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# How to Calculate Safe Secondary Load for Step Up Transformer

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#### pnielsen

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I would like to use a 240V to 12V "standard" power transformer in reverse for voltage step-up.

I plan to connect the 12V side across the output of a 10W audio amplifier module. My purpose is to drive a resistive load at higher voltage from the 240V side. A 10W 8 ohm resistor will be wired in series with the 12V side to properly load the amp. in the absence of the speaker for which it was designed. The fact that a power transformer is not ideal for audio frequencies is not a concern for this application.

How do I calculate the minimum series resistance that can be connected across the 240V side so that the amp is not overloaded or the signal distorted?

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A 10W 8 ohm resistor will be wired in series with the 12V side to properly load the amp.
That's not needed with modern solid-state amps.
They are fine with no load.
How do I calculate the minimum series resistance that can be connected across the 240V side so that the amp is not overloaded or the signal distorted?
Depends upon the VA (power) rating of the "standard" transformer.

Vin/Vout, 240:12 =20:1 or the inverse of that 12:240= 1:0.05
That is to say you will have a 20 to 1 change in voltage.
The impedance will change by the turn ratio squared. 20*20 = 400.
If you want the audio amp to think it is working into a 8 ohm load you can put a (8*400)=3200 ohm resistor on the output of the transformer.

That's not needed with modern solid-state amps.
They are fine with no load.
Depends upon the VA (power) rating of the "standard" transformer.

So, for example, I can dead short the output pins of a TDA series amp IC and still obtain a clean signal? For how long? If the chip is rated at 10W, what is limiting its output current? Or, are you assuming the transformer primary will provide enough resistance, even if much lower than 8 ohms?

Given the specs in my original question, I think a VA of between 10 and 15 would suit.

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Vin/Vout, 240:12 =20:1 or the inverse of that 12:240= 1:0.05
That is to say you will have a 20 to 1 change in voltage.
The impedance will change by the turn ratio squared. 20*20 = 400.
If you want the audio amp to think it is working into a 8 ohm load you can put a (8*400)=3200 ohm resistor on the output of the transformer.

At 100% efficiency, those figures indicate loading of 75mA or 18W. Under real conditions, I thus assume the 3K2 resistor will be an acceptable match for the amp's previously stated 10W rating.

To what extent would the 20*20 calculation be influenced by frequency? For example, 100Hz vs. 10KHz sine wave.

calculation be influenced by frequency
In a perfect world, with a perfect transformer, no effect by frequency.
BUT-1
The inductance of the transformer will effect the low end frequency response. I don't know what the inductance is but at DC the impedance approaches zero. The transformer was designed for 240V 50 to 60hz on the 240V winding. So I think this project will work down to 50hz or maybe lower.
BUT-2
The transformer may saturate at low frequencies and high voltage. (voltage * time) = problem.
BUT-3
The transformer was not designed for high frequencies. I don't know what will happen. The wires inside the transformer make capacitance. This capacitance will appear across the windings and will short out the audio amp at high frequencies. I am certain it will work fine at 500hz. (done that) I do not know about 10khz.
BUT-4 etc. There are some more things but too much data for today.
If I had the transformer and did some simple tests ……..

So, for example, I can dead short the output pins of a TDA series amp IC and still obtain a clean signal? For how long? If the chip is rated at 10W, what is limiting its output current? Or, are you assuming the transformer primary will provide enough resistance, even if much lower than 8 ohms?

You caused a little 'confusion' by saying the 10 ohm was to 'load' the amplifier, and as the reply said you don't need to 'load' it, a transistor amp is fine with no load.

However, what you actually meant was to 'limit' the output current.

But as has been posted since, the secondary load is reflected to the primary, so it depends entirely on what the secondary load is - and if correctly designed a limiting resistor wouldn't be required - and any such resistor would waste most of the power from the amp.

But as always, the vague nature of the question (no exact details) means any replies can only be vague.

If I understand correctly, taking ronsimpson's example, if I insert a 3K2 (or similar) series resistance on the secondary (240V) side no current limiting resistor would be needed on the primary? I assume that is the point Nigel is making.

However, that raises a question in my mind. Why does the primary side of PA transformers incorporate 4 and 8 ohm windings, apparently to match the amp's output impedance, if the current is intrinsically limited by the secondary?

no current limiting resistor would be needed on the primary
I am only saying that a primary 8 ohms or a secondary 3k look the same to the amp.
To protect the amp from shorts it does not matter.

OK. For me that is the main take-away from this thread. Thanks to everyone for their replies.

If I understand correctly, taking ronsimpson's example, if I insert a 3K2 (or similar) series resistance on the secondary (240V) side no current limiting resistor would be needed on the primary? I assume that is the point Nigel is making.

However, that raises a question in my mind. Why does the primary side of PA transformers incorporate 4 and 8 ohm windings, apparently to match the amp's output impedance, if the current is intrinsically limited by the secondary?

For the same reason as above - if you change one side impedance, you have to change the other to keep everything correct. So in the PA case the secondary is 100V line (impedance doesn't 'matter', as it's specified for the voltage level - which is the point of 100V line), if you change the impedance of the amp powering it, then you need to alter the primary tap to match, so the ratio stays the same.

Not so important with transistor amps, as they don't have an output impedance (as such - just VERY low - no matching required), but the wrong turns ratio on the transformer will reduce the power, just as using too high an impedance speaker would (which is what you're actually doing).

You need to realise that it's not a 'voltage' transformer it's an 'impedance' transformer, even though it's the exact same component (could be a 'current' transformer as well).

So, for example, I can dead short the output pins of a TDA series amp IC and still obtain a clean signal? For how long? If the chip is rated at 10W, what is limiting its output current? Or, are you assuming the transformer primary will provide enough resistance, even if much lower than 8 ohms?
A transformer does not provide a short to AC.
Its open circuit impedance is high, due to the primary inductance (which is reflected to the secondary).
Of course this impedance goes down with frequency and becomes just the transformer winding resistance at DC.

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