Flyback Transistor

[dknguyen]

The #614096 Ranging Transformer, and the #619391 (6500) Ranging Transformer are both used the same way.
The input is listed as a DC voltage due to the way it is used in the circuit. If you look at the 6500 Ranging Module schematic, one side of the transformer (T1) is tied to +5V DC, while the other side is pulsed using a transistor (Q1). The input to the transistor can be a burst of 1 to 20 pulses at a 50 KHz rate. There should also be a time of rest between the bursts of pulses. When the pulses are stepped up by the transformer, the output is a 400Vpp burst of 50 KHz that will drive the electrostatic transducer. The zener diodes (CR1-4) placed on the output of the transformer will protect the transducer from possible damage due to excessive voltage.
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THis is what Senscomp told me about their transformer. THey have also told me it's a flyback transformer. Now...if V=L*di/dt is true, doesn't that mean the output voltage is dependent on how fast I the transistor shuts off if I use an open-collector/source transistor to drive current through the primary?They use some BJT in their circuit but I want to use a MOSFET and there is very little additional information other than what I have just said. Like if the transistor shuts off twice as fast, won't the flyback votlage be twice as high? Or am I missing something here?

 

[ronsimpson]

I may be looking at a different schematic…but
I do not know if the flayback is operating in resonance or non-resonance mode.

non-resonance (no capacitor): Then the transistor is on, current builds up in the primary of the transformer with time and voltage divided by inductance. That is to say the longer the time is the more current (power) is stored on the transformer. Then the transistor opens up the voltage on the collector will go up until it finds a limit somewhere. (load, or the 400 volt zener, or the breakdown voltage of the transistor) Running slow will slam more power into the load and likely increase the voltage depending on what the load is.

resonance (capacitor across the load): In this mode there is a capacitor somewhere. Maybe the load is a capacitor, there is a capacitor inside the transformer (self resonant frequency), in the case of a CRT monitor the capacitor is placed across the transistor. Transistor=on, current builds up with time. Transistor=off, current stored on the core inductance will move onto the capacitance as voltage. I^2L/2=V^2C/2 Running slow will increase the voltage!

 

[dknguyen]

I may be looking at a different schematic…but
I do not know if the flayback is operating in resonance or non-resonance mode.

non-resonance (no capacitor): Then the transistor is on, current builds up in the primary of the transformer with time and voltage divided by inductance. That is to say the longer the time is the more current (power) is stored on the transformer. Then the transistor opens up the voltage on the collector will go up until it finds a limit somewhere. (load, or the 400 volt zener, or the breakdown voltage of the transistor) Running slow will slam more power into the load and likely increase the voltage depending on what the load is.

resonance (capacitor across the load): In this mode there is a capacitor somewhere. Maybe the load is a capacitor, there is a capacitor inside the transformer (self resonant frequency), in the case of a CRT monitor the capacitor is placed across the transistor. Transistor=on, current builds up with time. Transistor=off, current stored on the core inductance will move onto the capacitance as voltage. I^2L/2=V^2C/2 Running slow will increase the voltage!

THe capacitor is the load in this case. It is a 50kHz capacitive ultrasonic transducer. In the app note, inductance of the transformer and the capacitance of the transducer seem to form a 50kHz band-pass filter. I always though this was just to "filter" out the harmonics in the square wave for a slightly cleaner 50kHz drive signal, but you are saying it may be operating in resonance?

I've never heard of the thing you described where the capacitor goes across the transistor, but I don't know too much about flyback circuits.

Page 6, right side (once turned CW)
https://www.electro-tech-online.com/custompdfs/2008/04/650020module20spec.pdf

I find it quite strang ethey don't have a snubber (like diode in series with a resistor across the primary). My understanding is that the BJT would die, but that's the circuit they seem to use in their rangers.