[AMRadio] Audio Driver Transformer advice

Paul Baldock paul at paulbaldock.com
Fri Jan 16 10:53:49 EST 2015

Thanks for your input.

I finished up using two step up transformers with the 8 ohms sides in 
parallel and the 500 Ohm sides in series to simulate a center tap. So 
that is effectively 8 Ohms to 2000 Ohms, center tapped. Testing in to 
a 1K resistive load the output looks reasonably distortion free,

- Paul

t 10:45 PM 1/15/2015, Donald Chester wrote:
>----- Original Message -----
> >>>>From: Paul Baldock <paul at paulbaldock.com>
>I'm building a modulator with 2 x 813's in pushpull (strapped as
>triodes) driven by a 20W Solid State audio amplifier. I need a driver
>transformer to go between the 813 grids and the 20W amp. The 20W amp
>can drive about 20V peak to peak. I think I need about 200V peak to
>peak for the 813's, so a 10:1 turns ratio ratio would be about right.
>So based on that an audio transformer designed to go from 600Ohms
>center tapped to 4 or 8 ohms would probably do the job. Have I got
>the math right? Does something like this exist?
>Paul >>>>>
>For best results, use the least amount of step-up possible to just barely
>get the required peak-to-peak audio voltage, with a little bit of additional
>head-room. Ideally, the driver amplifier will go into saturation (flat-top)
>at the same instant that the modulator is driven to its maximum capability.
>IOW, the driver and modulator stage reach saturation at the same time. I
>would design in a little extra  margin, so that the driver is capable of
>driving the modulator slightly beyond saturation, just to make sure you get
>the maximum undistorted audio power out of the modulator before the driver
>starts clipping.
>A typical class-B grid impedance varies widely over the audio cycle, from
>infinite ohms (on the negative half of the cycle when the tube is not
>conducting, down to a minimum of about 500 ohms at the crest of the positive
>peak. The grid voltage / grid impedance curve will likely be non-linear. The
>total impedance of the whole mid-tapped winding would be something like 4
>times 500 ohms, or 2000 ohms. Using the 8-ohm tap, the impedance ratio would
>be 8 ohms to 2000 ohms, or a 1:250 Z ratio. That would be equivalent to a 1
>: 15.8 turns ratio.  The 4-ohm tap would require more step-up, 1 : 22.7.
>I doubt that a 600 ohms to 8 ohms transformer would give enough voltage
>swing. It may be hard to find an output transformer with a primary impedance
>as low as 2000 ohms at that power level, although I'm sure such a thing
>exists somewhere. In the  real world, I would look for a good quality hi-fi
>output transformer rated for at least 20 watts, and preferably 30 watts or
>more, with as low a plate-to-plate impedance on the  primary as you can
>find, but nothing much below 2000 ohms.
>A (backwards wired) transformer with more step-up will deliver more
>peak-to-peak voltage to the grid, but the effective internal resistance of
>the source voltage will be increased, giving more distortion as the audio
>source voltage regulation is degraded. Putting it another  way, using too
>much step-up, the grids of the tube will cause the audio voltage output from
>the transformer to sag. Ideally you might find a transformer with multiple
>taps so you could experiment to find the optimum turns ratio, but you'll
>have to make do with the best you can find.
>Also, a transformer rated for higher power (and lower primary impedance)
>will tend to have lower winding resistance, which would reduce the audio
>voltage sag.
>Once you have found what looks like a suitable transformer, try both the
>4-ohm and 8-ohm tap and see which one gives the most satisfactory results
>with the least distortion.
>Don k4kyv
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