Voltage drop across JFETs? - diyAudio
 Voltage drop across JFETs?
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 14th July 2010, 02:56 PM #1 diyAudio Member   Join Date: Mar 2009 Voltage drop across JFETs? Trying to incorporate the circuit from the schematic hereThe Rat Bypass in to my projects and just trying to work out the value of the dropping resistor for the LED on an individual basis for different colours etc (using this calculator, as it happens Calculate Resistors for LEDs). Just wondering if anyone knows if there's a drop, how much it would be etc I know the diodes knock off 0.7V each anyway. Any help on this would be greatly appreciated, I tried googling "JFET voltage drop" and can't seem to find a definitive simple answer
 14th July 2010, 03:14 PM #2 diyAudio Member     Join Date: Sep 2003 Location: Upstate NY The jfet in that circuit is used as a switch, notice that when the switch is in position 1, the gate is tied to the drain, turning it on fully. When on it acts as a resistance. Look in the data sheet of your chosen jfet for Rds(on). It will probably be fairly insignificant compared to the 2K2 drop.
 14th July 2010, 06:40 PM #3 diyAudio Member   Join Date: Mar 2009 Yeah I know it's a switch, it's being used to activate the LED but I'm not sure what voltage is coming from the Source of the JFET as I don't know if there's a voltage drop through the component like a diode. I found a datasheet here but it doesn't seem to mention 'Rds'. I am specifically asking about effects on the voltage to the LED: is it just 9 volts, minus the voltage drop of two intervening diodes therefore equalling 7.6 volts, going through the component (i.e. what voltage would be measured at the anode of the LED)? Sorry, confused!
 14th July 2010, 07:23 PM #4 diyAudio Member     Join Date: Sep 2003 Location: Upstate NY Substitute a J111, which lists Rds(on) at 50R. You have 1.2V across the LED, 1.4 V across the diodes, leaving you with 7.4V across 2200 R + 50R. This gives you 3.3 mA, meaning that you lose .16V in the JFET. It's not like a diode drop, it depends on the current. If you ignore the JFET Rds(on), you'd come up with 3.4 mA through the LED, a 3% difference. The higher the voltage used to drive the LED circuit, the less significant Rds(on) will be. You can also look at the Vds curve on page 3 of the data sheet. Which is really what you are asking. Look at around 3 mA and in that circuit when on your Vgs will equal Vds. Notice how much steeper the curves get with more positive Vgs., I'd guess that at 3 mA, you'd be around .5V or less. Again, a small fraction of the total drop and would lead to fractions of a mA difference in LED current. It usually takes a few mA to make a noticeable difference in LED brightness. The long and short of it is that you can calculate the voltage drop, but as a practical matter in that circuit it doesn't matter much. If you are going to run your LED from a 15V rail, it matters even less, since the current limiting resistor will drop even more. Looking at the data sheets, I think that the reason the 2N5457 was chosen was it's lower pinch off voltage requirement. The J111 should work, as long as the volume pot is a reasonable value (not megohms).

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