If you have the metallic enclosure/panel, then it is needed to ground it NOT from the ground pin of the pot 🙂 I would try to connect the metallic enclosure with GND on the Veroboard somewhere (maybe directly to the GND wire from the power supply).
Thanks, I will do that tomorrow. I will just try the back panel first. If that doesn't remove the Hiss. I will ground the shaft as well. Since the volume knob is connected to the Shaft ultimately.
edit: Stupid question (1 more on my list). Say i am a little stuck in moving the board around freely now. Can throwing the wire into the ground pin of the input signal be enough? Because the ground pin is a connector in my board, which I can just screw it in, instead of soldering, or bad consequences awaits if I do that?
Edit 2: Oh better yet. The ground wire connection from the power supply is also a connector. I'll the add the wires from the back panel to the power supplies ground pin connector.
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Alright I will do that, thanks a lot.
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I would like to re-cap on questions asked a while ago.
The volume pot again & TL494
I would like to know if my understanding is correct. 😱
Firstly the volume pot
It's basically a voltage divider as Eva and Ledmania mentioned and the 10uF capacitor I use is to block the DC component and only allow AC. So the voltage value going into Pin 15 changes as you change the volume knob (pot).
But this voltage value is the AC components and the Duty cycle of the TL494 is dependant on DC. So whatever the voltage value you set in Pin 16 is what sets the Duty Cycle of the Error Amplifier?
Furthermore by feeding back the "F/B" Pin to Pin 15, your inputting about 2V into Pin 15, which is DC and not AC.
So in terms of DC component, the voltage values on both Pin 15 and 16 are roughly the same.
However the AC component is altered through that volume pot and that is what you hear in the speaker (The altered AC gain).
So you mucking around with the voltage level from the volume knob doesn't affect the duty cycle but it does affect the AC gain.
But why use 3.3uF capacitor which is in series with the 2.2K resistor?
Now the TL494
You can't leave the other error amplifier's input in Open Circuit (Why?), so to disable this error amplifier you pull one pin up to 5V and the other to ground via 2k resistor (Can it be any resistor value and not 2k?).
You parallel the two output collectors and pull it up to 12V via 680 Ohm resistor.
And as 81bas said, through Ohm's Law. By using a 680 Ohm resistor at 12V. I get an output current of about 17mA. While the Tl494's recommended output current is around 200mA.
Is that why the switching will still be hard even if current is reduced by increasing resistance. Because my output bipolar transistors are configured in common collector configuration?
------------------------------------
I would like to re-cap on questions asked a while ago.
The volume pot again & TL494
I would like to know if my understanding is correct. 😱
Firstly the volume pot
It's basically a voltage divider as Eva and Ledmania mentioned and the 10uF capacitor I use is to block the DC component and only allow AC. So the voltage value going into Pin 15 changes as you change the volume knob (pot).
But this voltage value is the AC components and the Duty cycle of the TL494 is dependant on DC. So whatever the voltage value you set in Pin 16 is what sets the Duty Cycle of the Error Amplifier?
Furthermore by feeding back the "F/B" Pin to Pin 15, your inputting about 2V into Pin 15, which is DC and not AC.
So in terms of DC component, the voltage values on both Pin 15 and 16 are roughly the same.
However the AC component is altered through that volume pot and that is what you hear in the speaker (The altered AC gain).
So you mucking around with the voltage level from the volume knob doesn't affect the duty cycle but it does affect the AC gain.
But why use 3.3uF capacitor which is in series with the 2.2K resistor?
Now the TL494
You can't leave the other error amplifier's input in Open Circuit (Why?), so to disable this error amplifier you pull one pin up to 5V and the other to ground via 2k resistor (Can it be any resistor value and not 2k?).
You parallel the two output collectors and pull it up to 12V via 680 Ohm resistor.
And as 81bas said, through Ohm's Law. By using a 680 Ohm resistor at 12V. I get an output current of about 17mA. While the Tl494's recommended output current is around 200mA.
81bas said:
Is that why the switching will still be hard even if current is reduced by increasing resistance. Because my output bipolar transistors are configured in common collector configuration?
I try to answer your questions, but this will produce only further many questions, I think... 🙄
To make the volume pot NOT changing your 'default' ('zero output') duty cycle (50%)
Because this will produce unpredictable output voltage of this error amp, where we need to make it zero, to disable this error amp.
Yes, there can be any resistor, which will limit the input current of the disabled error amp to the allowed value.
First, it is needed to define, what means 'hard switching'. 🙂 If your teacher meant, that 17mA is 'hard' to switch for TL494 (this is how I understand it), then 5-10 mA are relatively 'hard' too 😀
The 'recommended output current' 200 mA is a recommended MAX output current, as I understand it from datasheet. So there is NO need to run at such high (200mA) output current.
But why it is not recommended to lower these 17mA, because the open collector output usually causes the saturation of the bipolar output transistor. And opened saturated bipolar transistor still conducts the current between collector and emitter very effectively for short time, even when the voltage between base and emitter legs is zero! This time to recover from the saturation state can be shortened by using of higher output current (17mA in our case).
Hope these explanations are correct and help you to understand the processes in your schematic 😉
P.S.: I am sorry, but in my post the sentence 'the output bipolar transistors of tl494 are saturating in common collector configuration' needs to be corrected to 'the output bipolar transistors of tl494 are saturating in open collector configuration'... 🙁
To make the volume pot NOT changing your 'default' ('zero output') duty cycle (50%)
Because this will produce unpredictable output voltage of this error amp, where we need to make it zero, to disable this error amp.
Yes, there can be any resistor, which will limit the input current of the disabled error amp to the allowed value.
First, it is needed to define, what means 'hard switching'. 🙂 If your teacher meant, that 17mA is 'hard' to switch for TL494 (this is how I understand it), then 5-10 mA are relatively 'hard' too 😀
The 'recommended output current' 200 mA is a recommended MAX output current, as I understand it from datasheet. So there is NO need to run at such high (200mA) output current.
But why it is not recommended to lower these 17mA, because the open collector output usually causes the saturation of the bipolar output transistor. And opened saturated bipolar transistor still conducts the current between collector and emitter very effectively for short time, even when the voltage between base and emitter legs is zero! This time to recover from the saturation state can be shortened by using of higher output current (17mA in our case).
Hope these explanations are correct and help you to understand the processes in your schematic 😉
P.S.: I am sorry, but in my post the sentence 'the output bipolar transistors of tl494 are saturating in common collector configuration' needs to be corrected to 'the output bipolar transistors of tl494 are saturating in open collector configuration'... 🙁
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So time to recover from saturation increases when reducing current, like say 2.5mA (4K7 resistor is what he suggested).
What happens as recovering time from saturation increases? More Distortion?
Secondly I need to go way back into elementary electronics here. Error amplifier from reading on wiki just compares a ref voltage to the output voltage that is fed back.
So in our case pin 16 is the reference voltage Pin, while Pin 15 you could say is the output voltage that is fed back.
What I don't get is, if the voltage in Pin 15 is around 2V because of that feedback. Then the music input sent to Pin 15 itself is gain adjusted and hence has a voltage value.
Am I right in thinking that the voltage value of the music signal is AC while the voltage value of Pin 15 DC?
Because I hear clipping when I max out the volume, which I suspect means that the music signal amplitude is greater then the sawtooth magnitude (0-3V), and hence why it's clipping. But this would suggest that it's a DC component and not AC.
So what dominates in determining the voltage of Pin 15? The feedback from the Error Amplifier or the 10K volume Pot?
Edit:
Because I am getting Lost in seeing what they say about the Error Amplifier in the "Designing Switching voltage regulators with the TL494" document to what Ledmania does.
What i get (limited knowledge) is that ref voltage is being compared with fed back output voltage to match the output of the error amplifier to design specification (make sure it's 50% duty cycle).
I don't really understand how the input music comes in all of this and how it operates the error amplifier as such.
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Sorry I've just drawn something up quick
http://img38.imageshack.us/img38/9264/potoh.png
So if Va node we set to approx 2V and If i follow conventional Op-Amp theory
Va and Vb are the same.
So how is the voltage altering music signal operating in all this?
because that circuit without the music signal I understand. It compares the ref voltage to output fedback voltage to set the designed output voltage of the error amplifier.
The gain chagning music signal is confusing me.
Edit: Ohhh shoot control theory (BANGS HEAD). The input music and fed back signal (Negative feedback) is added together (BANGS HEAD SOME MORE) before it is sent into the Error Amplifier.
http://img38.imageshack.us/img38/9264/potoh.png
So if Va node we set to approx 2V and If i follow conventional Op-Amp theory
Va and Vb are the same.
So how is the voltage altering music signal operating in all this?
because that circuit without the music signal I understand. It compares the ref voltage to output fedback voltage to set the designed output voltage of the error amplifier.
The gain chagning music signal is confusing me.
Edit: Ohhh shoot control theory (BANGS HEAD). The input music and fed back signal (Negative feedback) is added together (BANGS HEAD SOME MORE) before it is sent into the Error Amplifier.
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I mean Like this
http://img511.imageshack.us/img511/1112/controlloop.png
So actually the 2V reference from Pin 16 is being compared to the error signal produced.
And what's being sent into Pin 15 is actually the error signal.
http://img511.imageshack.us/img511/1112/controlloop.png
So actually the 2V reference from Pin 16 is being compared to the error signal produced.
And what's being sent into Pin 15 is actually the error signal.
Still don't understand why I hear clipping at max volume. Is the output of the error amplifier above 3V at this point?
Wait actually is that diagram I drew correct as far as the summing junction goes. The feedback signal is negative? I am thinking of negative feedback and theory from that says the it's negative at the summing junction.
Otherwise are they both positive at the summing junction?
Otherwise are they both positive at the summing junction?
Simply check it with oscilloscope...
Also, your amp has an sensitivity about 0.5 V peak, if the internal sawtooth of TL494 is 5V peak. If the saw is 3V peak (as you said), then the sensitivity is 0.3V at all. And the usual PC soundcard output produces about 1 V at peak. So you should not wonder, why the clipping occurs 😉
If you want to avoid this clipping, simply reduce the sensitivity of your amp: change the 2,2K input resistor (after 3,3uF cap) saying to 4.7K...
Yes, they are both positive at the summing junction... The feedback becomes negative, when it goes into negative input of the op amp 😉
And actually, when the input music wave rises, the output signal of the op amp falls at this moment, i.e. the music wave is amplified and inverted simultaneously.
Alright thanks for clarifying that. Yeah so the error signal from that summing junction is what's compared with the reference voltage set through the trimpot.
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Actually Va and Vb are having the tiny difference and the opamp tries to reduce it via the negative fb always.
When the AC music voltage goes via input resistor (2,2k) to the negative input, then it increases the difference between Va and Vb. And the opamp reacts on this change by altering of it's output voltage. Then this output voltage is feed back via the fb resistor (22k) to compensate the voltage difference between Va and Vb. And now, using the same Ohm's Law, we can calculate, at which output voltage the difference between Va and Vb will be eliminated. So, if the AC component before the input res 2,2k is about 0.3V, then it is needed to produce about 3V AC component (x10) at the opamp output to compensate the Va/Vb difference via the 22k fb resistor...
Wow. Producing about 10V via the F/b resistor if the music signal is 1V peak like from the PC. No wonder clipping occurs. Comparing a 3V sawtooth to a 10V signal. It's overmodulating.
Closer and closer the music signal get to the sawtooth's magnitude, the closer it is to clipping because of that 0.3V sensitivity.
The Amp does a good job. I'll give it that. I can clearly hear a input music signal of about 50mV in magnitude at the output speaker. It's much louder then listening to that same volume in Earphones.
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So if you do reduce output current of the TL494, by increasing the pull-up resistor, 81bas, what happens. Do you get more distortion?
You already mentioned it takes longer to recover from saturation (the bipolar transistors). But because of this increase in recovery time, what happens?
You already mentioned it takes longer to recover from saturation (the bipolar transistors). But because of this increase in recovery time, what happens?
Also 2 more things
Firstly: Can the role of the 3.3uF cap be replaced by a 10uF (to maintain consistency among component values)?
I am a little paranoid he's gonna start picking out components from the circuitry in the TL494 and ask me why this, why that, in terms of component value.
The 3.3uF cap and 680 Ohm is what really makes me paranoid at the moment.
If he does pick em out and asks me questions. He's got me. I have no reply as to why this component value and not something else.
Heck I will say i used 680 Ohm resistor because it gave me a output current of 17mA, which is low. But if he asks why didn't you reduce that output current further. I don't understand what's the resulting conclusion of increasing recovery time from saturation on the bipolar transistors. So can't come up with a decent answer.
And if he sees the 10uF cap as input DC coupling capacitor, but a 3.3uF cap in series with the resistor and asks why not another value. Again, he's got me. No answer for that.
I used the 10uF capacitor as input coupling, having followed IRS2091 application note, so I can reference to that. But the 3.3uF :S
The rest of the circuit I just followed the IR2110 application data sheet, except for the bootstrap diode, which I received help from here.
Secondly: Let's see if my Power Calculation are correct 🙄
Vpk = 20V
Vrms = 20/sqrt(2) = 14.14V (rms)
Ipk = 2.5-3.0A (let's say 2.5A)
Irms = 2.5/sqrt(2) = 1.77A (rms)
So P = V * I = 14.14 * 1.77 = 24.999999 W
In other words I have a power rating of 25 Watt average power. 😱
Firstly: Can the role of the 3.3uF cap be replaced by a 10uF (to maintain consistency among component values)?
I am a little paranoid he's gonna start picking out components from the circuitry in the TL494 and ask me why this, why that, in terms of component value.
The 3.3uF cap and 680 Ohm is what really makes me paranoid at the moment.
If he does pick em out and asks me questions. He's got me. I have no reply as to why this component value and not something else.
Heck I will say i used 680 Ohm resistor because it gave me a output current of 17mA, which is low. But if he asks why didn't you reduce that output current further. I don't understand what's the resulting conclusion of increasing recovery time from saturation on the bipolar transistors. So can't come up with a decent answer.
And if he sees the 10uF cap as input DC coupling capacitor, but a 3.3uF cap in series with the resistor and asks why not another value. Again, he's got me. No answer for that.
I used the 10uF capacitor as input coupling, having followed IRS2091 application note, so I can reference to that. But the 3.3uF :S
The rest of the circuit I just followed the IR2110 application data sheet, except for the bootstrap diode, which I received help from here.
Secondly: Let's see if my Power Calculation are correct 🙄
Vpk = 20V
Vrms = 20/sqrt(2) = 14.14V (rms)
Ipk = 2.5-3.0A (let's say 2.5A)
Irms = 2.5/sqrt(2) = 1.77A (rms)
So P = V * I = 14.14 * 1.77 = 24.999999 W
In other words I have a power rating of 25 Watt average power. 😱
The tl494 should be ok but not great
Th output is limited to 500Khz max and the op amp is slow.
The biggest problems are 1. non linear sawtooth sample waveform
and an unbalanced output. max dutycycle = 97% so high side will be less then full rail value before clipping.
I agree with Eva that there are far better (and even cheaper) chips that will
perform much better.
Th output is limited to 500Khz max and the op amp is slow.
The biggest problems are 1. non linear sawtooth sample waveform
and an unbalanced output. max dutycycle = 97% so high side will be less then full rail value before clipping.
I agree with Eva that there are far better (and even cheaper) chips that will
perform much better.
It's just for a project. I had no interest in having a high performance or high fidelity amplifier. I have it working now. It's just a matter of covering my understanding now when he asks questions relating to it.
Namely the TL494 part. Because that part is more or less the most suspicious looking part in my circuit. No schematic diagrams which I reference it to (Sorry Ledmania) from any application note, yet I have it working somehow.
He'll definitely pick it out and chew me up on it before he's satisfied.
http://pdf1.alldatasheet.com/datasheet-pdf/view/26888/TI/CD4070BE.html
Wait.. what??..lol
Absolute Maximum rated DC input current is 10mA. But mines 17mA out of the TL494 chip.
How's it working at all.lol
Wait.. what??..lol
Absolute Maximum rated DC input current is 10mA. But mines 17mA out of the TL494 chip.
How's it working at all.lol
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