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Add DC Offset to Audio Signal After Amplifier Output

pnielsen

Member
I have a 10W audio amp module that uses a TDA2050 IC operated at 12VDC. Being push'pull, the 10Vpp output is centered on the zero line. I would like to shift this up 5V so that the signal is entirely above the zero line. In other words, so it swings between 0 and +10V. I have a separate 5VDC supply that can provide the necessary current. How should I connect this to shift the signal as described above?
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Why do you want to? - it seems a strange idea?.

However, you give too few details anyway - you say operated at 12V - is that single ended 12V?, in which case the output is at 6V DC anyway, and swings negative towards 0V and positive towards 12V. For speaker use you use a large coupling capacitor to remove the DC.
 

pnielsen

Member
Power supply is +12VDC. The output presently swings from about -5V to +5V. The amp output is push/pull and AC coupled via a cap. I want to use an external biasing voltage to shift this to 0-10V swing. I know this can be done by removing the cap, but then the amp itself is providing the added DC current. This causes overheating.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
I don't really see any easy way other than removing the capacitor? - if it overheats, then you need better heatsinking or a better amp (which is probably what you would be adding to do the required shift anyway).

What exactly are you trying to do?.
 

Ylli

Active Member
What does the ultimate load look like? If you apply +5 volts to the load, does it draw any significant current? May be as simple as pulling up the amplifier output line to + 5 volts though a resistor (the value of the resistor dependent on the load current requirements) or a high value inductance.
 

pnielsen

Member
What does the ultimate load look like? If you apply +5 volts to the load, does it draw any significant current? May be as simple as pulling up the amplifier output line to + 5 volts though a resistor (the value of the resistor dependent on the load current requirements) or a high value inductance.
I was thinking of something along those lines. This relates to an experimental project I have in mind, and for now I am just using a 100W 22 ohm resistor as a load across the amp's output.

Given the specs in my first two posts, how would I calculate the value of the pull up resistor and its required wattage?
 

unclejed613

Well-Known Member
Most Helpful Member
I am just using a 100W 22 ohm resistor as a load across the amp's output.
if you are successful, DON'T try connecting a speaker.
 

Les Jones

Well-Known Member
Most Helpful Member
If it was being driven with a continuous signal that gave an output of 10 volts peak to peak amplitude you could just use a diode to ground on the output side of the capacitor. (Anode to ground.) The neagive peak og the output waveform would not be quite ground. It would be about -0.6 volts. This may not be suitable for your application. You have been asked what the purpose is. An answer to this question would give us more chance of providing a solution.

Les.
 

pnielsen

Member
If it was being driven with a continuous signal that gave an output of 10 volts peak to peak amplitude you could just use a diode to ground on the output side of the capacitor. (Anode to ground.) The neagive peak og the output waveform would not be quite ground. It would be about -0.6 volts. This may not be suitable for your application. You have been asked what the purpose is. An answer to this question would give us more chance of providing a solution.

Les.
Thank you for your suggestion. Ideally, the output signal should not cross below the zero line. The purpose is to drive an resistive load between 0 and +10V with an amplifier output that swings +/-5V. I am just looking for a simple solution without having to build an amp from scratch or place a high current load on the one I have by bridging the cap, etc.
 

Ylli

Active Member
If your DC load is 22 ohms, then you would have to use a high value inductance to isolate the audio from the DC. You could try using the primary of a mid-sized filament transformer.
 

pnielsen

Member
If your DC load is 22 ohms, then you would have to use a high value inductance to isolate the audio from the DC. You could try using the primary of a mid-sized filament transformer.
What you are suggesting seems to be the same as at the bottom of the thread linked below.


Also given is the diode clamp described by Les. I will try that first since I do not have a suitable inductor on hand.
 

Ylli

Active Member
With a 22 ohm DC load, a simple clamp is going to be rather problematic - but certainly worth a try.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Thank you for your suggestion. Ideally, the output signal should not cross below the zero line. The purpose is to drive an resistive load between 0 and +10V with an amplifier output that swings +/-5V. I am just looking for a simple solution without having to build an amp from scratch or place a high current load on the one I have by bridging the cap, etc.
Well that's not the purpose - that's how you're trying to do it - it would be FAR more helpful if we knew what you're trying to do, rather than how you're trying to do it.
 

pnielsen

Member
You are asking for a practical application, but there is none. It is an experimental setup to drive air coils. Audio frequency modulation of a steady state, unidirectional magnetic field.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
You are asking for a practical application, but there is none. It is an experimental setup to drive air coils. Audio frequency modulation of a steady state, unidirectional magnetic field.
I would still suggest better heatsinks or a bigger amp, with the coupling capacitor removed - suggestions such as a big pullup resistor probably isn't really going to be much help.

But now you've actually told us what you trying to do, it's essentially amplitude modulation - and the more 'modern' method would be as I've described, powering the 'device' from the output of the amplifier.

A much older method (back in the valve transmitter days) is to use a transformer in the DC feed to the 'device' - so the audio amplifier feeds the primary, the secondary connects to the positive supply, with the other end of the winding feeding the device, which goes to ground. The issue with this is the transformer, which you would have to design and build yourself - hence the far simplar modern method of using an audio amplifier to feed the 'device'.
 

Les Jones

Well-Known Member
Most Helpful Member
After looking at the data sheet for the TDA2050 do do not see any problem with with connecting a 22 ohm resistor between the output of the TDA2050 and ground. If you bias the amplifier to give s static + 5 volts DC output then the current through the 22 ohm resistor will be 0,23 amps. The power dissipated in the top output transistor will be 7 x 0.23 = 1.61 watts. The worst case would be with the output biased at + 6 volts. The current then would be 0.27 amps and the power dissipated in the top output transistor would be 0.27 x 6 = 1.62 watts. The peak current when the amplifier output was at + 10 volts would be 0.46 amps.
The power dissipated in the top transistor would be 2 x 0.46 = 0.92 watts. None of these values come close to the maximum power dissipation ( 25 watts) or the maximum current (5 amps) ratings. Are you trying to run the amplifier without a heatsink ?

Les.
 

pnielsen

Member
I was just trying to keep things small scale. Yes, I think you are right. Bigger heatsink and an aluminum clad load compensating resistor in series with the low resistance coils.

But it is always good to learn a few other techniques which I may try just for fun. I have seen a transformer used to add DC offset to function generator output. There is a Keysight publication that describes it.

 

pnielsen

Member
After looking at the data sheet for the TDA2050 do do not see any problem with with connecting a 22 ohm resistor between the output of the TDA2050 and ground. If you bias the amplifier to give s static + 5 volts DC output then the current through the 22 ohm resistor will be 0,23 amps. The power dissipated in the top output transistor will be 7 x 0.23 = 1.61 watts. The worst case would be with the output biased at + 6 volts. The current then would be 0.27 amps and the power dissipated in the top output transistor would be 0.27 x 6 = 1.62 watts. The peak current when the amplifier output was at + 10 volts would be 0.46 amps.
The power dissipated in the top transistor would be 2 x 0.46 = 0.92 watts. None of these values come close to the maximum power dissipation ( 25 watts) or the maximum current (5 amps) ratings. Are you trying to run the amplifier without a heatsink ?

Les.
Thanks for pointing that out. Your figures are what I am seeing on the lab power supply read out. The present heatsink is probably too small, but I can fit a larger one.

The 22R resistor is 10W ceramic but it gets too hot to touch. Given that the current is less than one watt, is there an explanation for this?
 

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