Ok I am looking on how to delay the turn on of a transistor, or any other component. The precision of the timing circuit doesnt matter, it just needs to slow down the saturation of the transistor for safety purposes; to allow a digital component (a Tinylogic XOR Gate to be precise) to activate before the transistor does. If I were to use a resistor in series with a capacitor attached to ground, and the base of the transistor attached to the resistor, It will not only delay the turn on, but it the saturation of the transistor will constantly be going up and down. This is what I want to avoid, the transistor saturation going up and down. I just want it to delay and then hold at a specific saturation.
Delaying the saturation of a transistor
Aug 19, 2006
13 Replies
I don't think you means saturation. Why do you ?
Graham
But what is the minimal delay required? If it is in miliseconds or parts of it then a 5V relay shorting the base to ground and activating on the power rail would create such delay. Reed relays may be too fast and mechanical relays may draw too much current, so give us some info as to what are the limits in which to move.
Have fun
Stanislaw Slack user from Ulladulla.
In message , dated Sat, 19 Aug
2006, Eeyore writesMr. J D may not be a native English speaker. He means 'collector current'.
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
I suspect you're on the right track but if so why doesn't he want the collector current to go up and down ?
Maybe he wants a delayed swich on with a 'snap' action ? In which case a comparator or dedicated reset type circuit would be appropriate.
Perhaps he can advise.
Graham
You can prevent the transistor from going into deep saturation by connecting a diode with low voltage drop from base to collector; cathode to the collector. You can trim that even more by adding a low value resistor between directly in series with the base of the transistor. The ON voltage of the transistor will be slightly raised.
Tam
Baker Clamp !
Graham
At least you recognized that wrapping that delay into the single transistor will also slow the rise/fall times. The R+C method shunts the high frequency transistor drive and slows it down. The only way to prevent this is to Schmitt trigger the transistor input, with either a tiny logic Schmitt using an RC on its input, or another transistor. A tiny logic Schmitt is probably the cheapest, most compact, and most consistent performer of your choices. Of course, this answer is given in isolation and without knowledge of your specific circuit. If "safety" is critical, the *best* way is to use the XOR output as an input to whatever eventually drives the transistor. This may get complicated with an XOR and I won't waste time on conjecture without additional information.
Thanks for all the replies. I am a native english speaker. To clear up what I meant: I only need a few milliseconds of delay time, and then stay at the peak. Like a capacitor but with only one rise, and no fall. Someone mentioned using a diode to do this, can you explain more in detail?
In message , dated Sat, 19 Aug 2006, Mr. J D writes
Apologies. The use of the word 'saturation' instead of 'collector current' has echoes of French terminology.
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
You could do it logically. See...
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For your case just use the "TAU" output and drop the XOR portion.
...Jim Thompson
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Thanks for the schematic. I don't want to do it logically, as it require an excess of parts, I have very limited space. I was almost out of space when I decided to add an XOR gate.
Can't you use a gate with schmitt inputs ?
Graham
The diode suggestion has to do with limiting the excess minority carrier charge in the transistor base by shunting base drive current around the base and directly into the collector at small BC forward bias voltage, which is why a Schottky diode is best and sometimes this combination is called a Schottky transistor. It is a speedy switch because the storage time, or roughly delay in time after removal of base drive until collector current begins to fall , is a function of the excess minority carrier base charge. This has little to do with your problem. You're not going to beat a tiny logic Schmitt trigger used to drive the transistor: View in a fixed-width font such as Courier.
. . . . dual .---> OUT . schmitt | . .--|--+--[R]--+----| o--| o-[Rb]-| . | |/ |/ |>
. === | | | . | C | | | . gnd1---------+-----+----' --- . gnd2 . . . RxC~ 1.5ms . . . .
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