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Topic: Opamp setup doubt (Read 10307 times) previous topic - next topic

Marciokoko

#60
Jul 16, 2016, 03:13 am Last Edit: Jul 16, 2016, 03:20 am by Marciokoko
polymorph,

Does Vin keep the cap, the 5k r to Vcc+ and the 100k resistor to ground?

Cause I just burnt one of my transistors.  Its my first time burning a component :-)

OOOOooohhhhhhhh!  I just noticed the bottom bjt is a 2n3906!?  I thought both were 2n3904 so I replaced them both with 2n2222 I had!

That would do it, right?

Grumpy_Mike

Quote
But Im only getting 0.48V at Vout in the circuit
Why is this a problem?

Polymorph- that circuit is running on a split supply so there is AC in the speaker.

Southpark

That's the history. Digital files don't smudge.
True....although... a bad printout could have a missing dot.

Southpark

#63
Jul 16, 2016, 10:51 am Last Edit: Jul 16, 2016, 12:59 pm by Southpark
Here are my LTSpice results:
Put the Vcc on the right hand side of the coupling capacitor....not to left of it. The capacitor is meant to let the source AC signals through to the amplifier and stop dc from getting to the source.

Also.... speaker coils don't like dc current flowing through them.

The push pull scheme mentioned by polymorph is a good approach. Checkout amplifier types... like class AB..... class A .... class C. and class D.

Marciokoko

Southpark

You mean in my original circuit?

Grumpy_Mike

@Southpark
Did you miss the fact that the circuit is using a split supply as well?

Southpark

#66
Jul 17, 2016, 12:23 am Last Edit: Jul 17, 2016, 12:25 am by Southpark
@Southpark
Did you miss the fact that the circuit is using a split supply as well?
Yeah.... I just checked again. He/she has got split supply.... so ok. Thanks Grumpy Mike.

Southpark

#67
Jul 17, 2016, 01:32 am Last Edit: Jul 17, 2016, 03:49 am by Southpark
Southpark

You mean in my original circuit?
Yeah mate...... the resistor goes after the AC coupling capacitor. See the last figure of this link here...

Also note the low leakage capacitor in series with the resistor at the '-' terminal. That blocks DC, which sets the output voltage to be 4.5 V DC (when there's no AC input)..... click here......

The above is for single voltage supply.... eg a battery .... 9V for Vcc, and 0V for ground.

But since you have dual supply +9V and -9V, then you won't need to add any DC offset voltage to your op-amp output. But... in any case..... for the original diagram that you had... the capacitor should be on the left-hand side of that resistor (that connects to +9V). Also, the resistor values 5K for top resistor and 100K for bottom resistor are significantly different.... so make sure to think about why those particular values were chosen or used there.

Marciokoko

I got those values off a afrotechmods YouTube video.  I just read in your article that the resistors to Vcc+ and Gnd should be similar.  No problem, I can change that.  I just realized the guy at the electronics store gave me 100 ohm R instead of 100k!!!  I checked with the ohm meter but I guess I glanced over and saw 98 and assumed 100k instead of 100 Ohms.

I'd like to understand the capacitor a but more.  Capacitors block DC but not AC as it charges and discharges.  So we want it there to stop the DC from Vcc+.  So why would we want the cap before the Vcc+?

Southpark

#69
Jul 17, 2016, 06:53 am Last Edit: Jul 17, 2016, 06:56 am by Southpark
So why would we want the cap before the Vcc+?
I sometimes like to think of it as principle of superposition. The DC voltages set up the amplifier circuit in a way so that the output voltage sits at a particular DC voltage (forgetting about output AC coupling capacitor for the moment). The output DC voltage usually sits somewhere around half the maximum and minimum values of supply voltage. Eg... if 9V battery..then Vcc is 9 Volt and ground is '0' Volt. So the output needs to be configured (or designed) to be somewhere in between .... eg 4.5 Volt. This is the DC condition.

Then..... when you start adding an AC voltage to the input, output voltage will certainly change...... and the change will be relative to that initial DC voltage. So applying an AC signal to the input will make the output have AC too..... but that output AC signal will be centred (vertically) around the initial DC voltage.

But adding an AC coupling capacitor prevents DC voltages from getting through it (eg.... an ideal capacitor be considered to be an open circuit for DC signals). So the other end of the capacitor will be AC signals....but no DC. This is on the OUTPUT side of things.

But.... on the INPUT side of things.... if you don't put an AC coupling capacitor, then the DC voltage considerations would then need to include whatever is connected to your amplifier (on the input side). If you have a capacitor on the input... then there is 'DC' isolation. So the circuit just before the amplifier won't mess with the DC voltages of your amplifier.

Marciokoko

I didn't understand.  If the cap is first and then the Vcc+, won't all the DC go "into" the opamp instead of if the cap is after the Vcc+?

Southpark

#71
Jul 18, 2016, 01:39 am Last Edit: Jul 18, 2016, 01:43 am by Southpark
I didn't understand.  If the cap is first and then the Vcc+, won't all the DC go "into" the opamp instead of if the cap is after the Vcc+?
In the last figure at ..... click here ..... if R3 = R4, then you'd get (+Vcc/2) at the '+' terminal due to the voltage divider.

Due to usual op-amp operation, the '-' terminal DC voltage also becomes +Vcc/2. The '-' terminal is very high impedance... so no current flows into it (ideal hypothetical case). No DC current flows through R1 due to the capacitor C2. All this amounts to no DC current through R2. So if no DC current flows through R2, then the output DC voltage becomes identical to the DC voltage at the '-' terminal.... ie +Vcc/2. That's because no current flow through R2 means that R2 has no voltage drop across it.

Consider it as principle of superposition, there's a DC component and also an AC component. The DC voltages are generally considered first. The AC component will later be ADDED (summed) with the DC component. So if you apply a pure sinewave signal at the left-side of C1, then the output will have a sinusoidal component that rises and falls (above and below) Vcc/2 (instead of above and below zero Volt).

However, if you add an AC coupling capacitor at the output, then the DC component gets blocked, and you would end up with the typical sinusoid waveform that goes above and below zero Volt.


polymorph

So, is the center point between the two 9V batteries connected to circuit ground?
Steve Greenfield AE7HD
Drawing Schematics: tinyurl.com/23mo9pf - tinyurl.com/o97ysyx - https://tinyurl.com/Technote8
Multitasking: forum.arduino.cc/index.php?topic=223286.0
gammon.com.au/blink - gammon.com.au/serial - gammon.com.au/interrupts

Marciokoko

#73
Jul 18, 2016, 11:25 pm Last Edit: Jul 18, 2016, 11:25 pm by Marciokoko Reason: image
Yes, the circuit point is ground.  I just got the 100kΩ resistors and Ill try out the circuit.  FTR, Im trying my original circuit which is this one:


Except that I have a tl071 opamp, the transistor is a 2N2222 and I modified the capacitor by moving it behind the 5k & Vcc+.

Southpark

#74
Jul 19, 2016, 02:15 am Last Edit: Jul 19, 2016, 02:20 am by Southpark
the 5k & Vcc+.
Watch the 5K resistor though, since it will form a voltage divider (5K and the 100K) below it. To get half-voltage at the mid-point of the divider, the two resistors should be about the same value.... eg 100K top, 100K below.

Actually.......  since you're using dual supply voltage, I don't think you even need that top resistor at all. No voltage divider needed, since you got dual supply. Just keep the 100K there, and remove the top (ie. 5K) resistor altogether.

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