Calculating the base and the collector value of the transistor with base voltage restriction

Thread Starter

Ashad@27

Joined May 28, 2021
5
Hi,
I'm having trouble working out the value for a base and collector resistor with an NPN transistor for my circuit. How can I adjust the base and collector resistor so that at
1.3V at the base, NPN transistor should be open
5V at base, NPN transistor be short

For your reference, I am attaching my schematic
IMG_20210528_211752.jpg
 
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Audioguru again

Joined Oct 21, 2019
3,199
You need an accurate comparator and reference voltage to do what you want.

Each BC847B transistor will have a different base-emitter threshold voltage and the voltage changes when the temperature changes.
 

Papabravo

Joined Feb 24, 2006
16,160
The voltage at the base will never be more than about 0.7V. As it approaches that level, the transistor will turn on and current will flow from the 3.3V supply through the 10K resistor and drop Vce(on) across the collector-emitter junction. As AGA has mentioned there is no way to make this circuit reliable and repeatable. Every transistor will be different.

Here is a simulation which shows the transistor threshold in the neighborhood of 590mV to 605mV and Vce(on) at about 6mV. I can change the base resistor and the voltage won't change much.
 

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Papabravo

Joined Feb 24, 2006
16,160
A 1K resistor from base to ground added to your circuit should do the trick.

Bob
That is not what the TS asked for. He asked for 1.3 volts at the BASE. Your change will turn the transistor on when the input voltage is approximately 3.0Volts, and the voltage on the base approaches 0.667V. If that is what he actually wanted - then great. He might want a different value so we should tell him how to calculate it.
 

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Papabravo

Joined Feb 24, 2006
16,160
So the first thing to consider on this circuit is that 3.3V divided by 10K gives a maximum collector current of 330 μA which is pretty tiny by any standard. If we decide that we want a forced beta of 10 (this is what we do in a switching circuit), that requires a base current of 33 uA. This amount of base current, small as it is, will turn the transistor on enough for the maximum collector current to flow. If we have 33 uA of base current and 600 μA of current to ground through R3, that means there must be 633 μA coming through R2. So 4.32K x 633 μA = 2.735V voltage rise from the 600 mV on the base giving us a threshold of 600 mV + 2.735 V ≈ 3.33 V on the input from the voltage source.

I don't know if this is a low power, battery operation or not, but the collector current is a bit low for comfort. I guess it can't be considered really low power if R2 is consuming almost twice the current of R1
 

BobTPH

Joined Jun 5, 2013
3,316
That is not what the TS asked for. He asked for 1.3 volts at the BASE. Your change will turn the transistor on when the input voltage is approximately 3.0Volts, and the voltage on the base approaches 0.667V. If that is what he actually wanted - then great. He might want a different value so we should tell him how to calculate it.
No, but it is likely what he meant.

Bob
 

Papabravo

Joined Feb 24, 2006
16,160
No, but it is likely what he meant.

Bob
I think maybe he wanted 1.3Volts on the Resistor R1 in my drawing. That would be 1.3-0.6 =0.7 across 4.32K for 162 μA. 33 μA into the base and there's not enough current for R3 the 1K to GND. There is a problem there, or maybe I missed something.
 

BobTPH

Joined Jun 5, 2013
3,316
Not the way I read it at all. I think he wants 5V on the right end of the 4.3K resistor in his drawing to turn the transistor on, and 1.3V to turn it off.

Bob
 

Papabravo

Joined Feb 24, 2006
16,160
Not the way I read it at all. I think he wants 5V on the right end of the 4.3K resistor in his drawing to turn the transistor on, and 1.3V to turn it off.

Bob
He said, and I quote: "so that at 1.3V at the base, NPN transistor should be open"

1.3V at the base is IMHO pretty unequivocal. If he somehow managed to get 1.3V at the bas the transistor would certainly not be open. You speculation sounds reasonable but since the transistor requires so little base current I don't know how you would guarantee that it would be off with 0.7 volts across the resistor. I guess -- tell me what values you would use.
 

click_here

Joined Sep 22, 2020
194
The way I read it was that they want the NPN to be "off" at an input to the base of 1.3V and in saturation mode at 5V - They used the words "open" and "short" which tells me that they want it to behave like a switch.

With this in mind, I'd be tempted to just add 2 diodes in series to the 4k3, which means that the voltage will need to be over 0.7V x 3 before current starts flowing through the base/emitter

So at 1.3V it will be "open" and 5V "short"
 

Papabravo

Joined Feb 24, 2006
16,160
The way I read it was that they want the NPN to be "off" at an input to the base of 1.3V and in saturation mode at 5V - They used the words "open" and "short" which tells me that they want it to behave like a switch.

With this in mind, I'd be tempted to just add 2 diodes in series to the 4k3, which means that the voltage will need to be over 0.7V x 3 before current starts flowing through the base/emitter

So at 1.3V it will be "open" and 5V "short"
That's a good point, but aren't we adding components to make a flawed idea work rather than asking "what is the best way to meet the requirements"? We are now at six components and counting.
 

click_here

Joined Sep 22, 2020
194
That's a good point, but aren't we adding components to make a flawed idea work rather than asking "what is the best way to meet the requirements"? We are now at six components and counting.
Absolutely :)

...But you never know, they might already have a PCB made and they might need a "hacky" way of getting this small part of a larger circuit working.

That being said, they might be still on the drawing board, so alternative ideas is what they need

If the later is the case, I'd agree with AG and say that a comparator circuit is what they are after :)
 

BobTPH

Joined Jun 5, 2013
3,316
The clue for me is that he shows a resistor between the base and 5V, which is one of the input voltages he mentions.

Why would you take someone literally when he talks about having 5V on the base? We know he cannot mean that.

Bob
 

Papabravo

Joined Feb 24, 2006
16,160
The clue for me is that he shows a resistor between the base and 5V, which is one of the input voltages he mentions.

Why would you take someone literally when he talks about having 5V on the base? We know he cannot mean that.

Bob
I give up. what values would you use to achieve the TS's goal?
 

click_here

Joined Sep 22, 2020
194
The clue for me is that he shows a resistor between the base and 5V, which is one of the input voltages he mentions.

Why would you take someone literally when he talks about having 5V on the base? We know he cannot mean that.

Bob
I think that we all agree that "the base" means via the resistor, as we can see where the 5V is being applied to in the circuit diagram
 

sparky 1

Joined Nov 3, 2018
544
The output on the simulation is very nice.
Rushing out the door to work friday morning, actually the pins are almost same made a mistake if 1K to ground would work.
the load test page 8, resistor load and capacitor load for a sc70 an output of 50Ω staying under 30 degrees C
is having a good feel for current handling.
 
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