Originally posted by green
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BIPOLAR ALTERNATIVE TO H BRIDGE
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Thanks, could you describe your coil to get 300uH, .2 ohms and 100p. Diameter, number of turns, wire size, winding method(bundle, spiral, ?)and cable length.Originally posted by moodz View PostThe relatively high resistance of your coil is causing the ramp ... increasing the capacitance only makes the ramping worse in accordance with an RC constant.
You are losing more than half a watt in your coil due to I2R losses.
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The relatively high resistance of your coil is causing the ramp ... increasing the capacitance only makes the ramping worse in accordance with an RC constant.Originally posted by green View PostI tried stepping the capacitance C1, 100p 200p 400p in spice polar2D(reply #30). Appears ramp is steeper with increasing capacitance.
You are losing more than half a watt in your coil due to I2R losses.
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I tried stepping the capacitance C1, 100p 200p 400p in spice polar2D(reply #30). Appears ramp is steeper with increasing capacitance.Originally posted by Altra View PostIf I understand your question right. You can add more parallel capacitance across the coil. This will widen the half sine, slowing the change in current. This will also lower the peak flyback voltage. Tfly = 3.14*sqrt(LC).
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If I understand your question right. You can add more parallel capacitance across the coil. This will widen the half sine, slowing the change in current. This will also lower the peak flyback voltage. Tfly = 3.14*sqrt(LC).Originally posted by green View PostIs there a way to modify the circuit to make the ramp less steep?
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The polar drive circuit is interesting, would like to try it. My Tx coil resistance is closer to 3 ohms. Tried a spice circuit with Tx=3 ohms and all MOSFETS STW11MN80. Get about 18A/sec ramp at switch point. Is 18A/sec acceptable? Is there a way to modify the circuit to make the ramp less steep?Attached Files
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Some thoughts, maybe doesn't make sense or not correct. Current ramp or tilt causes a X signal. Important when sampling signal.Originally posted by moodz View Post...the coil resistance in the breadboard circuit is just over 1 ohm, with a 1.24 volt power supply and the pulse rate slowed to 1 pulse per second the "ramp" or "tilt" on the waveform was under 1 ma / second.
Once the coil reaches "terminal" current according to ohms law the only thing that can vary current will be power supply fluctuation or resistance change due to thermals in the FETs etc.
Power consumption was 600 milliwatt for a 800 ma current excursion at 500 volt peak flyback.
Tried a couple things with spice. 1A change with polar circuit. Square wave amplitude adjusted for 3.3A/second. R signal 10,000 times greater than X signal. Maybe allowable ramp could be greater than 3.3A/second?Attached Files
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...the coil resistance in the breadboard circuit is just over 1 ohm, with a 1.24 volt power supply and the pulse rate slowed to 1 pulse per second the "ramp" or "tilt" on the waveform was under 1 ma / second.
Once the coil reaches "terminal" current according to ohms law the only thing that can vary current will be power supply fluctuation or resistance change due to thermals in the FETs etc.
Power consumption was 600 milliwatt for a 800 ma current excursion at 500 volt peak flyback.
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Maybe a constant current source, similar to this circuitAttached Files
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Larger (external) source resistance and voltage can reduce those variations at the cost of a poorer efficiency.Originally posted by Carl-NC View PostThis circuit will be sensitive to components and temperature. The FET resistances and body diode characteristics will matter so you may be looking at differences in Spice models. Drop in e.g. IRF740s and see what happens. Ideally you would add a current-monitoring resistor to the low-side switches and throttle the supply voltage to achieve a particular current.
A ramp will induce an offset in the receiver. How much your circuit tolerates would depend on the noise level and the sensitivity of the amplifier.Originally posted by green View PostThanks
Does anyone know how flat the current needs to be(Amps/second)? I'm guessing maybe 3A/second.
In my simulations 0.2A/s is about right.
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ThanksOriginally posted by Tinkerer View PostI meant: use the resistance and capacitance of your own coil, including cable and shield. This circuit is very sensitive to capacitance and resistance.
Does anyone know how flat the current needs to be(Amps/second)? I'm guessing maybe 3A/second.
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I meant: use the resistance and capacitance of your own coil, including cable and shield. This circuit is very sensitive to capacitance and resistance.Originally posted by green View PostNot sure what you mean. I assumed .2 ohms and 100p was coil resistance and capacitance.
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Teleno posted some mosfet models here -> https://www.geotech1.com/forums/show...656#post226656Originally posted by green View PostBeen playing with circuit in spice. Spice doesn't avalanche the Mosets so I added back to back diodes across the coil to avalanche at 600V. Peak current not flat if avalanches. Increasing coil resistance causes a problem also. Does anyone know how flat the current needs to be(Amps/second)?
that handle the avalanche mode correctly.
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Been playing with circuit in spice. Spice doesn't avalanche the Mosets so I added back to back diodes across the coil to avalanche at 600V. Peak current not flat if avalanches. Increasing coil resistance causes a problem also. Does anyone know how flat the current needs to be(Amps/second)?
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