About TDA7294

[Dragon Tamer]

To answer your questions in order;

The non inverting input is the input that doesn't invert the audio signal.
The inverting input does invert the audio signal.
Vs is the supply voltage
The differential input voltage is the minimum voltage needed between the inverting and non-inverting input for the output to change it's amplitude (I'm not 100% on that one)
The peak output current is the amount of current that the IC can supply with a spike in the audio signal.
The total power dissipation is the amount of power the IC draws from its supply when it is maxed out. This is also the amount of heat that the IC will give off while it's running.
The junction temperature is the max lead temperature when soldering the IC to it's board.
Storage temperature is the temperature range that the IC can be stored in while it is not running (no power applied).
Input bias current, the current needed on any given input to drive the IC.
Slew rate defines the speed of the IC, the higher the slew rate the better the frequency response.
Input noise current is the amount of current needed for the IC to detect noise (noise is bad) the same with the voltage.
Input resistance is how much of a load the inputs present for the source.
Thermal shut down will shut the IC down if it gets too hot so that it will not destroy itself. The IC will turn back on once it has cooled down enough.

Hope this helps.

 

[Jackdaniel]

To answer your questions s..............................

...Hope this helps.

sure it helps mate..n Thanx alot dude

 

[Jackdaniel]

lol i learned about "voltage divider" last night, i was a little bit confused bcause someone posted here that a resistor can't be used to reduce voltage.. Lol

 

[audioguru]

lol i learned about "voltage divider" last night, i was a little bit confused because someone posted here that a resistor can't be used to reduce voltage.. Lol

Correct.
A resistor cannot be used to reduce voltage, it reduces the current. Two resistors reduce voltage when they are in a voltage divider that divides a voltage.

 

[Jackdaniel]

Correct.
A resistor cannot be used to reduce voltage, it reduces the current. Two resistors reduce voltage when they are in a voltage divider that divides a voltage.

why do the resistors get very hot when i use it to control a 12v 200ma fan from a 12v 5amps transformer? Should i increase watts of the resistors? Why stands "watts" for resistors? Do "watts" reduce current or voltage?.. Oh a load of noob questions lol

 

[Reloadron]

why do the resistors get very hot when i use it to control a 12v 200ma fan from a 12v 5amps transformer? Should i increase watts of the resistors? Why stands "watts" for resistors? Do "watts" reduce current or voltage?.. Oh a load of noob questions lol

Well this is going off the original topic but since it is your thread what the hell?

Attached is an image for later viewing but for now, as to resistors. Resistors limit current and the unit of resistance is the Ohm (Ω). In simple terms resistance is the opposition to current flow. The unit of power is the Watt so power is expressed in Watts. So no, Watts do not reduce current flow.

If for example I apply 10 volts across a 10 Ohm resistor Ohms Law (I suggest you get friendly with Ohm's Law) will tell us the current flowing through our simple circuit is 1 Amp. Ohms law states I the current is equal to the voltage divided by the resistance so 10 volts / 10 Ohms = 1 Amp.

There are several formulas for power but in keeping it simple the power is equal to the Voltage times the Current so we say P = E * I. So in this case we have 10 Volts * 1 Amp = 10 Watts.

While the resistance of resistors is stated in Ohms (the unit of resistance) the amount of power they can dissipate is stated in Watts. So for the little circuit example we would need a resistor rated for at least 10 watts. Actually more as we will be dissipating 10 watts of power in the resistor. I can buy a 10 Ohm resistor rated for 1/8, 1/4, 1/2, 1.0, 2.0, 5.0, 10.0, and up to hundreds of Watts. They would all have a value of 10 Ohms but they would be rated to dissipate higher and higher power. If I use a 1 Watt resistor in this application it will get real hot for a real short time and likely burn up. Thus when using resistors it is important not only to select the correct ohmic value but the correct power rating. Got it? Make sense?

Attached are four examples of simple voltage divider circuits. Each of them will output 6 volts. However, the power dissipated by each will be considerably different. Think about it. Additionally, using a simple divider like this to actually power anything has a major flaw or drawback. Note the lower resistor in each divider. Any load I place on Vout will effectively be in parallel with that lower resistor. This effectively changes the resistance and any worth of my divider goes down hill real fast. Think about it.

As to your fan. Attached is also a table based on a 12 volt everyday generic computer type fan. Actually if you look at the Voltage, Current and Speed (RPM) it is pretty linear. Note the current at 12 Volts.

You mention your fan at 12 Volts has a current of 200 mA. So let's say you want to reduce the fan speed to 3/4 full speed or run the fan at about 9 volts. We get something like Vsupply - Vfan / 200 mA or 12 - 9 = 3 / .2 = 15Ω and that gets our current down to about 150 mA. The 15Ω resistor will have a current through it of .15 Amp so that gives us the voltage 9 Volts * .15 Amp = 1.35 Watts. You would want at least a 2 Watt rated 15 Ohm resistor. Make sense?

Ron

 

[audioguru]

A series resistor is a stupid old way to slow down a DC motor:
1) The motor needs a high current to start running but a resistor limits the current so the motor might not start.
2) The resistor wastes a lot of power by making heat.

Instead, pulse-width-modulation is used to control the speed of DC electric motors, something like a light dimmer in your home.
High frequency pulses at full voltage are applied to the motor but are narrow for low speed and are wide for high speed. Narrow pulses have low average power but lots of torque to start the motor to run. Wide pulses have high average power for the motor to run at full power and speed.
But the transistors that control the width of the pulses switch on (low voltage across them so the heating in them is low) and switch off (no current in them so the heating in them is nothing).

 

[Reloadron]

A series resistor is a stupid old way to slow down a DC motor:
1) The motor needs a high current to start running but a resistor limits the current so the motor might not start.
2) The resistor wastes a lot of power by making heat.

Instead, pulse-width-modulation is used to control the speed of DC electric motors, something like a light dimmer in your home.
High frequency pulses at full voltage are applied to the motor but are narrow for low speed and are wide for high speed. Narrow pulses have low average power but lots of torque to start the motor to run. Wide pulses have high average power for the motor to run at full power and speed.
But the transistors that control the width of the pulses switch on (low voltage across them so the heating in them is low) and switch off (no current in them so the heating in them is nothing).

Dear Mr. AG (aka Scrooge)

While I fully agree with everything you posted I tried to answer the poster's question and explain a little along the way. In the case of a basic generic computer type fan they start without a problem with 7 volts applied.

Now looking back at the original poster's post history and questions do you think bringing PWM into the equation is really a good way to go?

Ron

 

[Jackdaniel]

Well this is going off the original topic but since it is
Ron

thanx ron, now it makes some sense to me.. So can i light up an LED rated for 1.5v from 230v AC? what kinda resistor might i need?

 

[Reloadron]

thanx ron, now it makes some sense to me.. So can i light up an LED rated for 1.5v from 230v AC? what kinda resistor might i need?

Likely the best answer in my opinion and just my opinion is none. You don't light a LED from 230 volt AC. I am not saying it can't be done, I am saying in my opinion it is unwise to do so. However, if lighting a LED off 120 or 240 VAC trips your trigger you can start here
then there is also this variation as well as a dozen other flavors. Regardless of the flavor they all use capacitors.

Ron

 

[audioguru]

I think an IR LED (it is invisible) works at only 1.2V to 1.5V. A red LED works at 1.7V to 2.2V. A blue or white LED works at around 3.5V.
An ordinary LED that is 5mm in diameter works well at 20mA. Since it uses only DC then you can rectify and filter the 230V AC into 324V DC.
The current-limiting resistor to drop the current from the 324V to 20mA at 2V is 324/0.02= 16200 ohms. Use 18k which is a standard value. It will heat with (322V squared)/18k= 5.76W so if it is a huge 10W size then it will get very hot.

 

[Jackdaniel]

the TDA7294 is quite expensive so i am frightened i may fry them, can anyone post a good circuit schematic? Would be vewy vewy helpful

 

[audioguru]

The datasheet has some good schematics.

 

[Dragon Tamer]

Use the schematic from page 10 of the data sheet for your circuit. Just label the pin numbers with the pin description from page 2 of the data sheet.

I will admit the schematic is a bit much for someone who is new to electronics but at the same time this chip is a bit much for someone who is new to electronics. Your best bet would be to start with some smaller, cheaper amplifiers to learn how they work before moving on to something this complex. A good place to start would be the LM386.

If you are still insistant on using the TDA7294, then pay close attention because this is important; In the schematic the supply lines labled +40V and -40V are the main power supply and this is what will power the IC. The supply lines labled +20V and -20V are the stand by power lines, when the IC is put into standby mode it gets it's power from these lines. These could be any value that you want as long as the standby power is lower than the supply power.

You will need to add a heat sink to this IC because it will get hot no matter what voltage you use to supply it. You don't need to worry about frying the IC with over voltage as long as you stay away from it's maximum supply voltage of +50V -50V.

On the note of your fan, you don't need a resistor to protect the fan (I'm assuming this is why you added the resistor in the first place). As long as you supply the fan with 12V, it doesn't matter what current you supply it with the fan will only draw the current that it needs. If you are trying to protect your power supply using this fan then chances are your power supply is not very well built. A good powe supply won't need any fans to cool it. A good example would be the power supply in your TV for instance. The only reason your computer has a power supply is because the enginers needed to get a lot of power out of a small space so the heat sinks are realatively small.

Before you attempt to build this amplifier then you should probably read up on some basics about electronics so that you don't destroy something expensive, start a fire or kill yourself.

https://en.wikipedia.org/wiki/Ohm's_law Read the sections about circuit analysis, temperature effects, and relation to heat condidtions as well as the introduction
https://en.wikipedia.org/wiki/Resistors Read this whole thing, it's all important
https://en.wikipedia.org/wiki/Capacitors Read the section about suppresion and coupling
https://en.wikipedia.org/wiki/Voltage Read this whole thing before you start playing with the AC mains
https://en.wikipedia.org/wiki/Electrical_current Ditto
https://en.wikipedia.org/wiki/Power_supply#AC.2FDC_supply Just skim through this one to get an idea of how they work and what types there are
https://en.wikipedia.org/wiki/Amplifiers Read through this one very carefully and take note of what it is talking about

There is a lot more to tell you but I don't have the time, so read those and see if you still want to try this. (I'm not being mean, I'm trying to help you stay alive)

 

[Ziddik]

UPDATED SCHEMATICheya jack, i got this schematic from google, hope it will be useful and someone will point out if there is something error
**broken link removed**

 

[alec_t]

It might be helpful if it had some component values. Plus the fact that the chip is labelled TDA7294 and the Fig.1 title includes TDA794 doesn't exactly inspire confidence! ;-)

 

[Dragon Tamer]

That's a good scehmatic but none of the parts are labeled with a value. The only thing that bothers me about this power supply is the fact that the standby power and the main power are from the same 30V dual rail supply. I think it would be better if the standby power was at most half of the total supply power.

I'll see if I can throw some part values out there to make building this a little easier.
C1: 470nF
C2: 10nF (this part is not completely necessary and is for noise suppression)
C3: 0.47uF
C4: 0.47uF
C5: 10uF
C6: 10uF
C7: 22uF
C8: 22uF
C9: 100uF minimum
C10: 100uF minimum

R1: 10k (not completely necessary and can be ignored)
R2: 13k
R3: 20k
R4: 20k
R5: 22k
R6: 1.1k

 

[Ziddik]

ok.. I will get another one

 

[Ziddik]

in the updated schematic the positive + supply connected to mute/stndby pins through a 10k resistor! Is that all right?

 

[Dragon Tamer]

Yes, they are specified as switches in the schematic. As long as the switch is not closed the amp will function normally.