Since a power transformer can work backwards, as long as the ratings are respected. I had always assumed that an OPT can work backwards.
However, I don't really know for sure. Would things like bandwidth be adversely affected?
However, I don't really know for sure. Would things like bandwidth be adversely affected?
As long as the ratio of driving impedance to rated winding impedance are the same,
the frequency response should be similar.
But, what will you use to drive an 8 Ohm secondary, a solid state amplifier?
Why?
Are you trying to get a solid state amplifier and an output transformer to run an electrostatic speaker?
Or, are you going to get a tube amplifier plus a second output transformer to run an electrostatic speaker?
One possible problem is if the solid state amplifier has a DC offset voltage output.
A typical 8 Ohm speaker may have a voice coil DCR of 6 Ohms.
A typical 8 Ohm output transformer may have a 0.8 Ohm DCR.
6 Ohms/0.8 Ohms = 7.5 times more current from the solid state amplifier into the output transformer than into the speaker.
the frequency response should be similar.
But, what will you use to drive an 8 Ohm secondary, a solid state amplifier?
Why?
Are you trying to get a solid state amplifier and an output transformer to run an electrostatic speaker?
Or, are you going to get a tube amplifier plus a second output transformer to run an electrostatic speaker?
One possible problem is if the solid state amplifier has a DC offset voltage output.
A typical 8 Ohm speaker may have a voice coil DCR of 6 Ohms.
A typical 8 Ohm output transformer may have a 0.8 Ohm DCR.
6 Ohms/0.8 Ohms = 7.5 times more current from the solid state amplifier into the output transformer than into the speaker.
For the regular OPT with 8 ohms secondary, it won't be very useful I suppose.
I have a pair of OPT meant for headphones, 10k:50ohm. I am toying with the idea to use a headphone amp to drive the 50ohm end and connect the input to a power tube. Edcor also has OPT with 600ohm output. But I admit, it only make sense if you have these lying around.
You made a good point about the the low DCR.
I have a pair of OPT meant for headphones, 10k:50ohm. I am toying with the idea to use a headphone amp to drive the 50ohm end and connect the input to a power tube. Edcor also has OPT with 600ohm output. But I admit, it only make sense if you have these lying around.
You made a good point about the the low DCR.
Same current in speaker.
A transistor amp's offset voltage is usually "solid", does not change with load. Adding the transformer just demands more current from the transistors.
A backward transformer is a classic way to get high voltage AC/Audio from a low voltage amp. Driving AC motors. Driving electrostat speakers. Bandwidth needs careful study. (Driving audio into capacitive loads, there is no Good Answer, only good-enough answers.)
to naviblue
please look here
OPT Characterization
We are playing since 2017 with this method; it is the best way to understand the OT itself.
I am preapring also a test set for s.e. OT with dc current; I am late due some "incident" 🙂
But you can understand exactly what will happen in a s.e. also without bias current.
Walter
please look here
OPT Characterization
We are playing since 2017 with this method; it is the best way to understand the OT itself.
I am preapring also a test set for s.e. OT with dc current; I am late due some "incident" 🙂
But you can understand exactly what will happen in a s.e. also without bias current.
Walter
Interesting idea. I think an actual interstage transformer would probably be a better choice. Keep in mind that OPTs are often gapped so they can allow for significant DC current to flow without saturating. You won't necessarily need that in an interstage transformer. This could mean that the transformer could be optimized differently.
That said, if you already have the transformers I don't see any harm in trying.
With 10k:50 the impedance ratio is 200. That means the turns ratio is sqrt(200) = 14.1. If used to drive, say, a 300B (170 V peak-to-peak swing) you'd need a 4 V RMS output from your headphone amp. That should be quite manageable.
If your headphone amp has high DC offset, you'll get some standing current in the transformer. You could eliminate that with a large capacitor (say 2200 uF) in series with the headphone amp output. But if the DCR of the transformer secondary is reasonably high (say a few ohm or higher) DC offset is not likely to cause any issues.
Tom
PRR,
Did I mis-state the action of DCR and DC offset correctly?
Perhaps I should have stated DC coupled solid state amplifiers, with negative feedback.
Take a solid state amplifier that has 50mV DC offset.
Connect an 8 Ohm speaker that has a 6 Ohm DCR.
50mV/6 Ohm = 8.3mA
Take the same solid state amplifier that has 50mV DC offset.
Connect an output transformer 8 Ohm secondary that has 0.8 Ohm DCR.
50mV/0.8 Ohm = 62.5mA
62.5mA is far more than 8.3mA.
62.5/8.3 = 7.53 times more current.
And 62.5mA just might be more than the quiescent standing current in the totem pole output stage.
I guess I did not explain it properly.
Did I mis-state the action of DCR and DC offset correctly?
Perhaps I should have stated DC coupled solid state amplifiers, with negative feedback.
Take a solid state amplifier that has 50mV DC offset.
Connect an 8 Ohm speaker that has a 6 Ohm DCR.
50mV/6 Ohm = 8.3mA
Take the same solid state amplifier that has 50mV DC offset.
Connect an output transformer 8 Ohm secondary that has 0.8 Ohm DCR.
50mV/0.8 Ohm = 62.5mA
62.5mA is far more than 8.3mA.
62.5/8.3 = 7.53 times more current.
And 62.5mA just might be more than the quiescent standing current in the totem pole output stage.
I guess I did not explain it properly.
A proper value resistor in series ( the actual nominal load 4ohm , 8ohm and so on ) should be added anyway so you don't short or overload the transistor amplifier . You will short an amplifier connected to a transformer secondary even without load on the other side , it is not like an impedance matching in a tube amplifier . The low inductance low resistance of the secondary is not limiting the current 😀 , DC or AC .
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I used a secondary model of 8 Ohms impedance.
With a DCR of 7.8 Ohms, the loss would be almost 6 dB.
I believe you meant a 32 Ohm or higher impedance secondary, if the DCR is as high as 7.8 Ohms.
With a DCR of 7.8 Ohms, the loss would be almost 6 dB.
I believe you meant a 32 Ohm or higher impedance secondary, if the DCR is as high as 7.8 Ohms.