Instability solved depth lost

Ha the 386 now that was a machine in it's time with the black and white monitor and in fact they were green I remember the screens that came back to the lab from the bank they had an image burned on the screen (Dad use to work for Olivetti) I knew every body over there and I could go where ever I wanted to, now those were the days LOL. I was thinking of changing that connector with one that I have that is also the screw type the pin's are a lot thicker and it wont move around like the clip one that I had in there I think that I will do that in the morning mind you if I wait a few more minutes it will be morning LOL (It is)

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Xavier

		Quote:			Posted by xavier
	4) Resistance of the coil = 1R6 resistor in parallel = 560R therefore R= 1.5944 (Practically unchanged ) Now if I could get the maths that will show me why this works I'd be a very happy guy   Idea LOL I guess that I should replace the poor Q plug measure the resistance then do the same with the good Q plug but I'm a bit short on time at the moment.

Resistance in parallel is calculated using the formula: 1/R^total = 1/R¹ + 1/R² + . . .
No matter how many you have, just keep going. Use this page to plug in the values and save time:

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When you add the coil resistance, remember that the formula does not apply when the circuit is on. Resistance in an AC circuit is an ever changing factor. When you add the resistor in parallel, the only time it is actually effective in terms of resistance is when the circuit is off. When it is on, the resistance acts as a "snubber" absorbing transients and spikes that occur when the field in the coil collapses after being switched off. You stabilize that small portion of the waveform that occurs near the "turn off" time of your TX pulse by absorbing any transients. The bad part is that there is a cost. Your TX power will be diminished.

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		Quote:			Posted by xavier
	I know how to do the basic calculations like capacitance and resistance also amps and watts the basic stuff but how would I know what resistor (Snubber) I should use?

The easy way? I always use the empirical method. I throw a 10R6 potentiometer into the circuit and adjust for the best response time without dampening the return. Measure the pot setting later and use that value or as near as you can get.

You can calculate it using a formula, sure, but the AC nature of the circuit makes it hard and time consuming. Keep a 10R6 (10 Megohm for the Americans) pot handy with wires and alligator clips on the wiper and one end and you'll be very much happier.

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		Quote:			Posted by xavier
	This sound like a great plan do I replace the damping resistor after then? and how do I know what will be the correct value for it? also why do you use a 10R6 pot? the resistor that I got in at the moment is only 560R

Use the design spec for the resistor. The snubber value might be as high as that 10R6 or lower. This covers all possibilities. If you want to fine tune the result, get a few different values. 1R6, 100R3 and so forth. Add clip leads and go for it.

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		Quote:			Posted by xavier
	Thanks GD that answers my question but will the damping resistor not have to be done in the same manner as the snubber resistor? I went out today to do some testing and the result is not disappointing but also not what I would like to have but have improved.

That resistor does dampen the signal but the value is determined not by the circuit reaction but by the selection of your MOSFET. If you change the MOSFET part to another part number, you must also change the resistor.

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