Trouble designing a Common emitter amplifier; base voltage is negative

Thread Starter

Wellks

Joined Jan 3, 2022
6
Hey !

I've been stuck on this problem for quite a while now so i finally ask for some help because i couldnt find any more information on forums or online courses.

I'm currently designing a common emitter amplifier but i don't understand why my base voltage is going negative, if you have any tips helping me for the design or explain to me why this voltage is negative it would be really appreciated.

I try to amplify a rectangle signal (50% duty cycle) 0-5V, this soucre cannot deliver more than 10mA.

I've been using this Instructable tutorial and i'm using LTspice to simulate.



If you have any question or need more information tell me i will give it to you asap.

Thank you for your help

The simulation file is linked to this post, don't pay attention to the components on the side (not on the screen).
forum_elec.png
 

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Thread Starter

Wellks

Joined Jan 3, 2022
6
I posted the worng screenshot, n006 and n007 are inverted, i don't find how to delete or modify the post i'm sorry...

Mod: image now corrected.E
 
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ericgibbs

Joined Jan 29, 2010
15,332
hi W,
A quick check of your circuit, I can see why you have a problem, almost every component value is way too low.

As this Homework.
Do you have any design calculations to post.?

E

Update.
What is the asc file for, it is not the same circuit
 
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Thread Starter

Wellks

Joined Jan 3, 2022
6
hi W,
A quick check of your circuit, I can see why you have a problem, almost every component value is way too low.

As this Homework.
Do you have any design calculations to post.?
Thank you for your response,

Here are my calculations, as I said this is based on the instructable linked in the first post :

R collector : I want the collector voltage to be at half the power supply voltage --> 5/2 = 2.5 and I want to have 2 A flowing inside this resistor, so by the Ohm's law we have Vcc/(2*Irc) = Rc
Rc = 5/(2*2)

R emitter : I assume the emitter voltage needs to be at 10 % of the supply voltage, so Ve = 0.5V
Its the same surrent that flows thought Rcollector (+- base current) so I used the Ohm's law : Re = Ue/Ic
Re = 0.5/2
Re = 0.25 ; really low value but i don't know the impact of this.

Base resistor : It's a voltage divider circuit, the voltage needed at the base is 0.65V (silcium transistor junction) + 0.5V (emitter voltage). So I need the base voltage to be at 1.15V, which is equal to 0.23 * supply voltage (5V)
I took arbitrarily the value of R1 as 2200 Ohm and then i computed the value of R2 to obtain the right voltage :
R2 = R1/((1/0.23)-1) = 657 Ohm

The bypass capacitor is computed with this relation, this is for a 150kHz signal :
Cb=1/(2π*f*Xc)
We take the impedance Xc equal to the emitter resistor -->Xc = Re = 0.25 Ohm
Cb=1/(2π*150 000*0.25)
Cb = 4.2 uF

For the input capacity I had computed something about 1uF but i red somewhere (true ?) that precision didn't really matter so i used the same value as the bypass capacitor.

I agree that my value are pretty low but I don't know what is wrong in my calculations or in my method.

Update

What is the asc file for, it is not the same circuit
Sorry i said that the components not on the screenshot should be ignore, but here I link a cleaned one, I should have created this one earlier.
 

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ericgibbs

Joined Jan 29, 2010
15,332
hi W,
Thanks for the update.

The reason for the negative at the base is that you are driving with a 0v to +5v pulse on one side of the input coupling capacitor, so on the other side
ie: the Base the voltage will swing from a positive through zero to negative level.
The zero point is set by the combination of the input resistors and the Base to Emitter voltage.

When the swing is positive the Base voltage will be approx +0.6V.
When the swing is negative the transistor Base/Emitter junction is reverse biassed
;ie: non-conducting, so the Base voltage can swing down to approx -4.4V.
That's not a problem.

E

This is your circuit, using a diode in place of the transistor BE junction, to give you and idea of what the action is.
EG 1203.pngEG 1203.png
 
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Thread Starter

Wellks

Joined Jan 3, 2022
6
Ok thank you so much I finally understand, is that a commom value 4.4V (like the 0.7V junction) when reversed or it just depends on the montage ?
So is my design correct ?
Because as a result I don't have what i calculated, is that linked to the fact you evocated earlier, that my values are too low ? I see how I can increase the value of the voltage divider R1 R3(should I ?)
 

ericgibbs

Joined Jan 29, 2010
15,332
hi W,
How exactly are you planning to use the circuit, what will it be driving,?
I finally understand, is that a commom value 4.4V (like the 0.7V junction)

As a check change the pulse input to say 4V and then say 6V, check the Base voltages, this will show the relationship.
E
 

Thread Starter

Wellks

Joined Jan 3, 2022
6
It's meant to be used to power a coil for wireless energy transfer. I used a resistor to try some circuit and train myself beacuse I knew I had some difficulties with that montage.
What still causes me some trumble is that when I increase the value of the collector resistor, the power drove through the charge increases (here I switched back to my coil)
1641312462485.png1641312518104.png

As a check change the pulse input to say 4V and then say 6V, check the Base voltages, this will show the relationship.
E
Oh yes I got it, it's just a simple mesh equation... Thank you
 

Thread Starter

Wellks

Joined Jan 3, 2022
6
Hi,
Thanks a lot for your help Eric and Bob, I will try to read about those class B amplifier to design one, it might indeed be more appropriate to my usage.
 
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