Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
I ran into a little sag I don't have any 3.6 volts zener diodes and from what I read the USB data lines need a 3.6 voltage level I was thinking a have some white leds couldn't I use them to clamp the line to 3.6
Option A, connect the resistor to 3.3V.
Option B, use just a resistor to 5V and rely on the other side to do the clamping to the 3.3V rail.
Option C, use a divider and hope it doesn´t get too loaded and stays above the 2.8V.
None of A,B and C will work the usb chip is bit bang Usb and has to run at 5 volts so I got no 3.3 volt rail. Im going to use the leds they look like they would work and I have some that should do the trick.
Iet me spend a little time here they say that the data - and + line needs to swing from .3 to 3.6 volts for low speed but all I can find said it's 0 to 3.3 volts I have 3.3 zenners I'll try them but now I kind of like the led idea too LOL
The main difference is that LED's are forward biased and cannot handle only 10uA and must not exceed >-5V ( reverse bias)
Whereas Zeners are designed to handle reverse current but LED's have same or lower ESR for same power level. e.g. 1W LED is ~1Ω @3V.
e.g. Maximum Zener Impedance [email protected] 1Ω typ 3Ω max at 1W rating for lowest voltage range.
Given 2 conditions, connected with 50 Ω load and no load, the variation in Vo will be great.
May I suggest two AlGaAs Red in parallel reversed in series to provide a 2V drop from 5V and achieve Vo= 3V max with an ESR of 25+15=40Ω is a decent match and will reduce any risk of latchup from overvoltage on the interface with a 3.3V chip and 3.6V limit for interface.
No R's required, but not short cct proof. to 5V you can add 10Ω in series. but this is an unlikely fault.
Otherwise using your schematic it's hard to balance max current, max voltage and matched impedance with connected and disconnected cable.
I think I'll try the 3.3 volt zener diodes I have. Im not thinking the best these days could you kindly show me. I'm thinking just replace the zener with two leds there not much current on the data lines they just don't need to reach 5 from the attiny
You dont need 2 leds to replace one zener, just one, if you put the led forwards into the circuit, ie not reversed like a zener would be it should be fine, another thing with leds is that the reverse max voltage will be lower than a zener, however 5v shouldnt be enough to do them any harm.
There are two conditions ;
50R termination connected
no load disconnected.
Requirements are 20mA max in LED and 3.6V max into data + or - and 2.8V min
As long as these are satisfied it is OK.
With two 2V reversed LEDs current is 60mA so scrap that idea.
Optimal solution requires a series R1 from driver and series R2 to LED such that no connection load drive current is 20mA or 100 Ohm's total (minus 25R+15R = 60 Ohm's shared by R1& R2 ) with 2V drop allowing 0.4V drop on R2 to LED yields 0.4V/0.02A= 20 Ohms and 40 Ohms to driver R1, the. when connected Vo = 5V* (50/(50+40+25)= 2.17V which is low but may work.
Otherwise use a 1W LED with 1Ohm ESR and calc R1=> 50/(50+R1+25)*5V=2.8V then R1=29 ohms and I (no load )= 3V/54 ohms =56 mA