Hello,

I have done an emitter bias circuit with an Re feedback resistor to minimize fluctuations generated by different transistor betas!

Please view the circuit in attachment before I go on!

Now, you can see that I have taken extra care to provide precise components in the circuit so I can match the theory as close as possible with the practical world.

Okay, so, when I measure VCe in the circuit attached below, I get very close to approximately 1.2VDC. So the transistor I am using in this very circuit had a beta of 104 according to my multi meter. As you can see we actually calculated for the real beta in reference to it's circuit being 135.

So I yanked out this transistor which had a beta reading 104 according to my multi meter and plugged in another transistor with a measured beta of 113 according to my multi meter.

With this new transistor, my VCe was 1.018 VDC. So I said to myself,

"okay the Re feedback resistor is having a little trouble holding the VCe at 1.2VDC, But hey it ain't that bad!"

Then I tried a transistor with a measured beta of 156 according to my multi meter and yyyiiiiiiiikes !!!!!!!! I nearly fell off my chair LOL!!!

VCe was a deceptive value of 0.318 VDC..... waaaaaay off even though I used the emitter bias method with an Re resistor!!!!

Now what ????

As you all have pointed out to me, I can't always assume that my transistors will have a beta between say 90 and 115... what happens if I get a transistor with an off the wall beta like this one... why doesn't the Re compensate for this????

Is it because Re has to be higher, lower... confused!!!!

any help is greatly appreciated!

PS. I noticed that the latter transistor was a KSP2222 instead of a PN2222... this should not be a problem ... right?

r

 

If you want Re resistor to "work" Ve voltage must be greater than Vbe. Do you know why?
This is why I use Ve > 1V in my example. Or Re = (0.1...0.3)Vcc/Ic.
Also don't forget about saturation.
And Vce is a voltage between collector and emitter not between collector and ground.

 

Also try this circuit

I think that this circuit will work batter at such a low supply voltages.

 

This is why I use Ve > 1V in my example. Or Re = (0.1...0.3)Vcc/Ic.

ha! i will use 1v across Re..... okay, I will try it tomorrow...

And Vce is a voltage between collector and emitter not between collector and ground.

ooops !! yes .... I meant Vc!
I will get back to you and let you know how it went!

thanks jony130

 

My simulation of the circuit gives a Vce of 1.5V with a beta of 156 and Vce = 1.94V wth a beta of 113. (See attachement below.)

Don't know why your values are so far off. Suggest you double check the circuit wiring and component values.

Incidentally you are showing Vce as measured to ground in the schematic when it should be measured from collector to emitter.

 

I think your base resistor value is too high and your emitter resistor value is too low.

The base bias when there is an emitter resistor should be a voltage from a voltage divider, not a current from a single high value resistor.
The emitter resistor should have a value of about 1/10th the collector resistor.

 

Hello guys...

My simulation of the circuit gives a Vce of 1.5V with a beta of 156 and Vce = 1.94V wth a beta of 113. (See attachement below.)

Don't know why your values are so far off. Suggest you double check the circuit wiring and component values.

Incidentally you are showing Vce as measured to ground in the schematic when it should be measured from collector to emitter.

I made the mistake of calling the voltage VCe instead of Vc. In any case if you still think my values are off I don't know what else to say... I know I get a Vc of 1.242 VDC and a Ve of 0.104 VDC with that very circuit.... However I am pretty sure if you build it you will come up with my values or at least very close when using a PN2222 that has a beta of 104 when measured by a multi meter that puts a 10ua current through it!!!!

Please let me know!!!!

The base bias when there is an emitter resistor should be a voltage from a voltage divider

But if instead of a voltage divider I have a power supply providing the 2.0 VDC, what's the difference? The circuit still works.... I just have to try it with a bigger Re.....

Actually while I am here... I will try that right away!

regards
r

 

Your multimeter does not measure the beta of the transistor at the same current as in the circuit.

A voltage divider provides a voltage to the base so the beta does not matter much.
Your circuit has a single high value resistor providing a current not a voltage so the beta is VERY important for it to work or fail to work. But you are guessing how much is the beta.

 

A voltage divider provides a voltage to the base so the beta does not matter much.

But explain one thing then.... in the attachment below, doesn't, the 40K do a Vdrop? And if so, doesn't the base have a voltage value when measured from base to ground? So the voltage at the base is 2V - VRb ... no?

According to KVL, VRb is approximately 0.3VDC, so the base should have about 1.7VDC ... no?


So here we go,

Any transistor I try on this circuit can now provide a vc of approximately 2.1VDC.. Rc is 2.2K (forgot to write it!)

However, the former circuit was able to swing from 1.2VDC to approximately 0.3VDC by varying Vs from 2.0VDC to 2.8VDC whereas now I can't do that any more ???

confused again!

r

 

When you varied Vs then it varied the current in your series base resistor which varied the base current which varied the collector current which varied the collector voltage.

Why can't you understand this simple stuff?

You can't do WHAT anymore?

 

It only "works" (as you claim) because the transistor is either near or actually in saturation. Frankly this is like trying to teach your grandmother how to suck eggs.

Suppose the HFE [β] was low enough to have the transistor somewhere in the linear region then assuming Vbe=0.65V the collector current for your schematic would approximately be given by ...

\(I_c=\frac{1.35}{1+\frac{41}{\beta}} \ mA\)

Suppose β=41 this would give Ic=0.675 mA [Ib=16.46uA] thereby giving Vc=3.3-2.2*0.675=1.815V

With Ie=Ic+Ib=0.69mA then Ve=0.69V this would mean Vce=1.125V so not in saturation.

 

You can't do WHAT anymore?

I apologize, perhaps I wasn't clear in previous posts! Let me try to explain the requirements again !

The first circuit below called "Re.jpg" was a circuit that was able to control Vc from 1.2V down to approximately 0.3VDC when Vs went from 2.0VDC to 2.8 VDC.

This was good, however, In this very thread I said that when I changed transistor the Vc of 1.2VDC was not as steady anymore due to different betas in different transistors.

So Jony130 suggested I try an Re in my circuit to have better Vc stability when going from one transistor to another.

I did this in the following attachment called "Re-Works" and it did work... but when I vary Vs from 2 VDC to 2.8VDC I don't have Vc swinging from 1.2 to 0.3VDC anymore

I am new to this stuff guys... I don't understand all the in's and out's like you guys do you know.

r

 

The circuit is weird.
What produces the Vs of 2.0V? A dead lithium 3.0V coin cell? Two dead carbon zinc or alkaline cells?
Then the battery voltage changes when the load current changes and most of the measurements are meaningless.

 

It only "works" (as you claim) because the transistor is either near or actually in saturation.

Yes... that's true!

r

 

I think you will learn more about transistors if your power supply voltage is much higher (farther away from the saturation voltage).

 

When you added the emitter resistor then you added negative feedback. When the base current increases then the collector and emitter current increases. But when the emitter current increases then it reduces the increase in base to emitter current.

 

When you added the emitter resistor then you added negative feedback. When the base current increases then the collector and emitter current increases. But when the emitter current increases then it reduces the increase in base to emitter current.

So, perhaps adding the negative feedback is no good for what I want to achieve!
I don't know anymore Audioguru ... it seems everything I try here does not amount to what I require as a circuit operation.

So let's see here, when my 2.0VDC at Vs rises to say 2.2VDC this increases my base current and then increases my collector/emitter current which also increases my VRc which should decrease my Vc which is what I want. (A ramp)

But then by the same token when the emitter current increased it reduces the increase in the base/emitter current, which obviously decreases my collector/emitter current whereby decreases my VRc thus increasing the Vc back to pretty much the same value I started with.

So all this didn't get me anywhere... I am back where I started.... Bof! How do you want my circuit to create a ramping voltage at Vc using an Re feedback configuration of this sort???

Again, I don't know where to go from here.... I am still reading up on transistors and theories but I am stuck... I wish I knew what to tell you!

thanks anyways
regards
r

 

Jony130,

Also try this circuit
I think that this circuit will work batter at such a low supply voltages.

How would I control it.... I mean, where would my Vs of 2V go?

Thanks
r

 

You never said before that you want the output to ramp.

The output will ramp if the input ramps. But it sounds like your input suddenly switches high.
Then the transistor must be an integrator that makes its output ramp.
An integrator has a capacitor that makes negative feedback.
An integrator is simple when made with an opamp.

 

Then the transistor must be an integrator that makes its output ramp.
An integrator has a capacitor that makes negative feedback.
An integrator is simple when made with an opamp.

I am not knowledgable enough with op amps...
So, I was wondering if you would know where I
can find a circuit example which uses a transistor to do
this and also explains the theory and calculations
in respect to such a circuit.

I need an input to the transistor of 2.0VDC ramping to
3.3 VDC with an output ramping from 0.4VDC to
approx. 1.2VDC respectively.

The reason i am trying to do this with the circuit I showed in
this thread is because I am trying to do the ramping
operation using the active region of the transistor.

this worked with the very innitial circuit in this post, but
there was no feedback resistor. and now adding the feedback
resistor, well I can't seem to ramp.

I don't think I saw any theory about transistor integrators
in this site's tutorials ... but I will check again.

thanks