The load resistance is reflected from the secondary to the primary via the square of the turns ratio. For example, if the primary to secondary turns ratio, which for an ideal transformer is the same as the voltage ratio, is 1:3), Then R ohms on the primary looks like 9R ohms at the secondary. I'm assuming that your transformer has a 1:1 turns ratio. So, with a 600 ohm source resistance, and a 600 ohm load, the output voltage will be 1/2 the source voltage. . When a transformer is described as a 600 ohm transformer, this simply means that the transformer is design to work in systems that have 600 ohm source resistances and 600 ohm loads. This makes it easier to calculate power distribution. It has nothing to do with the winding resistance of the transformer. Regards, Jon
I may have mislead you in my previous reply. I meant that if you applied a voltage from a 600 ohm source to the unloaded transformer, and then applied the 600 ohm load, the voltage would drop to 1/2 its unloaded value. Of course, if you adjust the primary voltage with the load applied, then the secondary voltage will equal the primary voltage. Regards, Jon
I'm experimenting with a small 600 ohm phone line transformer. I have yet to understand how a transformer reacts when its output is not loaded with the specified load or not loaded. For instance, if I'm driving the primary with a 1KHz 1Vp-p signal through a
600 ohm resistor (the source impedance is considered very low) and the secondary is loaded with a 600 ohm resistor, I assume I will get also a
1Vp-p at the output (less some minimal losses). But what happens if the output is loaded with 5K (like a potentiometer) or not loaded at all? Is there any way or rules of thumb to determine the output voltage under those conditions?
It is a transformer, so the voltage out is the same as the voltage in (or almost) as long is it is operating with voltages, currents and frequencies within its ratings. The 600 ohm in and out specs give some guidance about what sort of sources and loads it operates with while appearing pretty much like an ideal transformer. It does not specify the impedance of the transformer, itself. In other words, if the source has a 600 ohm impedance, and the load is a 600 ohm impedance, then the output voltage will be almost the same as the input voltage over the widest frequency range (may be something like
300 to 3000 Hz).
Other source and load impedances will allow less ideal transformation (more loss) land probably over a narrower frequency range.
So, in this particular case, it is like a resistor divider. No load, you read the source voltage even if there is a (reasonable) resistor is in series. With a load equal to that resistor, the output drops to half.
When the load goes from 600 ohm to 5k ohm you are drawing less current thru the transformer and the resistive losses will be less causeing the output voltage to increase. If the 600 at the input is in series then with the input winding then that is the primary cause of output voltage changing with load.
Dan
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Dan Hollands
1120 S Creek Dr
Webster NY 14580
585-872-2606
dan.hollands@gmail.com
www.QuickScoreRace.com
The relationships between the primary and secondary windings for a tightly coupled ideal transformer are --- Np/Ns = Ep/Es = Is/Ip = square root (Zp/Zs).
If you measure the primary voltage and the unloaded secondary voltage, their voltage ratio is very nearly the turns ratio. If the transformer is ideal, the power in the primary is equal to the power in the secondary. In a 2 to 1 stepdown transformer, if we have 1 volt at 1 amp in the primary then for the power to be equal in the secondary we would have to have 1/2 volt at 2 amps in the seconday. Using ohm's law , the primary looks like 1 ohm and the secondary looks like 1/4 ohm. Plug those numbers into sqrt(Zp/Zs) and as if we didn't know better, the Z ratio is 4. The 1/4 ohm load in the secondary looks like a 1 ohm load in the primary.Whatever load is connected to the secondary will appear 4 times larger in the primary. If you reverse the transformer and use the secondary as the primary, it becomes a 1 to 2 step up transformer. Apply 1 volt at 1 amp to the primary, and you would have 2 volts at 1/2 amp in the secondary. The primary looks like 1 ohm and the secondary looks like 4 ohms. Still we have one watt in the primary and one watt in the secondary. The 4 ohm load connected to the secondary appears as 1 ohm in the primary. Whatever load is connected to the secondary will appear 4 times smaller in the primary.
Part 2 - The secondary load appears in the primary multiplied by the square of the impedance ratio. For 1 to 1 , they are equal, for step down, the primary Z is larger than the secondary load and for step up, the primary Z is smaller than the secondary load. Whatever load is reflected to the primary, it is in series with the source Z, and this forms a simple voltage divider. You can use ohms law to figure out what voltage the primary will see. Transformers can transform pure resistance as well as reactive values.
The difference in performance between a loaded and unloaded microphone transformer, when used with an SSB transmitter, is hardly noticeable. Over the middle frequency range the unloaded output voltage increases by a few dB.
An increase in the audio HF response is suppressed by the side-band Xtl filter and is unheard. Save yourself the cost of a terminating resistor.
The microphone impedance and its frequency response is a matter of guesswork anyway. =========================================
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