Calculation output impedance and coupling with amplifier stage - Page 1
Forum section: electronics / beginners

author: giorgik
author_rank: Contributor
timestamp: April 20, 2015, 02:14:14 pm
content: Hello everyone. I made the classic sine wave oscillator to network lag with an OP-AMP TL081.Everything works fine. Connecting it to an amplifier B. F. (using a double LM380 8 pin bridge-connected as shown on datasheet) is known as the amplitude of the sinusoidal signal output is considerably lower (by about 11V is passed to 5V). Both circuits are powered with dual voltage +/- 12V. How can I eliminate this effect? What is the formula of the output impedance of the oscillator?

author: jlmoon
author_rank: Unknown
timestamp: April 20, 2015, 02:34:11 pm
content: Greetings, how about a schematic of what you're trying to do. Would be of great help to those who can help you.

author: giorgik
author_rank: Contributor
timestamp: April 20, 2015, 05:23:45 pm
content: you are right jlmoonsee attachment as a sine wave oscillator scheme, where:C = 22nF, R = 2,2k?, RP = 100k?. IC : TL081, Vcc = 12V. The frequency thus obtained is 1.34kHz.Amplifier B. F. is is the second attachment, where:U1, U2 = LM380 (8 pin); R1 = 100k? logarithmic; R2 = 100?; C1 = 22uF 25V; C2 = 4700pF ceramic; C3, C4 = 100nF ceramic; C5 = 10nF ceramic; C6 = 220uF 25V; AP = 4? 4.5W music.

author: TimFox
author_rank: Super Contributor
timestamp: April 20, 2015, 09:37:38 pm
content: I don't understand the connections to the variable R1 in the second attachment.Normally the top end goes to the driver, the bottom to "ground", and the wiper to the load.Also, what biases the inputs to the two amplifiers in the second drawing? The amps are connected between +12 and 0 V, so the bias should be around +6 V at the input. If the bias is internal to the amps, then the electrolytic capacitor is connected with reversed polarity since your generator has 0 V DC level.

author: Fank1
author_rank: Regular Contributor
timestamp: April 21, 2015, 12:08:29 am
content: You don't need to calculate it, you can measure it.Measure the output with no load.Then put a potentiometer on the output and adjust it so that the output is exactly half of the previous measurement.The output impedance of the circuit is the same as the resistance of the load.

author: Melt-O-Tronic
author_rank: Frequent Contributor
timestamp: April 21, 2015, 12:38:44 am
content: The "Great And Benevolent W2AEW"has bestowed upon us a terrific instructional video for exactly this situation.I've used the technique several times (as recently as 2 days ago) with great success in audio as well as RF frequencies (10 MHz & 144 MHz).

author: giorgik
author_rank: Contributor
timestamp: April 21, 2015, 10:02:16 am
content: thanks TimFox and Fank1.The amplifier circuit to B.F. is taken from the datasheet of the LM380 of the National which is identical to my circuit.

author: macboy
author_rank: Super Contributor
timestamp: April 21, 2015, 01:13:27 pm
content: If your oscillator is generating an 11 V signal (without load), and you are connecting that to an amplifier with a fixed 34 dB (50 V/V) gain, then you need to take a step back and look at what you are doing.

author: giorgik
author_rank: Contributor
timestamp: April 21, 2015, 06:07:59 pm
content: you are right macboy, only now I realize that I have made quite a mistake, I have not given the gain of the B.F. . I started to want to build a circuit amplifier BF for general use in the laboratory, thinking they can use various types of sources with different impedances. I was in the house of the LM380 and so I decided to build the amplifier using the circuit presented in the datasheet for the LM380. Then I wanted to build a simple sinusoidal oscillator using the oscilloscope to measure its output: 11V and freq. 1.34kHz. Then I immediately connected the oscillator output to the input of BF and I measured the output of the oscillator (with the oscilloscope) a value of 5V and freq. 1.34kHz. The amplifier of B.F. has a gain of 50, but has a voltage regulator input (see R1) that I adjusted to the minimum possible value. Then taking into account that the output impedance of the oscillator (source) is greater than the value set by R1 (input impedance amplifier B.F. or load for oscillator) explains the lowering to 5V. What you should do is put a voltage follower between oscillator output and input of B.F. . In this way should see the source output to a load much higher than the equivalent resistance of the oscillator itself and a lower impedance to the amplifier input of B.F. . So far you agree?