Both assumptions correct.I assume you mean the BJT since you said base. I also assume you mean "base resistor" to be a series resistor.
Well, I was thinking about a base protecting (i.e., current limiting) resistor. But I guess because the Op-Amps can't source much current, a base resistor isn't required?It simply isn't needed. The output of the opamp provides a voltage that turns on the BJT when it gets high enough, and sticking a resistor in there wouldn't accomplish anything. Or did you mean something else?
Ok, what about just 0 to 2V? I.e., why is the limit 0 to 1.2 specifically?2. You could amplify the output with another opamp. If you're powering it with 5V, you'd need one capable of rail-to-rail output swing.
1: You bring up a good point. Looking at the circuit it would seem that the base-emitter junction would clip the output at the base-emitter voltage of about 0.7V. The output could never get to 1.2V as shown. I believe you do need a resistor in series with the base, but not sure what value. You may have to experimentally try some different values to get the output voltage you want...................
Question 1: Why doesn't the transistor have a base resistor?
Question 2: What needs to be changed in the circuit to make it output another range, say from 0 to 5V?
Don't understand you comment about polarized capacitors. Did you put them in backwards? Of course it won't work without the capacitors.I've put the circuit together finally. I was struggling with it because of my own mistake (of course). It seems that the capacitors are really polarized. Without them it didn't work. I'm attaching a picture of what the output and input looks like (bottom input, top output). As you can see there's some distortion there. Is that OK for this circuit?
Also, interestingly enough putting the probe on either the collector or the gate of the JFET produces a straight line of 6V (that's my power supply). Is that normal? I was expecting to see a wave of some sort. Or does it mean that that part of the circuit is not working at all?
Lol, I meant I used ceramic non-polarized capacitors instead of electrolytic polarized ones.Don't understand you comment about polarized capacitors. Did you put them in backwards? Of course it won't work without the capacitors.
J176What JFET are you using?
I thought it was strange as well. But the output voltage does stay at 1.2-1.3V. I'll check tonight.There should be no DC voltage on the drain (JFETs don't have collectors) of the JFET. Sounds like you have a miss-wire or C1 is in backwards. That likely explains the distortion.
Non-polarized caps will work just fine if they are the required capacity (10μF ceramic caps are not common).Lol, I meant I used ceramic non-polarized capacitors instead of electrolytic polarized ones.
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I thought it was strange as well. But the output voltage does stay at 1.2-1.3V. I'll check tonight.
I might have miswired something. It definitely didn't work. Why do they specify polarized caps if it doesn't matter -- just curious?Non-polarized caps will work just fine if they are the required capacity (10μF ceramic caps are not common).
I was talking about the gate of the JFET and the collector of the BJT. But still, the drain is pulled up to the positive rail, doesn't it? Shouldn't it be at the positive rail voltage?The input DC voltage is blocked from the output by C1. Thus there should be no DC voltage on the drain of Q2 since there is no source for the voltage (unless your input has a DC offset).
Can you explain why exactly we need the C4? If we remove it completely and break the connection to ground, won't it work the same?These impulses are filtered by C4, decreasing the gate voltage on the pjfet
Without C4, you will get 5V pulses on the pjfet gate. You want a voltage with as little ripple as possible there.Can you explain why exactly we need the C4? If we remove it completely and break the connection to ground, won't it work the same?
Also, I couldn't find 240K resistors. I used 220s, is that OK?
Ok. What's the best way to test if the JFET + BJT part is working? The BJT collector voltage should change periodically and I should catch it with a scope -- is that right?Without C4, you will get 5V pulses on the pjfet gate. You want a voltage with as little ripple as possible there. 220k resistors are fine.
It won't work unless you build the entire circuit. The way to test it is to apply an input with varying amplitudes, and see if the output level is relatively constant. The proof of the pudding, so to speak.Ok. What's the best way to test if the JFET + BJT part is working? The BJT collector voltage should change periodically and I should catch it with a scope -- is that right?
They don't specify polarized capacitors because they are required, it's just that most capacitors above about 1μF or so are usually electrolytic types which are all polarized by their nature (but it's not particularly desired).I might have miswired something. It definitely didn't work. Why do they specify polarized caps if it doesn't matter -- just curious?
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I was talking about the gate of the JFET and the collector of the BJT. But still, the drain is pulled up to the positive rail, doesn't it? Shouldn't it be at the positive rail voltage?
EDIT:
Sorry, it's the gate that is pulled to +V. So, yeah, I was talking about the gate all along. Meaning that I don't see the voltage drop caused by the drop on the collector of the BJT. I must have messed up somewhere. I've checked the circuit a 100 times though -- will do again.
You might see a small blip but the capacitor is there to smooth the voltage change to it remains at near DC.Ok. What's the best way to test if the JFET + BJT part is working? The BJT collector voltage should change periodically and I should catch it with a scope -- is that right?
Finally started working correctly yesterday. After checking and re-checking everything, I noticed that once the pot was removed the distortion was gone. After an hour of scratching my head, I decided to put the 10K (R7) resistor in. And it did it. The output is completely fine now. crutschow said 10K was not essential, so I left it out. But without it and the pot in, there's distortion. There's always a possibility I messed something else up of course! crutschow, I'm not complaining or trying to disrespect you at all -- just recounting what happened.It won't work unless you build the entire circuit. The way to test it is to apply an input with varying amplitudes, and see if the output level is relatively constant. The proof of the pudding, so to speak.
You guys are so freaking good. Unbelievable. Hat's off to you all.The collector voltage on the BJT should go down as the input amplitude is increased. It won't change periodically unless you change the input amplitude periodically.
Still don't see why the 10kΩ output resistor is needed. Perhaps it's some quirk of the op amp. But certainly the pot is needed to provide a source for the negative charge to C1 to match the positive charge going into the base of Q1.Finally started working correctly yesterday. After checking and re-checking everything, I noticed that once the pot was removed the distortion was gone. After an hour of scratching my head, I decided to put the 10K (R7) resistor in. And it did it. The output is completely fine now. crutschow said 10K was not essential, so I left it out. But without it and the pot in, there's distortion. There's always a possibility I messed something else up of course! crutschow, I'm not complaining or trying to disrespect you at all -- just recounting what happened.
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One simple question: so now I have the output from the AGC. The circuit that actually analyzes the frequencies is on another breadboard controlled by an arduino and powered by it. How do I interface them together correctly?
I.e., would it be correct to just connect the output of the AGC to the input of the equalizer chip (which is on the arduino breadboard) and connect the ground from the AGC breadboard to the ground of the arduino breadboard?
The datasheet recommends a 10k pulldown on the output to get it to go within a few millivolts of the negative rail, and to eliminate crossover distortion.Still don't see why the 10kΩ output resistor is needed. Perhaps it's some quirk of the op amp.
Yes, just connecting the two grounds together should work.
I agree with this.But certainly the pot is needed to provide a source for the negative charge to C1 to match the positive charge going into the base of Q1.