if that's true.
w9cd7B5QRRo?t=7m15s
and also 5DoYuxDzczQ?t=2m45s
But I don't understand, is how is the voltage effected like that, before it even goes threw the emitter. How is the voltage effected, or how does the voltage drop like that, before it get's to that diode. So calculations are 4.3 or 4.4v, and not the full 5v.
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The procedure is okay. The assumptions are not. You assume a base voltage of 0.7 V. You assume gain of 50. You can't do much better on paper, in theory. Build a circuit with real parts and step away from theory to real world parts.
Do they like devices?
I think they should be the components or parts, a device should be made by many other components, as if cent is the component and dime is the device, and the dollar is the big device, what do you think ...
You will have to get down to using real world parts. You don't really know what the gain of the transistor is or if it is in saturation under these conditions.
I think you should start to study what is the linear and what is the logic.
1. Logic test -- Control the leds, relays, motor, working in saturated region is better, and the hFE is set to 10.
2. Linear Test -- Control the leds, relays, motor, working in linear region, and test the hFE by youself, adjust the base(in series with a VR100K and 1K fixed resistor) current and to measure the Ib and Ic current and Vce, through this testing then you will know why we using the logic method to control the load.
3. PWM Test - a new method that you worth to dig.
1. Logic test -- Control the leds, relays, motor, working in saturated region is better, and the hFE is set to 10.
2. Linear Test -- Control the leds, relays, motor, working in linear region, and test the hFE by youself, adjust the base(in series with a VR100K and 1K fixed resistor) current and to measure the Ib and Ic current and Vce, through this testing then you will know why we using the logic method to control the load.
3. PWM Test - a new method that you worth to dig.
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It is trying to be precisely that; concise.
Mouser and such are more oriented to manufacturing. Jameco is more friendly to non-manufacturing people.
Okay. A good definition.Do they like devices?
I think they should be the components or parts, a device should be made by many other components, as if cent is the component and dime is the device, and the dollar is the big device, what do you think ...
Re: 0.6 or 0.7 V
Okay for the classroom in theory. It doesn't really apply in real life. Simulations are only theory. Looking at a data sheet for a transistor we find gain can be anywhere from 50 to 300 under specific conditions. You really can't get precise. What does the part you actually have do in reality?
Components are not precise. Resistors can be 5% off. Capacitors 20% or more. In theory you might calculate precise numbers but when you get to the real world you will find it was only an illusion.
re: videos you reference
Good for basic theory. Looking at a data sheet for the transistor, 2N23904. At the base currents the videos suggest (1 mA to 5 mA) our base voltage could be as high as 0.9 V. At very low currents it might be as low as 0.47 V (per my exercise). On page 3 of the data sheet there is a chart of Base to Emitter On voltage. It could be as low as 0.4 V or as high as 0.9 V. Keep in mind the temperature is what is inside the transistor, surrounded by that plastic that keeps the heat in.
Gain, from the data sheet), can be a very wide range. The data sheet states expected gain at different collector currents. When designing a circuit make sure the gain is within the capability of the transistor at that collector current. You design for less than the rated values. A circuit designed for a gain of 20 will work fine for a transistor rated at 100. 5 mA base current, 100 mA collector current is a gain of 20. If the transistor is rated at 100 for that 100 mA your circuit will work reliably at a gain of 20 for any 2N3904 you use. If you design for a gain of 300 not all 2N3904s will work in your circuit.
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