A darlington has a high current gain, but the same voltage gain as an ordinary single transistor. Voltage gain has nothing to do with beta.
Beta affects the input impedance.

Yes, but current gain in this case translates as voltage gain, so to quote you...

I don't think so.

If you need a common emitter design with a high gain value, the transistor is the first place you start, followed by the Rc/Re ratio and the impedance of the load it is feeding. It is possible to have a common emitter design alone that will have a gain of 1000, but not with a conventional transistor with a beta of 300.

I haven't done much design work with common base, so I have some boning up to do.

Sorry for the double-post but I didn't see the 2nd page until now.

A BC107 transistor has a gain-bandwidth product of typically 300MHz. It will still have plenty of current gain at 30MHz. It will have plenty of voltage gain at 10MHz.

Gusmas doesn't want a cascaded common-base driving a common emitter because it has an extremely low input impedance.
Instead he wants a cascode pair:

Interesting, most of the designs I was used to seeing at Collins Radio was Common Base, followed by Common Emitter, followed by Common Collector. Low input impedance, low output impedance with reasonably high gain and frequency responce values. Most of the frequency ranges for these kind of amps were 10hz to 16Mhz, or 70Mhz amps.

Why do you think the other configuration is more desirable? Common emitter designs are usually meant to feed medium impedance loads, while the common base design definately has a low impedance input.

 

well i still need a set of correct values that will give me a gain of 10 for my common emitter so ye............ im screwed

 

You have a gain of 10, read this post.

Or am I missing something?

If you don't understand where I got the values, just ask. We really need to see the total design, as well as input and output sources and impedances to be of real help, but so far the numbers look good.

 

ye i got a gain of 10 but if i built it the gain not 10

 

well my formulas is correct or am i wrong?

 

Is this all simulated, or have you built something? If simulated up the value of the caps to 1000µF, or raise the frequency to 1KHz, and see what happens.

BTW, when you can let us see the total schematic.

 

i first want my common emitter to work i just changed what u said and still not working the circuit im using is @ the start of my post....... check if u can see something wrong there besides the input freq

 

Yes, but current gain in this case translates as voltage gain

No.
Enter Transistor Voltage Gain in Google. Many articles say what I said: beta has nothing to do with voltage gain.

The first link:
http://users.tpg.com.au/users/ldbutler/TransisVoltAmp.htm

"Current Transfer Ratio Hfe And Stage Gain
If you were to select a bipolar transistor for an amplifier to obtain maximum voltage gain, you might be tempted to select one with the highest current transfer ratio Hfe. In fact, this would be of no avail as voltage gain is essentially dependent on two factors, namely the emitter current (Ie) and the output load resistance (RL), but not Hfe. On the other hand, a high Hfe could increase the voltage gain of the previous stage."

 

"Current Transfer Ratio Hfe And Stage Gain
If you were to select a bipolar transistor for an amplifier to obtain maximum voltage gain, you might be tempted to select one with the highest current transfer ratio Hfe. In fact, this would be of no avail as voltage gain is essentially dependent on two factors, namely the emitter current (Ie) and the output load resistance (RL), but not Hfe. On the other hand, a high Hfe could increase the voltage gain of the previous stage."

OK, but I'll stick with the math. The base voltage sets the base current if there is an emitter resistor, if the beta is less than Rc/Re then the beta sets what the collector current will be. This turns out to be pretty close to the beta.

To Gusmas, we really need to see the whole circuit, since there will be interaction between the two stages. I'm also assuming the 10KΩ for the output load is pretty close to accurate, it matters. The loading is part of the gain equations.

 

No.
Enter Transistor Voltage Gain in Google. Many articles say what I said: beta has nothing to do with voltage gain.

The first link:
http://users.tpg.com.au/users/ldbutler/TransisVoltAmp.htm

"Current Transfer Ratio Hfe And Stage Gain
If you were to select a bipolar transistor for an amplifier to obtain maximum voltage gain, you might be tempted to select one with the highest current transfer ratio Hfe. In fact, this would be of no avail as voltage gain is essentially dependent on two factors, namely the emitter current (Ie) and the output load resistance (RL), but not Hfe. On the other hand, a high Hfe could increase the voltage gain of the previous stage."

I'd have to agree with audioguru on this one.
It seems strange at first but it is true.
To get a big gain out of a CE stage is a little bit more trickier than simply increasing RL. That's where an active load really becomes useful.

However back to the topic. I'll show you a circuit i quickly put together. I assume you wanted a common base circuit with voltage divider biasing.

I get a gain of ~200 with this circuit. Beta of the transistor is 100.

 

S_lannan what program do u use to simulate that circuit maybe? it looks better than the program im using?

 

I use Multisim version 10.
I ran your common emitter design with the component values you specified.

I find it runs as expected. You get a gain of 5, that's because the load is effectively in parallel with the collector resistor and reduces the gain.

An important consideration to take into account is the signal coming in.
I used a 1mV peak sine wave @ 1kHz.

This simulator gives a bc107b transistor a beta of 400.
As audioguru pointed out this really has more significance in DC biasing and the ability of the transistor to buffer the signal coming in. In this configuration,'Hie' which is the transistor's input impedance (beta*('re+RE1+RE2)). That will be over 200,000 ohms. This will not affect the voltage divider that sets the operating point of the transistor.

It's a sound circuit design but given that the load impedance at this time is an arbitrary value we cannot really comment. If the circuit was driving a 30kΩ load it would give a gain of ~10.

 

you said you want a gain of 1K? Most transistors have a beta (β) of between 10 to 100. This is the gain you will get, it is the max the transistor is capable of. You can gang them together as in a Darlington configuration, which will multiply the gains of the transistors together. Two betas of 50 will create a beta of 2500, but this is DC gain, and will roll off quite fast.

We were both wrong. A darlington has less voltage gain than a single transistor.

You said that the voltage gain is multiplied by the bets being multiplied.
I said that the voltage gain will be the same as a single transistor.
But the darlington has a voltage gain that is much less in my simulation that has the emitter completely bypassed for max voltage gain. The voltage gain is less with the darlington because its internal emitter resistance is doubled.

The circuits do not have any negative feedback so they are extremely distorted with the top of the waveform compressed. The voltage gain is a little higher when the transistor conducts more DC current so its collector voltage does not go as high. Then the voltage gain with a single transistor is 157 and the voltage gain with the darlington is 82.

 

Worth noting in the two examples given is that the base bias voltage is very nearly the same for each example but the effective Vbe in the darlington is twice the Vbe in the single transistor circuit.

A quick calculation of the base bias voltage is around 1.5 volts. This is very nearly the same as the effective Vbe of the darlington transistor. That would mean that the darlington transistor would be much nearer to cutoff than the single transistor amplifier.

Would it be helpful to adjust the darlington base bias voltage to account for the addition base voltage drop of the darlington?

hgmjr

 

What gain does your simulation give if C3 is reconnected so that it bypasses both R3 and R4? My quick calculation suggests the gain will be around 100.

 

There is a maximum gain that can be obtained from a bipolar transistor with a resistor load and a given supply voltage:

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

 

In my sim:
1) I reduced the input signal level so that the output is not so badly distorted.
2) I measured the collector average DC level for the transistor at 4.65VDC.
3) I set the bias on the darlington so that the collector voltage is almost the same as the transistor (4.6V).

The voltage gain of the transistor is 140 (R3 and R4 in the original circuit are both bypassed).
The voltage gain of the darlington is almost half at 73.

EDIT:
One transistor in the darlington was a 2N2222 by default. when it was changed to a 2N3904 and the bias set so the collector is 4.65V then the voltage gain is exactly half of the single transistor at 70.

 

ok thx alot guys for the help but this is what i came up with.... Acording to our lecturer a very small value for Re1 for the common emitter should be choosen .. i chose 100 ohm and 400 ohm for Re2 ,another thing he told us is that the common base Re= Re1. so when i finished simulating my cct it worked 100% correct i used 100uf capacitors and the total voltage gain was 1000... but when i built the actual cct on my breadboard i only got a max of 400 voltage gain.... and the cct i constructed on my breadboard was exactly the same as in the simulation so now i ask what will cause this to happen ????!

 

How about posting an image of your final circuit?

 

I haven't been too much help here because my computer has decided to puke out. I'm still fighting it, I'm currently using a backup computer that doesn't have any of my files. Looks like a new CPU/Motherboard may be in my future, and I'm working on transfering my stuff from my old hard disk to the new machine short term.

Anyhow, it's been said, but we still need to see a schematic. It doesn't have to be working, we'll help you tweak it until it is, and if there is something major wrong we'll be glad to show and explain where the problem is.

I brought the Darlington up as an example of a single stage solution, I didn't mean to focus on it. You're original configuration sounds workable, and what you have posted looks good, we just need to see the rest to work with it.