Transformer Problems

[3lectrokid]

I have got a transformer here : **broken link removed** and have wired up the primary like this:
.
however i need to convert the output into DC and have +25V, GND and -25V. How Would I do this with this particular transformer

 

[MikeMl]

The basic configuration will look like this. I have not looked at that circuit in detail, but the method of using the secondary CT and a full-wave bridge looks ok

 

[3lectrokid]

but wouldn't i have a problem because the secondary consisits of two 25v windings

 

[MikeMl]

The filter capacitors will charge to ~1.4*25 = 35V. At an output of 25V, the regulator(s) will be dropping 10V at whatever current you will drawing. The 10V drop across the regulator will manifest as heat in the regulators. Depending on how much current you want to draw off the supply, the regulators will have to be on a heatsink.

 

[3lectrokid]

So how big will my heatsink need to be if i had a 5v circuit drawing .5a. i cannot find any calculators on the internet to do this calculaton

 

[Hero999]

How much current do you require?

As you've purchased a 160VA transformer, I assume it's more than the 1.5A an LM317 can power?

Work out the power dissipation, using Ohm's law.

Calculate the maximum acceptable thermal resistance.

I'll explain the process in more detail, if you answer the above questions.

 

[3lectrokid]

i need around 2 amps of +25v and -25v unregulated for my amp ics which i've sorted but i also will need around 0.5a of regulated 5v which i need to sort out a heatsink for.

i'll be using a 7805 regulator due to the ease of use.

i dont understand ohms law for power dissipation so can you help me with this.

i'm not sure what you mean by thermal resistance but i will be using a 1.1°C per watt heatsink with a fan. hovever the heatsink will also be sinking 6 audio amps.

Thanks for your help

 

[Hero999]

The output voltage will really be about 35V with a 25V transformer.

The LM7805 has a maximum input voltage rating of 35V and the power supply voltage will be slightly higher when unloaded. I would recommend using an LM317 with an 8W 22R series resistor to help get rid of the excess heat.

What's the intended power output into what impedance?

Going by the supply voltage you've used, I'll assume it's a 50W amplifier so the transistors will dissipate need to 50W of heat.

Is the 1.1°C/W with or without the fan?

Assuming a thermal resistance of 1°C/W for the thermal pad and 0.5°C/W for the junction to case the junction temperature rise will be 92.5°C for a single amplifier but it'll be much higher for six amplifiers.

 

[3lectrokid]

looking at the datasheet i have just found out that the lm1875 can only deliver up to 30watts of power. if i am having 35v off a 25v transformer wouldn't that mean i would be giving 70v to the amp which has a max rating of 60v.

1.1°C/W is without the fan.

a datasheet is **broken link removed**

 

[Hero999]

Adding a fan will greatly reduce the thermal resistance of the heatsink.

Yes, you're right the IC has a maximum rating of 60V so you need to reduce the supply voltage. I'd recommend an 18V transformer for a rectified voltage of 25V which will also enable you to use the LM7805, again with a 22R series resistor.

 

[3lectrokid]

ok thanks. i have ordered a lower voltage transformer. i will send my other one back to rapid. can you please explain why putting capacitors in parallel with the output of the transformer increases the voltage

 

[Hero999]

The transformer voltage is specified in RMS (Root Mean Square), the peak voltage for a sine wave is √2 the RMS voltage. When a bridge rectifier and filtering capacitor is added to the output of the transformer, the voltage developed across the capacitor is equal to the peak voltage minus the rectifier losses which is just under 25V in the case of an 18V transformer.

 

[unclejed613]

short answer is take the AC voltage and multiply it by 1.414 (if you want to do a quick calculation, 1.4 will work), for the rectified and filtered voltage. so a quick calculation of 25x1.4=35Vdc for the rectified voltage is close enough. output power calculations can be ballparked as well. with rail voltages of +/-35V, the "ballpark" output power would be 35x0.707 (quick calculations use 0.7) then square the result, and divide by the load resistance, so 35x0.7=24.5, then square it, giving 600.25, divided by the load resistance (8 ohms for example), gives 75 watts into an 8 ohm load. with the 18V transformer:


18x1.414=25.452Vdc
25.452x0.707=18Vrms (seems like we were here before? you're right)
18^2=324
324/8=40.5W into 8 ohms

of course the actual power output will be a little bit less, since no amp actually has a rail-to-rail output, but clipping generally occurs between 0.7 and 1.5V shy of the rail voltages.

you will also need to know the current required for 6 amplifiers delivering full output power into their loads, so for 8 ohms, and each amp delivering 18V into it's load, the current per amplifier is 18/8=2.25A, multiply by the number of amplifiers 2.25x6=13.5A, and add 20% for operating current of various subcircuits and other "overhead" 13.5+20%=16.2A required from the power supply