# Baby's First OpAmp

C

#### CitizenRuth

Jan 1, 1970
0
Waaaanh! Electronics is hard.

I'm trying to become familiar with my 358n dual OpAmp chip. I am trying
to amplify an oscillator signal, an AR pulse from a 555. In general,
I'm trying to learn about synthesizer basics.

I'm using a non-inverting amp configuration with the R1/R2 =~ 1 for a
gain of approx. 2. But when I hook up the pulse, the output from the
OpAmp is actually significantly lower than the input.

I figured this chip is bad, because I got it on the street, but when I
use a straight DC voltage at the input instead of the output from the
555, the OpAmp circuit works as one would expect.

Can someone help me understand? It's not incorrect use of multimeter is
it? I have the $60 radio-shack one and I tried measurements with both AC and DC voltage meter (for the pulse circuit). Thanks much! J #### John Popelish Jan 1, 1970 0 CitizenRuth said: Waaaanh! Electronics is hard. I'm trying to become familiar with my 358n dual OpAmp chip. I am trying to amplify an oscillator signal, an AR pulse from a 555. In general, I'm trying to learn about synthesizer basics. I'm using a non-inverting amp configuration with the R1/R2 =~ 1 for a gain of approx. 2. But when I hook up the pulse, the output from the OpAmp is actually significantly lower than the input. I figured this chip is bad, because I got it on the street, but when I use a straight DC voltage at the input instead of the output from the 555, the OpAmp circuit works as one would expect. Can someone help me understand? It's not incorrect use of multimeter is it? I have the$60 radio-shack one and I tried measurements with both
AC and DC voltage meter (for the pulse circuit).

Thanks much!
No opamp is perfect and does exactly what you expect an opamp to do.
And the 358 is no exception. I think it might be time for you to
learn how to read a data sheet to anticipate if the imperfections in a
particular opamp will have significant impact on a particular application.

http://www.national.com/ds.cgi/LM/LM158.pdf

The fact that your opamp performs more ideally at DC than it does with
a pulse input may either have to do with voltage swing peaks or with
voltage swing time rate of change. Opamps have limits in both areas.

Take a look at the specs called Input Common-Mode Voltage Range,
Output Voltage Swing, the graph of Open Loop Frequency Response (gain
versus frequency), and the graph of Large Signal Frequency Response
(signal swing versus frequency).

If you have any questions what any of these specs mean to your
experiment, come back with those questions and the actual values of
supply voltage, resistor values and pulse voltage and frequency you
are experimenting with.

B

#### Bob Eldred

Jan 1, 1970
0
CitizenRuth said:
Waaaanh! Electronics is hard.

I'm trying to become familiar with my 358n dual OpAmp chip. I am trying
to amplify an oscillator signal, an AR pulse from a 555. In general,
I'm trying to learn about synthesizer basics.

I'm using a non-inverting amp configuration with the R1/R2 =~ 1 for a
gain of approx. 2. But when I hook up the pulse, the output from the
OpAmp is actually significantly lower than the input.

I figured this chip is bad, because I got it on the street, but when I
use a straight DC voltage at the input instead of the output from the
555, the OpAmp circuit works as one would expect.

Can someone help me understand? It's not incorrect use of multimeter is
it? I have the $60 radio-shack one and I tried measurements with both AC and DC voltage meter (for the pulse circuit). Thanks much! What are the power supply voltages that the 555 and 358 are operating at? Also what is the frequency of the 555 and pulse width? Typically the 555 will put out a pulse from near zero to within a volt or so of its power supply voltage. If that is 12 volts, the output might get to 11 or so. Similarly the 358 can only get with a volt or so of its power supply voltages. Now if the 555 is delivering a signal that is 0 to 11 Volts and you are trying to multiply it by two, non-inverting, the op-amp will have to deliver 0 to 22Volts, clearly beyond its range. So, that is the first thing to look for, what is the voltage you are trying to amplify and what is the available voltage that the amplifier can swing. Obviously you have to stay within that range by lowering the gain, lowering the 555 output or offsetting the amplifier so it stays within range; e.g., -11V to +11V or something similar. A second thing to look for is the frequency and pulse width of the 555. If these are too high or pulse width too short, the amplifier cannot respond. Furthermore the rise and fall times of a 555 are surely much faster than the slew rate of a 358 will allow. This will put slopes on the rise and fall of the wave form. A 358 is not an appropriate amplifier to amplify fast pulses and this sloping make keep the waveform from reaching its peak depending on how wide the pulse is. Thirdly, a multimeter is not the proper instrument to measure pulses with. You need a scope to see what is actually happening. Good luck. Bob B #### Ban Jan 1, 1970 0 Bob said: What are the power supply voltages that the 555 and 358 are operating at? Also what is the frequency of the 555 and pulse width? Typically the 555 will put out a pulse from near zero to within a volt or so of its power supply voltage. If that is 12 volts, the output might get to 11 or so. Similarly the 358 can only get with a volt or so of its power supply voltages. Now if the 555 is delivering a signal that is 0 to 11 Volts and you are trying to multiply it by two, non-inverting, the op-amp will have to deliver 0 to 22Volts, clearly beyond its range. In the data sheet is written single supply, 5V to 32V. If you want the circuit to function, you will need at least a +24V supply for the opamp(in case the 555 is working on 12V). Another possibility would be to operate the 555 on 5V and then use the 358 with a +12V supply. since the OP got already a timer running, he or she will be also getting this opamp stage to work. When measuring with a multimeter, it will show only half the expected value, because half of the time the output is at 0V and half at the double voltage, if the duty cycle is 50%. In this case a measurement directly at the output of the timer can be compared to the one on the opamp output. It will show the desired ratio 2:1 when the supply voltage is high enough. B #### Bob Eldred Jan 1, 1970 0 Ban said: In the data sheet is written single supply, 5V to 32V. If you want the circuit to function, you will need at least a +24V supply for the opamp(in case the 555 is working on 12V). Another possibility would be to operate the 555 on 5V and then use the 358 with a +12V supply. since the OP got already a timer running, he or she will be also getting this opamp stage to work. When measuring with a multimeter, it will show only half the expected value, because half of the time the output is at 0V and half at the double voltage, if the duty cycle is 50%. In this case a measurement directly at the output of the timer can be compared to the one on the opamp output. It will show the desired ratio 2:1 when the supply voltage is high enough. True, but if the 555 is running a several hundred KHz then the 358 will not even respond at least to the those voltages. There is no mention of frequency or pulse width and I don't think you can assume 50% duty cycle. In short we don't have enough information to conclude much of anything. Bob P #### Peter Bennett Jan 1, 1970 0 Waaaanh! Electronics is hard. I'm trying to become familiar with my 358n dual OpAmp chip. I am trying to amplify an oscillator signal, an AR pulse from a 555. In general, I'm trying to learn about synthesizer basics. I'm using a non-inverting amp configuration with the R1/R2 =~ 1 for a gain of approx. 2. But when I hook up the pulse, the output from the OpAmp is actually significantly lower than the input. I figured this chip is bad, because I got it on the street, but when I use a straight DC voltage at the input instead of the output from the 555, the OpAmp circuit works as one would expect. Can someone help me understand? It's not incorrect use of multimeter is it? I have the$60 radio-shack one and I tried measurements with both
AC and DC voltage meter (for the pulse circuit).

Thanks much!

When you feed DC into the op-amp, and measure the output with the
meter, the meter is reading DC, and should indicate the expected
voltage.

When you measure the output voltage with the pulse input, the meter,
when set to DC volts, will read the average voltage. If the pulse
duty cycle is 50%, the meter should read about half the maximum value.
If the pulse has a much shorter positive period than zero period, the
meter will read well below half.

With the meter set to AC, it may read the RMS value, which will be
less than half the peak-to-peak value.

C

#### CitizenRuth

Jan 1, 1970
0
CitizenRuth said:
Waaaanh! Electronics is hard.

I'm trying to become familiar with my 358n dual OpAmp chip. I am trying
to amplify an oscillator signal, an AR pulse from a 555. In general,
I'm trying to learn about synthesizer basics.

I'm using a non-inverting amp configuration with the R1/R2 =~ 1 for a
gain of approx. 2. But when I hook up the pulse, the output from the
OpAmp is actually significantly lower than the input.

I figured this chip is bad, because I got it on the street, but when I
use a straight DC voltage at the input instead of the output from the
555, the OpAmp circuit works as one would expect.

Can someone help me understand? It's not incorrect use of multimeter is
it? I have the \$60 radio-shack one and I tried measurements with both
AC and DC voltage meter (for the pulse circuit).

Thanks much!

Thanks for all the input, everybody - very helpful. Lots of things I

By the way, I meant to write AF, not AR, i.e., Audio Frequency. AR is a
CSound term . . . .

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