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Subject: Re: relays through NPN transistor

Date: 08/29/03 at 4:55 AM
Posted by: Pieter Hoeben
E-mail: pieter[AT55]hoeben.com (replace [AT55] with @ to reply)
Message Posted:

In Reply to: relays through NPN transistor posted by sheen on 08/28/03 at 2:09 AM:

Hi, the diode should at least be able to handle the current that flows through you relay coil at 5V. Preferrably use a diode that can handle more.
The base resistor should be chosen so the transistor will fully saturate when you drive the relay. I do not know the coil resistance, but lets say it was 50 ohms. The current would be 5 V / 50 ohms = 0.1 A. Then you divide it by the DC gain of the transistor. I do not have that either, lets guess a HFE of 50. So your base needs 0.1 A / 50 = 2 mA. But that would just drive the transistor, so you need more. I would go for a 4 mA or more to make sure the transistor saturates. Assuming that the PIC gives as high 4V at 4 mA load, the base resistor would be:
( 4V (PIC out) - 0.6 V (Ube) ) / 4 mA = 850 ohm. Use a standard value below that. Note: I guessed all values here, please recalculate it with: correct coil resistance, transistor gain, PIC loaded high output voltage.


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