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Negative Swing

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Electronman

New Member
Hello guys,

To see how a speaker cone does vibrate on a circuit with a single ended power supply I designed the bellow to see if the cone does the negative vibration.
The strange thing is that there is a NEGATIVE swing at the load (RL) too??!
I am wondering how it does happen while there is no any split power supply or divider?
I thought maybe the collector of the transistor and the R1 are creating a voltage divider but soon I noticed the voltage divider is not able to case a negative voltage at RL because the other pin of the RL goes to the ground. I mean the RL is only able to swing between the positive and ground.

Any idea please?
 

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Nigel Goodwin

Super Moderator
Most Helpful Member
Hello guys,

To see how a speaker cone does vibrate on a circuit with a single ended power supply I designed the bellow to see if the cone does the negative vibration.
The strange thing is that there is a NEGATIVE swing at the load (RL) too??!
I am wondering how it does happen while there is no any split power supply or divider?
I thought maybe the collector of the transistor and the R1 are creating a voltage divider but soon I noticed the voltage divider is not able to case a negative voltage at RL because the other pin of the RL goes to the ground. I mean the RL is only able to swing between the positive and ground.

Any idea please?
It's connected via a capacitor, so has no DC voltage across it, so of course it swings both negative and positive.

No need for a negative supply, unless you are using DC coupling.

Voltages are relative, not absolute - it's a question of what it's relative to.
 

Electronman

New Member
It's connected via a capacitor, so has no DC voltage across it, so of course it swings both negative and positive.

No need for a negative supply, unless you are using DC coupling.

Voltages are relative, not absolute - it's a question of what it's relative to.
Thanks,

How the capacitor does or cases so???
 

Electronman

New Member
But the capacitor must reach to NEGATIVE rates of voltage to generate the negative voltage for the load. How it does reach?
 

crutschow

Well-Known Member
Most Helpful Member
Let's say there's no signal. The amplifier output at the capacitor input will be at a DC level of about 1/2 the positive supply voltage, and the output of the capacitor will be at 0V (through the speaker impedance to ground) since the capacitor does not conduct DC, only AC. Now we generate a negative going signal the causes the amp to drive the cap input in a negative direction from 1/2 the supply voltage to ground. This signal is coupled through the capacitor which drives the speaker in a negative direction from 0V to -1/2 the supply voltage, thus generating the negative signal you see at the speaker terminal.
 

ericgibbs

Well-Known Member
Most Helpful Member
Can you show it by a picture or simulator please?
Your explanation makes sense somewhat.
hi,
Look at this plot.

Note the current flow [green] of the Load.
This charge/discharge of the 100uF, due to the sinewave voltage on the transistor collector, causes the current the the load to change direction. As it changes direction so does the polarity across the load.

OK.?:)
 

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Electronman

New Member
hi,
Look at this plot.

Note the current flow [green] of the Load.
This charge/discharge of the 100uF, due to the sinewave voltage on the transistor collector, causes the current the the load to change direction. As it changes direction so does the polarity across the load.

OK.?:)
It does not make sense yet!

At the bellow pic why there is a negative signal (bellow zero) for the signal source and the output load while there is a DC single ended power supply??
I seem to be too confused with these circuits.
Suppose a dual split power supply with load connected at its center pin (I.e The ground), The other pin of the load is variable between the positive pin of the power supply and e negative pin. Now we can tell that this load is swinging at the positive and the negative voltage. We can generate that state virtually but becuse one pins of the load at the first pic I posted is connected to the ground of the single ended power supply (I.e at the negative pin of the power supply while there is no any real or virtual ground) I am not able to see how that negative voltage is created? the capacitor at the first pic must reach to negative rates of the voltage (if we consider the negative pin of the power supply as the ground (i.e virtual center pin of a split power supply) of the load).
 

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ericgibbs

Well-Known Member
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hi e,

Look at this simplified circuit and the Cap and R4 load are transposed.

When the switch is in the 'a' position the cap is charged to V/2, the current flowing into the cap thru the R4 load.

If the switch is then set to 'b' the cap discharges thru the R4 load, the current now flows the opposite direction thru R4.

The voltmeter 'V' will show a positive voltage while the caps charges and negative voltage while the cap discharges.
 

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Nigel Goodwin

Super Moderator
Most Helpful Member
Think of the capacitor as a battery.

It's charged up to half the supply rail, so has 5V on the +ve end, and 0V on the negative (it's like a 5V battery).

When the collector of the transistor swings fully low, you're effectively connecting the +ve of this 'battery' to the 0V line, so it's -ve end is now -5V below 0V.

It's that simple.
 

andy257

Member
I have a question regarding the posters original attachment. The common emitter circuit has a base resistor attached to the supply rail. Now this will supply the base current but is this a good idea as the circuit is designed around a particular HFE and changing the transistor to another of the same type will not give the exact same HFE?

someone correct me if this is wrong.

Thanks
 

ericgibbs

Well-Known Member
Most Helpful Member
I have a question regarding the posters original attachment. The common emitter circuit has a base resistor attached to the supply rail. Now this will supply the base current but is this a good idea as the circuit is designed around a particular HFE and changing the transistor to another of the same type will not give the exact same HFE?

someone correct me if this is wrong.

Thanks
hi,
You are correct.
The collector may not be biased to approx +V/2 and so cause distortion.

Connecting the base bias resistor back to the collector and not +V gives the required bias over a range of transistor Hfe's.
 
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Nigel Goodwin

Super Moderator
Most Helpful Member
hi,
You are correct.
The collector may not be biased to approx +V/2 and so cause distortion.

Connecting the base bias resistor back to the collector and not +V gives the required bias over a range of transistor Hfe's.
But also reduces the gain of the stage considerably - swings and roundabouts!.
 

ericgibbs

Well-Known Member
Most Helpful Member
Having said that, an added bonus is the lower distortion and better frequency response.

Might not be a 'free' lunch, but you get free gifts with it :D
If you want your cake and eat it, you could get a bigger portion if you split the 200K into 2*100K and capacitive decouple the junction to 0V.:rolleyes:
 
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