Hi there,
I do have a pair of Haufe t42 transformers laying around and i would like to use them to perform I/V conversion on a I2S dac with a single TDA1543 i also have on my shelf.
For what i have found in internet these transformers do have a 1:20 ratio.
I measured the windings :
primary @ 3,8 Ohm
secondary @ 3,1 KOhm
Can someone please help me out with calculating Rsec value for a setup like this:
I'm simply too dumb to do it and after reading tons of posts on forum and articles on internet I'm totally lost.
I really will appreciate help of someone who is smart,
Best Regards
Bartek
I do have a pair of Haufe t42 transformers laying around and i would like to use them to perform I/V conversion on a I2S dac with a single TDA1543 i also have on my shelf.
For what i have found in internet these transformers do have a 1:20 ratio.
I measured the windings :
primary @ 3,8 Ohm
secondary @ 3,1 KOhm
Can someone please help me out with calculating Rsec value for a setup like this:
I'm simply too dumb to do it and after reading tons of posts on forum and articles on internet I'm totally lost.
I really will appreciate help of someone who is smart,
Best Regards
Bartek
You can calculate yourself. The rule is that the secondary load is transformed to the primary by 1/n^2. If the turns ratio is 1:20 (which is probably too much) and the secondary load is 1k then the primary will see 25R. The full scale output of the TDA1543 is 2.3mA, this will generate 2.3mA x 25R = 57.5mV. This voltage will be transformed to the secondary by 1:20, so there will be 1.15V peak-to-peak. The RMS value is Vpeak/(2 x sqrt2), about 400mV.
In reality it will be lower due to primary and secondary resistance.
You can change the secondary load and recalculate.
BUT: you have to consider the lower corner frequency, which can be calculated from the primary (or secondary) inductance, and the primary (or secondary) external load. Let's suppose the secondary inductance is 10H. You can measure it with open primary. Then T = L / R, that is 10 / 1000 = 0.01s. ω = 1/T, that is 100. f = ω / (2 x π), f = 15.9 Hz. Looks good.
In reality it will be lower due to primary and secondary resistance.
You can change the secondary load and recalculate.
BUT: you have to consider the lower corner frequency, which can be calculated from the primary (or secondary) inductance, and the primary (or secondary) external load. Let's suppose the secondary inductance is 10H. You can measure it with open primary. Then T = L / R, that is 10 / 1000 = 0.01s. ω = 1/T, that is 100. f = ω / (2 x π), f = 15.9 Hz. Looks good.