Creating circuit to get half of voltage to Collector in Transistor

[Music Manic]

Hi I've setup a simple circuit on my breadboard (with the help of a friend to test my transistor) for amplification.

**broken link removed**

I have the:
Emitter going to ground
Base receiving a 2m2 resistor
Collector receiving 5k6 resistor(which is supposed to halve the voltage. I seem to be getting a 8.7V reading on my multimeter.
Why? What am I doing wrong?

I am using a nine volt battery as the charge.

Thanks

 

[keith1200rs]

That isn't the best circuit diagram I have seen, but as far as I can see you have only connected two of the three legs of the transistor so I am not sure what you expecting to happen.

Keith

 

[Music Manic]

That isn't the best circuit diagram I have seen, but as far as I can see you have only connected two of the three legs of the transistor so I am not sure what you expecting to happen.

Keith

I'm trying to Bias the Collector to half the P.D. Voltage.
Sorry about the pic, it was a bit rushed.

Thanks

 

[alexan_e]

Both resistors are connected to the same positive supply, a divider can not work like that , it needs the positive voltage to one end and the ground to the other end.

May I suggest you download a software simulator and experiment a couple of circuits before trying with a real one.

You can also try an online simulator https://www.falstad.com/circuit/e-index.html

 

[Music Manic]

Here's my friend's setup and he IS getting 4.5Volts Bias with both resistors attached to the positive terminal:

**broken link removed**

From left to right:
Collector
Base
Emitter

Thanks

 

[keith1200rs]

That has all three pins of the transistor connected which makes more sense. If you want to understand electronics you need to start drawing circuit diagrams, not photographing breadboards.

Keith

 

[jeffrey samuel]

your friends design is more similar to voltage divider bias where as you have tried the same with an open emitter your design is not feasible in real time for a tranny

 

[krixen]

If you have access to some resistors i have come up with a solution.

if you use the base resistor of 50 Kohm and a collector resistor as 2.5 Kohm it will bias half the voltage assuming the Beta value of your transistor is 100.

9 volts / base resistance = base current.

base current * beta = Collector current

9 volts - ( collector current * collector resistor ) = voltage drop from collector to emitter (in your case you want 4.5 volts).

 

[Music Manic]

That isn't the best circuit diagram I have seen, but as far as I can see you have only connected two of the three legs of the transistor so I am not sure what you expecting to happen.

Keith

Ah! I forgot to put the Emitter to ground.

All working now.

Thanks

- - - Updated - - -

If you have access to some resistors i have come up with a solution.

if you use the base resistor of 50 Kohm and a collector resistor as 2.5 Kohm it will bias half the voltage assuming the Beta value of your transistor is 100.

9 volts / base resistance = base current.

base current * beta = Collector current

9 volts - ( collector current * collector resistor ) = voltage drop from collector to emitter (in your case you want 4.5 volts).

Hi Krixen,

Could you give me some more info on how each part operates in conjunction with each other.
I understand what you're saying but would each transistor need different conditions to work in a certain way?
I'm working on NPN class A amp basics at the moment, just so I know exactly what is going on with the transistor.

Thanks

 

[Weylin]

Take a look at this: https://www.falstad.com/circuit/e-voltdivide.html
It shows an example voltage divider. Mess around with the resistances too and you'll see the pattern. The voltage is determined by the resistor ratio rather than the actual value.



I'm still not entirely sure what you are trying to do, though. You want the collector voltage to be half of what the base voltage is? Or half of the emitter?

 

[krixen]

The only difference between BJT's which you are using is the beta value. nPn is what i assumed you have so that will work.

so the current in the base can be written as: Base Current = [9 volts (from battery) - 0 volts (emitter is grounded) ] divided by base resistance.
now the collector current is : base current x beta value. usually beta is 100, sometimes it is 50 depending on the transistor. the ratio between
collector current and base current is beta. ( Ic/Ib ).

since now you have the collector current, if you want to find the voltage drop between the collector and emitter (V out) you can do
9 volts (from battery) - voltage drop on collector resistor (ie collector current * collector resistance) = V out

- - - Updated - - -

Just going to make a quick formula summary here, i read over my post and it can be a little wordy:

Battery will be called Vcc now:

Ib = (Vcc) / Rb
Ic = Beta * Ib
Beta = Ic/Ib (this is to find your beta value to see if its 100 or different)
Vout = Vcc - (Ic * Rc)

I forgot about the Diode voltage drop from the base to emitter, base current should be :

Ib = Vcc - .7v / Rb (usually is .7, should be easy to measure the voltage drop between base and emitter with a volt meter)

 

[jeffrey samuel]

Just going to make a quick formula summary here, i read over my post and it can be a little wordy:

Battery will be called Vcc now:

Ib = (Vcc) / Rb
Ic = Beta * Ib
Beta = Ic/Ib (this is to find your beta value to see if its 100 or different)
Vout = Vcc - (Ic * Rc)

excellent work pal but remember either the Ic or Beta value must be known if both are unknown calculation becomes more tedious

 

[FvM]

A basic disadvantage of th epresent bias circuit is that you are driving the transistor in perfect saturation if you assume a too low current gain. E.g. the transistor in the original example has apparently a B of about 200, if you calculate the circuit for B = 100 or an even lower value, the transistor saturates. Small signal transistors like BC547 cover a current gain range of 100 up to 800.

Changing the circuit to a collector-base bias resistor avoids at least saturation.

 

[krixen]

If you can give me the value for your new Vout when you hooked up the ground back to the emitter i can tell you what your beta value is for your circuit and then we can easily come up with resistor solutions for your design to get 4.5 volts across the collector-emitter

 

[LvW]

Recommendation to Music Manic:

I didn't participate in this thread (up to now). Why not?
I am too lazy resp. not enough motivated to translate your verbal circuit description (and all the following verbal contributions) into a circuit diagram.
You should know that it is common practice (for my opinion: a necessity) to support question like yours with a circuit diagram.
That is my recommendation for the future.
LvW

 

[krixen]

Its such a simple circuit that in my opinion its easily understood. Its a BJT amplifier with only two resistors.
He also seems to be new to electronics and I am sure it would be tough to make the circuit diagram.

 

[LvW]

He also seems to be new to electronics .......

Yes, most probably...and therefore my recommendation.

 

[Music Manic]

If you can give me the value for your new Vout when you hooked up the ground back to the emitter i can tell you what your beta value is for your circuit and then we can easily come up with resistor solutions for your design to get 4.5 volts across the collector-emitter

4.73V

I don't need it to be exact at the moment. I'm just being taught how to set things up at the moment and learn how it works. Thanks for the equation. It shows me how the parts interact.

- - - Updated - - -

Yes, most probably...and therefore my recommendation.

Here you go:

 

[jeffrey samuel]

It is an ordinary fixed bias ckt what is the problem here if you vary the Rb value automatically the op can be varied to suit your need

 

[krixen]

your beta value should be ~202 if my calculations are correct,
now you can control the collector current to give the Vout you need