# Transformer impedance match

I have a question about impedance matching with transformers.
If I have a source impedance of 20 K ohms, to step up the voltage does the primary have to match 20K ohms with a secondary of 40K ohms for a step up of 1:2. Or can I change it so the primary is 50 Ohms with a secondary of 40K ohms for a step up of 1:28.
I find this impedance matching very confusing

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I have a question about impedance matching with transformers.
If I have a source impedance of 20 K ohms, to step up the voltage does the primary have to match 20K ohms with a secondary of 40K ohms for a step up of 1:2.
What load impedance do you want to match, 40K ohms?
When you say a step up of 1:2 are you talking about impedance or voltage. The change in impedance is the square of the change in voltage. If you want to match 20K to 40K you will need a turns ratio of 1:1.41.

Or can I change it so the primary is 50 Ohms with a secondary of 40K ohms for a step up of 1:28. I find this impedance matching very confusing
How do you get a ratio of 1.28 from an impedance ratio of 50:40K?

vk6kro
I have a question about impedance matching with transformers.
If I have a source impedance of 20 K ohms, to step up the voltage does the primary have to match 20K ohms with a secondary of 40K ohms for a step up of 1:2. Or can I change it so the primary is 50 Ohms with a secondary of 40K ohms for a step up of 1:28.
I find this impedance matching very confusing
To match impedances, get the ratio of the impedances and take the square root of that figure.

eg 40 K / 20 K = 2
Square root of 2 is 1.414 so you need this turns ratio

eg2 40 K / 50 = 800
Square root of 800 is 28.28 so you need this turns ratio. (as you suggested)

Note, though, that you can't just have one turn and 28 turns. The reactance of the transformer primary has to be considerably larger than 50 ohms at the frequency in use. Otherwise, there will be a lot of signal current flowing in this reactance which will be wasted.

So if I get this right, if I have a source impedance of 40K and a load of 50. I need a turn ratio of 1:28.28 to get the same power from the source to the load. So if I have 10 volts in from the source I would have 10 volts out. Is this right?

uart
So if I get this right, if I have a source impedance of 40K and a load of 50. I need a turn ratio of 1:28.28 to get the same power from the source to the load. So if I have 10 volts in from the source I would have 10 volts out. Is this right?
No you'd need a turns ratio of 28:1 (more turns on the high impedance side ok)

uart
So if I get this right, if I have a source impedance of 40K and a load of 50. I need a turn ratio of 1:28.28 to get the same power from the source to the load. So if I have 10 volts in from the source I would have 10 volts out. Is this right?
No you'd get approx 10/56 of a volt out.

That's assuming you did it correctly (as per my above post). Doing it the way you've stated here you'd get less than half a milli-volt out.

So how do you step up voltage with an audio transformer. wouldn't you have to have a larger impedance on the secondary winding compared to the primary.

vk6kro
An audio transformer steps up the voltage or steps it down, just like any other transformer, depending on the turns ratio.

If you have more turns on the secondary, the voltage will be stepped up.

If the transformer is 100 % efficient, the power out will equal the power in.

If the input looks like 40000 ohms and there is 10 V RMS across it, the power in will be
10V * 10V / 40000 or 2.5 mW.

If the output is 50 ohms and there was 0.354 volts across it, the power would be
0.354V * 0.354V / 50 or 2.5 mW .......(I calculated the 0.354V separately)

So, you can see that the voltage across a smaller impedance (50 ohms) is less than for the higher impedance for the same power.
Also note that the 0.354 volts is just 10 volts divided by 28.28.

You can reverse the transformer and start with 0.354 volts RMS and get 10 volts out if you like.