Strange idea that wasn't much but I decided to ask anyways.

[keantoken]

I just had an interesting idea. I believe that if you connect the bases of two complementary transistors then a small change in current in the first transistor would create a change in voltage in the complementary transistor while not deteriorating the original current and with it being very linear if the transistors are complementary enough (e.g. 2N3904 and 2N3906). I believe that this could be used as maybe a biasing scheme. I haven't really thought much about this so deal soft blows, ok? . Here it is:

[anatech]

Hi keantoken,
You need to bias the bases on more nicely, the input will be low impedance for sure. Q1 has no reverse collector base bias and will behave like a diode.

Where are you taking the signal out?

-Chris

[keantoken]

I am using a simulator . in it I am using a .01V 1kHz sine as the input and the transistors are 2N3904/2N3906. The supply is 1v
and it currently looks more like an amplifier. biasing? I am not too good at biasing yet... I thought that a little more biasing might be needed... are you talking about putting a resistor hanging or balancing on the base connections? I think that this would destroy the linearity of the circuit. But putting one between the bases would work if that was the place. Are you talking about balancing a resistor on the collector of Q1? I think that this would work. You might have to be more specific on the biasing part because I am not quite sure about biasing yet. I recognize that "Q1 would just be a diode" means that something doesn't have current running through it like it should. Cheers to not saying "it won't work".

[anatech]

Hi keantoken,
I am no expert. Especially with simulators!

0.01V is too small to cause a junction to conduct. But the colector of Q1 should have a positive supply and an AC path to your circuit common (whatever you choose that to be). Your signal source has no return with regard to the rest of the circuit, it's floating.

Could you please indicate where you are taking your amplified signal from?

Cheers to not saying "it won't work".

Well, I've drawn some silly things in the past. Glass houses.

-Chris

[keantoken]

... I didn't think about this one... in order for .01V to alter a junction, there must already be a somewhat sufficient voltage running through it, so I suppose connecting a resistor between positive supply and Q1's collector would be essential. Sorry, I didn't understand what you meant by "where are you getting your output from" in your last post, I thought you meant what the input I was using was. I am getting this from the emmitter current on Q2. Thanks!

[keantoken]

Ok, bump that supply up to 2V. I think Q1 might not be getting enough base voltage. Hmmm, This could be tricky (for me). If the input voltage is high enough, then it may start reverse-biasing Q1. I would have to make sure there was enough voltage running through Q1 as to counteract this without disrupting the linearity of the circuit.

[keantoken]

I believe that the voltage drop on Q2 is causing Q1 not to get as much voltage and is disrupting the linearity (can't you tell I love that word) of the circuit. I should have thought of this earlier...

[keantoken]

I have altered this circuit and it seems that it is self-adjusting so that there is barely any output! I wonder if this may be a useful property...

[keantoken]

The battery in series with the input generator is to turn the AC into DC as a getaround for the "Q1 is a diode" problem. I made it's series resistance 100k.

[anatech]

Hi keantoken,
The first problem I see is that you are varying the current through C-E an bit. The current is not referenced to anything else, so possibly the base current of Q1 varies by Ie / beta = (a very small amount). Not very useful at this stage.

-Chris