I'm having a bit of a brain cramp here. See the attached schematic. It is a tube preamp with an input transformer. The transformer is a step up as shown and there is way too much gain. I could put an attenuator between the tranny and tube, but I found one resistor across the windings (resistor that is circled) works quite well, and brings the level down almost perfectly. What I am concerned about is input and output impedances of the transformer. What's the math. Please note I cannot make wholesale changes. The gain solution must be simple.
Input Transformer Impedance
- Thread starter audiopro
- Start date
That attenuator is wrong as drawn.
You are shorting the high impedance secondary, input impedance becomes very low and undefined, attenuation is probably happening in the generator itself thanks to its internal impedance, a mess.
You need a 2 resistor solution, either at the input or at the output.
Current transformer ratio ("gain") is 6.7:1 , nominal impedances 600 ohm:27k ohm
To reduce actual gain to half (-6dB) you can connect, say, 2 x 15k resistors in series across the secondary and tube grid to their junction.
To reduce to 1/4 (-12dB) you can use 22k in series with 6k8
In all cases the proper 600 ohm input impedance will be preserved.
You are shorting the high impedance secondary, input impedance becomes very low and undefined, attenuation is probably happening in the generator itself thanks to its internal impedance, a mess.
You need a 2 resistor solution, either at the input or at the output.
Current transformer ratio ("gain") is 6.7:1 , nominal impedances 600 ohm:27k ohm
To reduce actual gain to half (-6dB) you can connect, say, 2 x 15k resistors in series across the secondary and tube grid to their junction.
To reduce to 1/4 (-12dB) you can use 22k in series with 6k8
In all cases the proper 600 ohm input impedance will be preserved.
I would add a small capacitance in parallel to both resistor in order to make a compensated attenuator, because the triode's input capacitance will form a low pass filter.
Osvaldo de Banfield,
A 970pF capacitor is 8.2k at 20,000Hz. That would be -3dB at 20,000 even if the top of the transformer secondary impedance was infinite. And 485pF would be -1dB at 20,000Hz. Or, just looking at the 22k resistor and the grid capacitance as a low pass, with the 8.2k removed, the Grid capacitance would have to be 180pF to be -1dB at 20,000Hz, not usual for most tubes.
What grid input capacitance is 485pF, or 180pF (including the Miller Capacitance). Putting capacitors on the resistive divider resistors might do one thing . . . reduce the transformer resonance frequency . . . into the audio band. No need for capacitors here.
A 970pF capacitor is 8.2k at 20,000Hz. That would be -3dB at 20,000 even if the top of the transformer secondary impedance was infinite. And 485pF would be -1dB at 20,000Hz. Or, just looking at the 22k resistor and the grid capacitance as a low pass, with the 8.2k removed, the Grid capacitance would have to be 180pF to be -1dB at 20,000Hz, not usual for most tubes.
What grid input capacitance is 485pF, or 180pF (including the Miller Capacitance). Putting capacitors on the resistive divider resistors might do one thing . . . reduce the transformer resonance frequency . . . into the audio band. No need for capacitors here.
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Any attempt to reduce the gain of the first stage will likely increase its noise. Its better to reduce gain later in the circuit after the signal has been boosted well clear of noise floor. This could be as simple as reducing the anode load resistor.
Given the transformer is impedance-matching the front end its probably carefully designed to minimize the noise already.
Given the transformer is impedance-matching the front end its probably carefully designed to minimize the noise already.