# Amplifier Class

Discussion in 'General Electronics Chat' started by KCHARROIS, Sep 3, 2012.

1. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I'm looking to design an amplifier, class A amps are great but you can only get a max voltage swing of half of VCC. Does anyone no an amplifier class that amplifiers the first positive half of the signal and another amplifier that amplifies the negative half of the signal to get a full swing?

2. ### #12 Expert

Nov 30, 2010
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It's called, "class AB" or, "push-pull".

3. ### MrChips Moderator

Oct 2, 2009
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Not sure I know what you are asking.

Basically the power output of an amplifier is calculated as V*V/R where R is the load and V is either peak voltage or RMS voltage depending on which power you are calculating.

Hence to increase the power output you either have to reduce R or increase V.
An amplifier driven from a single supply voltage can only swing its output through V volts peak-to-peak.

Where as an amplifier with dual supply of +V and -V will be able to swing its voltage by 2V volts peak-to-peak i.e. four times the output power.

4. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Class AB push pull does not amplify your signal though its ment to give out power. If there is a way of giving amplification how do you achieve that?

5. ### #12 Expert

Nov 30, 2010
16,665
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You're asking about a power output stage that has voltage gain and outputs voltage near the maximum supply available, right?

Where's AG when you need him?

6. ### MrChips Moderator

Oct 2, 2009
12,624
3,451
Is this similar to the post on Sound Amplifier?
You have to make the distinction between the gain of the amplifier and the power output.
To provide amplification or gain, simply use an op-amp or a transistor.

Driving high power into low impedance load does not require amplification, just high current output.

7. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
OP amps do give you a voltage swing almost equal to VCC and VEE how could I do that with transistors? I looked at the schematic of op amps but dont quite get how they achieve it.

8. ### #12 Expert

Nov 30, 2010
16,665
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The rail-to-rail opamps use 2 MOSFETs to get a low impedance to the voltage supplies. Still, this question is kind of vague. Are you trying to get the last tenth of a volt out of a dual 70 volt supply? Are you trying to get a microphone to drive 50 milliwatts into an earphone? Are you chasing a theory and don't really know what you want?

9. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
When you bias a bjt amplifer you want your quiesent value to be half of VCC. So if I have a 15V supply and my quiesent value is 7.5V my signal cant be more then 15Vpp. Now what I want is an amplifier that has a VCC of 15V and a VEE of -15V so that my signal can have a signal a swing of 30 Vpp

Nov 30, 2010
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11. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Yes but I need one that can amplify voltage that final stage does not amplify the voltage signal only gives me the current power i need.

12. ### #12 Expert

Nov 30, 2010
16,665
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So, what you are saying is that a voltage amplifier and a current amplifier can both be done in one stage, and be bipolar, too. I don't think I've met that one.

13. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
Ok fine let me try and make this simpler. How does an op amp give that voltage gain that can almost reach its VCC and VEE?

14. ### #12 Expert

Nov 30, 2010
16,665
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It has several stages. An opamp has a voltage gain in the hundreds of thousands, and after you limit that gain with feedback resistors, it applies the resulting voltage to a pair of MOSFETS which have very low resistance when they are turned on.

You should look at a datasheet of an opamp and see how many stages it has inside!

15. ### KCHARROIS Thread Starter Member

Jun 29, 2012
292
1
I have but still dont see how they get there positive and negative voltage swing almost egual to there supplies.

16. ### #12 Expert

Nov 30, 2010
16,665
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I have a terrible feeling that I'm not geting through to you.
When a mommy MOSFET and a daddy MOSFET love each other very much...
No. That's not it.

When a P-channel Mosfet is connected to the positive voltage (25 volts today) and a bipolar transistor pulls its gate down by 5 or 10 volts, the MOSFET becomes a very low value resistor, a tenth of an ohm or so. The current from the positive supply flows through it to the speaker terminal, out to the speaker, and then to ground. If the speaker has 8 ohms and the MOSFET can pass at least 3.0864 amps, the speaker experiences 24.69 volts on its input terminal. That's the math if the MOSFET can become a tenth of an ohm. Now, turn the P-channel MOSFET off.

All the while, the N-channel MOSFET was turned off. Now, a bipolar transistor raises its gate by 5 or 10 volts and the N-channel MOSFET conducts. At this point, the speaker experiences negative 24.69 volts and 3.0864 amps flows the other way.

Suddenly you realize that the exact voltage required to control the MOSFETs varies from transistor to transistor. It is the feedback loop of the opamp that tells the input when the speaker has enough voltage. The voltage amplifier stages are controlled into stopping at the right drive voltage to get the right signal to the speaker.

Am I acting a fool?

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17. ### Audioguru New Member

Dec 20, 2007
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An audio power amplifier using ordinary transistors can have a fairly high voltage swing at its output when a bootstrap capacitor is added.

Here is an audio power amplifier with a differential input like yours but instead of coupling capacitors, it is DC-coupled.
Instead of an emitter-follower driver transistor without any voltage gain like yours, mine uses a common-emitter transistor as the driver that has plenty of voltage gain in addition to the voltage gain of the differential input.
My amplifier has negative feedback to reduce its very high voltage gain and reduce its distortion.
My amplifier uses a single-polarity supply but it can also use a dual-polarity supply like yours with a few simple changes.