mauifan wrote:AoE seems to suggest that the quiescent current through the transistor [collector] should be 1mA. Is there a particular reason why they like this value, or is it somewhat arbitrary? Can I choose whatever value I want, assuming of course that my choice results in power and current levels below datasheet max values?
mauifan wrote:1) I believe that I am more concerned with output voltage right now. Since I don't believe that I need to drive anything with my circuit, I chose 1mA as the quiescent current.
mauifan wrote:2) Given that I am using Vcc=6V, I want the quiescent value of OUT- to be at 3V. Therefore I need a resistor Rc that will give me a 3V voltage drop at 1mA. Ohms Law: Rc = (Vcc - Vc) / Iq = (6V - 3V) / 1mA = 3k. As was the case in the op amp version of this thread, I used a potentiometer set at about 3k in my test circuit.
mauifan wrote:3) Although this step probably wasn't necessary given that Rc=3k, I chose Re=100 ohms.
mauifan wrote:The datasheet for the 9014 I am using indicates that its maximum collector current is 150mA. That means that it needs Rc to be a minimum of 40 ohms. The smallest resistor I have is 100 ohms. Given that I will likely vary the pot I used for Rc to test gain, I just figured that Re=100 ohms would help protect the transistor from myself.
mauifan wrote:4) Given that Re=100 is small compared to Rc=3k, I assumed that about 1mA would flow through Re in quiescent state. The voltage drop through Re is therefore Ve = Iq * Re = 1mA * 100 = 0.1V.
mauifan wrote:5) The datasheet doesn't mention it, but I assumed that the forward voltage drop between base and emitter is .6V. Adding in the drop across Re, the quiescent base voltage needs to be about Vb = 0.7V.
mauifan wrote:6) I now need to select a voltage divider R1 and R2 in a manner that results in output Vb = 0.7V. On top of that, R1 || R2 needs to be much less than the impedance looking into the base of Q1. If Q1 is fully on, the impedance of Rc || Re is very close to Re, so I assumed that the transistor has a gain of about 100 and obtained the base impedance at 100 * 100 = 10k.
mauifan wrote:7) "Much Less" means that R1 || R2 needs to be about 1/10th of the base impedance, so I used a Thevenin equivalent power source of 0.7V with a 1K series resistor. My ElectroDroid calculator tells me that I need R1=4.7k, R2=1.6k, which I implemented in the form of a pot
firstname.lastname@example.org wrote:Hmm.. I show that pair of resistors having a center voltage of about 1.5v when the supply is 6v. My number crunching says you want the upper resistor to be about 7-1/2 times as large as the lower one ((6-.7)/.7)
Figuring the parallel resistance for that ratio and scaling to 1k says you want 8.5k and 1.13k. Adjusting that for values in the E6 series gives:
- 6.8k + 680 over 1k : center voltage =~ .707v, parallel resistance =~ 880 ohms
- 10k + 1.5k over 1.5k : center voltage =~ .692v, parallel resistance =~ 1.3k
either of which should work.
mauifan wrote:I am guessing that the answer is "buy a function/waveform generator"
mauifan wrote:Anyone have any recommendations that don't break the bank -- while at the same time remain usable?
email@example.com wrote:An Arduino, a DAC ( like this one: http://www.adafruit.com/products/935 ), and a handful of resistors and capacitors will take you a long way.
Zener wrote:I don't see any reason it would be inverted. It is a little out of phase which makes sense. The gain drops out with frequency which is typical. Why do you think it should invert?
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