sanstrian
New Member
I'm writing up a lab on signal multiplication that I obtained the data for ~2 years ago. The class is just an intro circuits class that I took as a physics major and a lot of this is very fuzzy to me. I'll have a few questions that I'd like to post in this thread as I progress through the writeup.
The gist of the lab is taking two signals, taking the natural log of each with log amps, adding the results together with an adder, and taking the exp of the result, which mathematically yields the product of the first two signals.
I.e., considering two signals A and B:
Question #1: The logarithmic and exponential amps are simple enough, but my book uses a transistor with a grounded base and the lab handout shows a diode. Are there any practical differences between these two? I understand that in essence a transistor grounded at the base is like a diode but something seems odd about this. Further, in the handout he mentions the following:
"Note that the signal generator's output impedance is too high to drive the diode when it is conducting (V > 0.6 volts). So use an opamp follower to drive the diode."
What the hell does he mean by that last sentence? This has confused me for some time now.
Thanks in advance.
The gist of the lab is taking two signals, taking the natural log of each with log amps, adding the results together with an adder, and taking the exp of the result, which mathematically yields the product of the first two signals.
I.e., considering two signals A and B:
ln A + ln B
exp(ln A + ln B) = exp(ln A)*exp(ln B) = A * B
exp(ln A + ln B) = exp(ln A)*exp(ln B) = A * B
Question #1: The logarithmic and exponential amps are simple enough, but my book uses a transistor with a grounded base and the lab handout shows a diode. Are there any practical differences between these two? I understand that in essence a transistor grounded at the base is like a diode but something seems odd about this. Further, in the handout he mentions the following:
"Note that the signal generator's output impedance is too high to drive the diode when it is conducting (V > 0.6 volts). So use an opamp follower to drive the diode."
What the hell does he mean by that last sentence? This has confused me for some time now.
Thanks in advance.