# Full-bridge output doesnt work as expected

I am making a half-bridge amplifier and successfully managed to amplify a sinusoidal wave (simulates a sound signal). When amplified it has almost 100V of amplitude. It has a inductor and a capacitor to create a low-pass filter.

+Sine waveform plot:

After that I duplicated the entire circuit and made some changes to act as the other side of a full-bridge. Basically I just inverted the signal.

It works fine and the -Sine signal has the same amplitude as +Sine but it is inverted. The following plot displays both outputs.

My plan was to put both outputs into a resistor (simulates a speaker) and it would have a 200V amplitude. I know those values might not be realistic but at the moment I just want to understand why my assumption doesnt happen. When I put both signals into a resistor the amplitudes goes to 120V instead of 200V.

If I plot +Sine and -Sine we can see that they no longer reach 100V.

They now reach 60V each.

Does anyone knows why this happens?

EDIT 1:

I tested a suggestion and deleted the 5ohm resistors. I also change the load resistance to 5ohm.

Unfortunately the voltage at the resistor still has a 120V amplitude.

EDIT 2:

Issue solved and explained by the solution comment from @aconcernedcitizen. I just needed to recalculate my filter values using half of load resistance.

• Did you delete the 5R resistors from the LC filters? Mar 12, 2021 at 12:33
• No, I used them to calculate the capacitance and inductance values so I thought if I delete them my +Sine and -Sine would be affected. Mar 12, 2021 at 12:35
• Your load is now the 8 Ohm one, not the 5 Ohm. If you left them there, then you have a resistive divider. You can easily test this with two sources and only the output filters, with and without the 5 Ohm resistors. Mar 12, 2021 at 12:36
• @aconcernedcitizen Tried that but and changed the 8ohm to 5ohm but it still goes to 120V only. Mar 12, 2021 at 12:43
• Add V or I (differential) probes to find where the voltage drop is occurring and why. I.e. compare point to point. Mar 12, 2021 at 13:28

You need to remove the previous loads and redesign your filter. The initial values of 39.8u and 1.6u are for a filter with Q=1 (note: underdamped), considering a load of 5 Ω (V(i)). Combining them into a bridge results in the inductors and capacitors being in series, thus the equivalent values will double for the inductors and halven for the capacitors. The output load will have to be doubled if you want the same response (V(a2,b2)).

If you leave the previous loads there while adding the 8 Ω load, the equivalent load will be the two 5 Ω loads in series, all in parallel with the 8 Ω, or about 4.44 Ω. Combine this with 2*39.8u and 1.6u/2 and you get a filter with a Q=0.446, which is well overdamped (V(a,b)).

If you remove the 5 Ω loads and leave the 8 ω, without modifying the LC values, you get a filter with a Q=0.8 (V(x,y)), and only if you change the load to be 10 Ω you will have the original Q=1 (V(a2,b2)).

Finally, if your intention is to have the load 8 Ω and the filter with Q=1, then you have to recalculate the LC values to be L=31.8u and C=2u (V(p,q)).

• I was thinking about changing the filter values but didn`t know that I should use half of the resistor value to calculate them. That is a very good explanation on what is going on and solves my problem. Thank you very much! Mar 12, 2021 at 14:07

The effect of the load (and the inverted feed) is to significantly increase the load current and, in turn, this may reduce the output voltage. Hence, you need to increase the drive signals to the MOSFETs. Maybe also consider using negative feedback so that this is done automatically when the load changes.

Previously you had 5 Ω to mid rail and now you have, in effect, 4 Ω to mid-rail (two lots make 8 Ω).