How can I simulate a bridged output of a speaker at 8 ohms?

Following on from my previous post on a single 5V rail amplifier, I'm trying to simulate a bridged output op amp set up to drive an 8 0hm load.

One of the solutions was:

Use a bridge output. This is two power amps with the signal inverted between the two. One drives one side of the speaker high while the other drives the other side of the speaker low the same amount at the same time.

• Input is a 400mV P-P Sinusoid

• Expecting a 10V P-P Output voltage swing

There wasn't any information I could find on google, so I attempted a design following the quote above (R5 is the simulated speaker load):

The output of it is the following: (Note: The top graph's reference node is the output of the inverting amplifier)

However, if I change R5 to 8ohms, the impedance of a normal speaker, I get the following output:

Is my understanding of a bridged output incorrect here? If not, why can't I simulate the correct circuit?

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Your first problem is your choice of power supplies. Because your opamps are run single-supply, they cannot provide a negative output when called for. For instance, if the input is positive, your inverter section must put out a negative voltage. Obviously it can't do that, and in fact what you see on your bottom trace is the classic behavior of a single-supply opamp being driven to the wrong polarity - phase reversal.

Second, I suspect that your simulator opamps are not rail-to-rail. This is why your outputs don't get near 5 volts.

Try this: most importantly, change your display horizontal scale to give you only 3 or 4 cycles of your signal. As presented, your data is really hard to see.

Second, instead of grounding your opamp V-, add a negative voltage source. Make both sources 15 volts instead of 5. This will allow you to see the qualitative behavior of the bridge.

Third, experiment with adding a resistor from the top of R1 to V+, and find a value (analysis will help here - hint, consider the effects of R2 and R8) that causes the DC level to reach 7.5 volts. At this point, U1 is pegged positive and U2 is pegged negative. No worries.

Fourth, use a resistive divider to bring the noninverting input of U2 from ground to 7.5, and you will see the inverter output become well-behaved. Likewise, connect the currently-grounded end of R4 to this divider, and U1 will start working properly.

Fifth, you can now figure out why the amplitude is slightly low, and adjust the gains accordingly. (hint - think about step 3 and the impedance of C1)

Sixth, reduce the voltage sources to 10 volts, then 5. Not knowing what opanps you have specified, I can't predict exactly what will happen, but it probably won't be good, especially at 5 volts. This says that you will need to find a replacement which can operate rail-to-rail.

If the circuit works fine at this point, you can check that all of the voltages are always above 0. If that is true, and only if that is true, you can eliminate your V- source.

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Is the analysis for finding the resistor value trivial or does it take some work? How does R2 affect the biasing since it goes to virtual ground? – tgun926 Jul 1 '14 at 12:27
It's pretty trivial. First, though, I altered steps 3 and 4 - my original version was wrong. Sorry for any confusion. R1 is connected to ground. R8, rather than R2, goes to a virtual ground. For the purposes of current flow, a virtual ground is indistinguishable from a real ground, so R1 and R8 can be considered to be in parallel. How much current does R2 carry? So does it affect the divider? – WhatRoughBeast Jul 1 '14 at 17:09
So, here's how you size a resistive divider. Let's say you have two resistors in series, R1 and R2, with the free end of R1 at some voltage V, and the free end of R2 at ground. Total resistance is R1 + R2. The current through each is the same, call it i. Then i = V/(R1 + R2), and the voltage across R2 is i R2. So the voltage is R2 x V/(R1 + R2), or VR1 = V x (R2 / (R1 + R2)). I assume you know how to calculate the value of two resistors in parallel. – WhatRoughBeast Jul 1 '14 at 22:28
I suspected R2 didn't affect the biasing, but the analysis on SPICE shows it does. To get a bias of 7.5V, you'll need a resistor equivalent to R1 || R8 = 820 ohms on top of R1 to Vcc, correct? However, when I simulate it, it gives me a bias of 7.4V. If I change the value of R2, the bias value also changes. Is this due to the op amp I'm using? It's an LT1677. – tgun926 Jul 1 '14 at 23:21
Well, 818 ohms, actually. using 820 ought to give a bias of 7.492 volts. I suspect that you've got your simulator only showing the first two digits. I wouldn't expect R2 to matter. The combination of 20 nA bias current and a Thevenin equivalent resistance of 410 ohms should only produce about 8 microvolts change. – WhatRoughBeast Jul 2 '14 at 0:01

Your simulation is with op-amps and if the model for the op-amp is accurate you can't expect "decent power" to be delivered to an 8 ohm bridged (or otherwise) load.

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I changed the parameters of the op amp to have an output current far greater than required (2A). However, looking at the current through the 8 ohm resister, it's still only 160mA P-P, and so the power is 102mW, when it should be 1.6W. Are there any other parameters that are affecting the circuit? – tgun926 Jul 1 '14 at 10:31
Tie the lower end of R1 to Vcc/2 - this will help because then your dc levels will also be correct - see whatroughbeasts answer. – Andy aka Jul 1 '14 at 11:37