1
\$\begingroup\$

I need to know what formulas or set of calculations would be used to figure out the base resistor and load resistors when hooking up a darlington pair as my datasheet for the ds2003 does not provide a hfe. The chip is 7 darlington pairs, and I am hooking it up to be controlled by my micro controller, to drive leds. If I know the current I need for my led to be saturated, how would I know what base resistor value I would need to put on the inputs to keep the base current under 25mA

Essentially I want to know what formulas are used or what is the common practice for designing such a circuit (led driver)

Single transistor calculations seem to be well published and I always have a hfe to work with but this is missing from the datasheet for these darlington drivers.

\$\endgroup\$

2 Answers 2

Reset to default
0
\$\begingroup\$

The DS2003 already has series and shunt base resistors that are okay up to 5V input, and should be fine down to much lower inputs.

http://www.ti.com.cn/cn/lit/ds/symlink/ds2003.pdf

\$\endgroup\$
2
  • \$\begingroup\$ Ok so I understand that you are saying that this chip has already got sufficient protection to handle a raw signal from a mcu with 5v as per the circuit diagram. This is great news. However in terms of darlington pairs in general, say if I was to design a custom darlington circuit using individual transistors is there a calculation used. \$\endgroup\$
    – meschael
    Commented Aug 10, 2014 at 22:17
  • \$\begingroup\$ The gain is the product of the two betas, so you don't need much current to saturate a Darlington. For example, if you allow 20 for each ("forced beta" << real beta), that's 400, so for 100mA on the collector 0.25mA is sufficient. The DS2003 allows for 1.4mA at 5V in. The 7.2K and 3K would typically be chosen to steal a small amount of base current so as to reduce the leakage when the input is low or open. In the case of the DS2003 its 83uA and 200uA (the main output transistor is driven by the other transistor so the base resistor can be less without affecting the required input current). \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Aug 10, 2014 at 23:56
0
\$\begingroup\$

You do not attempt to control the LED current by controlling the base drive of the transistor.

As Spehro indicates, these things already have appropriate resistors internally to allow them to be driven by 5 volt logic levels (I suspect they are ULN2003's in disguise). With 5 volts applied to the input, the transistors should be saturated, with the collector/emitter voltage about 1 volt for collector currents of 100 mA or so.

You control the LED current by adding a resistor in series with the LED.

\$\endgroup\$
1
  • \$\begingroup\$ This was useful in addition to Spehro's post but I could only accept one answer. I would still like to know about doing calculations in the future should I build a darlington pair from scratch but since I was not very clear in my post and made reference to the DS2003 that I am using this is an appropriate response. I thank you both. \$\endgroup\$
    – meschael
    Commented Aug 10, 2014 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.