It should be more than that shouldn't it? We should have a gain of 40. Red doesn't appear to be 10mV, your scope says 10mV per division so it is reading 20mV.

Yeh ,I fixed it.

I had the ground connected to the neg. term on function generator.

the picture on the last post is right now..

 

I'll try to design a CE amp with a gain of 20, my course books say that 20 is about max. for one stage, otherwise other problems creep in...ect...

using RC of 43K.

Then I'll simulate it and post the results for you and you can look at it and see what you think.

 

I'll try to design a CE amp with a gain of 20, my course books say that 20 is about max. for one stage, otherwise other problems creep in...ect...

using RC of 43K.

Then I'll simulate it and post the results for you and you can look at it and see what you think.

Sounds good! Although, they don't have a gain of 20 listed in the "On Characteristics." That's interesting that a gain of 20 would be max, this transistor can go up to a gain of 300.

 

Set values
VCC=9v.
RC=43K
Av.=20

calculated values
IC=104uA
RE=2K
VE=209mV.
VB=909mV.
ID = 45uA (divider current)




Now something to undwerstand about this, since the VB (base voltage) is around 1v. then this can only take a small signal less than 1v. pk. otherwise it would overdrive the base and cause clipping on the output.

So with the high RC value makes this a small signal amp.

I put a ammeter in the base circuit and am getting around 1.326 uA.

So this is close to your IC of .1mA and Ib of 2,5uA

 

That looks great! Although, it's too bad we can't go according to the specs. What if we kept the collector and base current the way we wanted it but made the emitter current so that the gain would be 20. So we would just keep that 2k resistor in there?

 

According to the 2N3904 specs, on the second page entitled "On Characteristics" you can see that with a hFE of 40 the collector current should be 0.1mA.

The proper procedure is not to look up a particular HFE and then take the collector current to be the particular value given on the data sheet for that HFE. The HFEs of a bunch of BJTs of the same part number vary all over the place. One 2N3904 will have a high HFE and another will have a low HFE for the same collector current.

What you should do is select your collector current to meet your desired performance goals as constrained by the available collector voltage and other factors. Then look up the minimum HFE at that collector current.

Knowing your collector current and minimum HFE, you know the maximum base current. Select your bias divider resistors so that the current in the divider is 10 times the maximum base current. You could use a different multiplier, but 10 times is a good rule of thumb value.

Select the ratio of the divider resistors to give you the desired voltage at the base. The emitter resistor will then set the emitter (and collector) current. You may have to bypass the emitter resistor with a capacitor to get enough small signal voltage gain.

The emitter current is set by (base voltage - .7)/emitter resistance.

Don't make the emitter resistor so large with respect to the collector resistor that you use up all the DC supply voltage across the emitter resistor. You want to have most of the DC supply voltage available for producing AC signal at the collector.

By doing this, you will have a biasing circuit that will give you your desired collector current with very little variation due to differences in the value of HFE.

Sounds good! Although, they don't have a gain of 20 listed in the "On Characteristics." That's interesting that a gain of 20 would be max, this transistor can go up to a gain of 300.

I would highly recommend that you add a word in some of your sentences. In this quote it appears that when you say "gain", sometimes you're referring to HFE. But, in some earlier posts when you said "gain", you were referring to the voltage gain of the stage.

Your sentence "That's interesting that a gain of 20 would be max, this transistor can go up to a gain of 300." seems to be using the single word "gain" to refer to two different things. I think you mean to say:

"That's interesting that a stage voltage gain of 20 would be max, this transistor can go up to a current gain of 300."

You should be aware that the voltage gain of a common emitter stage driven by a voltage source, even without an external emitter resistor, is not strongly dependent on HFE (actually I should say hfe, because now I'm referring to small signal voltage gain). The main thing that varies with hfe is the effective input impedance, so if you drive the stage with a low output impedance source, the voltage gain of the stage will not vary much with changes of hfe.

You can avoid confusing your readers and yourself if you say "voltage gain" when that's what you mean, and HFE (or current gain of the transistor) when that's what you mean.

 

Thanks for the tips! Do you have any comment about the voltage divider in my first post (second attachement)? How would I calculate the output current if I added R3?

 

Thanks for the tips! Do you have any comment about the voltage divider in my first post (second attachement)? How would I calculate the output current if I added R3?

The standard voltage divider formula, Vout = Vin*R2/(R1+R2), assumes that no current is withdrawn from the divider. If any is, then the standard formula doesn't apply.

In your second attachment back there, you've added R3. There's no need to do this. The Thevenin output resistance of the divider can include R3. You just select R1 and R2 to give the desired output resistance.

But, the important point is this: if the current drawn from a divider is a small fraction of the current through the resistors, then the voltage divider action is not upset very much. That's why you make the current in the divider resistors 10 times the (maximum) base current. Then you can effectively ignore the upsetting effect of the base current on the divider action.

There's no point in calculating the exact divider output for some particular value of base current (although it can be done without too much trouble), because some other transistor will have a very different HFE, and the calculation will no longer apply. You just make the divider current 10 times the maximum base current and ignore the effect of the base current (which could easily vary by 10 to 1 with different transistors).

 

So let me make sure I understand this:

1. hFE is current gain, which can also be known as β or beta.
2. HFE is voltage gain, which can also be known as α or alpha, or Au.

Do I have that right, anyone?

 

So let me make sure I understand this:

1. hFE is current gain, which can also be known as β or beta.
2. HFE is voltage gain, which can also be known as α or alpha, or Au.

Do I have that right, anyone?

Not exactly.

HFE is the DC current gain (DC β); it's the ratio Ic/Ib, where Ic and Ib are the DC currents.

hfe is the AC current gain (AC β); it's also the ratio Ic/Ib, but the currents in this case are the small signal AC currents.

Usually, HFE and hfe are similar in magnitude at a given operating point, and you can use one for the other for a good first order approximation.

The stage voltage gain is often denoted Av (think of it as "amplification, voltage")

The stage current gain is often denoted Ai. This current gain is different from HFE or hfe. HFE and hfe are intrinsic properties of the transistor by itself.

Av and Ai are properties of the transistor plus its surrounding bias resistors and load and emitter resistors, and any capacitors, if those capacitors' reactances are significant at the frequency of operation.

 

electronerd

Here is a good tip for you,

These posts that "ELECTRICIAN" posted,

PRINT them out and keep them in your design notebook, or hang them up close to your workbench, that is excellent information he gave you.

This way you won't forget which thread he posted them into, youll have it handy right with you in your work area.

 

Ok, I'll do that for sure. I'd like to go over it again.

 

The hFE is a minimum of 40 when the collector current is 0.1mA and its VCE is only 1V. It is almost saturated and is not linear.
The graph shows that the hFE is typically 230 (when the VCE is 5V) and changes a lot with temperature. Its max hFE is not shown.

You don't want and don't need the VCE to be only 1V. It is 1V in the example shown in the datasheet.

With a 9V supply and the base biased at 8.7V then the transistor is almost saturated and cannot amplify anything.

Why not look at a typical common-emitter amplifier transistor:
1) Its collector voltage is close to half of the supply voltage so it can swing equally up and down.
2) Its base current is about 1/10th the voltage divider's current so the different hFE of each transistor has a small effect on the divider's voltage.
3) The emitter resistor raises the base and emitter voltage a little to reduce the effect of a different VBE voltage for each transistor and with different temperatures.

 

Here's an interesting tidbit about BJT amplifiers.

What's the maximum possible voltage gain obtainable with a common emitter, resistor loaded, single transistor stage?

http://www.crbond.com/papers/ent21.pdf

 

But if we set the bias point to Vc=Vcc/2 then the maximum voltage gain is equal Av=gm*Rc= Vcc/(2*Rc*Vt)*Rc = Vcc/(2*Vt)
for Vt=26mV
Av_max≈20*Vcc

Voltage divider resistor can by calculate if we know IBmax and desire base voltage Vb
R1=(Vcc-Vb)/(11*IBmax)
R2=VB/(10*IBmax)

 

But if we set the bias point to Vc=Vcc/2 then the maximum voltage gain is equal Av=gm*Rc= Vcc/(2*Rc*Vt)*Rc = Vcc/(2*Vt)
for Vt=26mV
Av_max≈20*Vcc

The word "maximum" doesn't belong here; this is not a maximum voltage gain. It is just the voltage gain with Vc = Vcc/2 for a single transistor, single resistor loaded, common emitter (with no external emitter resistor) topology; the voltage gain can't be any other value for that very specific circuit (at room temperature, of course), for a reasonable range of collector currents.

Note that by using a larger collector current and lower value Rc, one can obtain a lower output impedance, and corresponding lower input impedance, with the same voltage gain.

All this assumes that the circuit is driven by an AC voltage source, with zero output impedance.

 

I thought it was obvious that I talk about this circuit.
http://www.crbond.com/papers/ent21.pdf
But of course you`re right

 

But a common-emitter transistor without any negative feedback driven from a low impedance at a fairly high level has such a high distortion (40% or more) that the output level cannot be measured!

 

But a common-emitter transistor without any negative feedback driven from a low impedance at a fairly high level has such a high distortion (40% or more) that the output level cannot be measured!

This is quite true, and worth mentioning to our friends ELECTRONERD and hobbyist, but nobody suggested otherwise.

I used the phrase "small signal" several times in my posts.

Drive the circuit with 1 mV and see if the voltage gain isn't near 200 (20 * 10 volts supply).

Of course, if one wanted to amplify a 1 mV signal, then Rc could be selected larger such that the transistor was nearly saturated, as CR Bond described in his paper, and the voltage gain would be even higher.

But, that's not the point. The point is to show that the voltage gain is 20*Vcc when Rc is selected to make Vc = Vcc/2.

 

The hFE is a minimum of 40 when the collector current is 0.1mA and its VCE is only 1V. It is almost saturated and is not linear.
The graph shows that the hFE is typically 230 (when the VCE is 5V) and changes a lot with temperature. Its max hFE is not shown.

You don't want and don't need the VCE to be only 1V. It is 1V in the example shown in the datasheet.

With a 9V supply and the base biased at 8.7V then the transistor is almost saturated and cannot amplify anything.

Why not look at a typical common-emitter amplifier transistor:
1) Its collector voltage is close to half of the supply voltage so it can swing equally up and down.
2) Its base current is about 1/10th the voltage divider's current so the different hFE of each transistor has a small effect on the divider's voltage.
3) The emitter resistor raises the base and emitter voltage a little to reduce the effect of a different VBE voltage for each transistor and with different temperatures.

You've said that with the 2N3904, the base current should be 1/10th that of the collector current in other posts as well. But, I thought that it was for saturation. I recall reading in your other posts that you needed this to saturate the transistor, is it supposed to help it be a Class-A amp? Also, you said "The graph" and I would like to know which one you're referring to. The current gain over collector current one?

As for the minimum hFE, thanks for reminding me about that, I overlooked it. Although, it's saying that you could achieve a minimum gain of 40, which is what you noted, but you could achieve a higher gain (which they don't specify). Why would the transistor be in saturation if I had Vce as 1V? What should Vce be then if It shouldn't be 1V? Also, you mentioned that hFE isn't important in saturation, where I think that it could be. Since transistors act like a switch in saturation, couldn't you use hFE to calculate how much current gain you get for your transistor switch? Just a thought, you can straighten me out here.