0
\$\begingroup\$

I have a water valve which should be operated between 6-12V. I want to control this with an "standard" arduino-transistor switch (see image below, change the motor to a valve).

enter image description here

When trying to calculate the base-resistor needed, I use these basic formulas

$$ I_b = I_c/\beta$$ and $$R = \frac{V_{cc} - V_{be}}{I_b} $$

Since the data-sheet for the valve says it needs 320 mA @ 12V, and Vcc = 5v (voltage from arduino) and Vbe = 1.2 (my bc337 transistor), I can calculate that R = ... (something)

However, these formulas does not say anything about the collector voltage. What would happen if I choose 1.5v, 5v, or 12v etc ? If I change the voltage supply to the valve, shouldent the collector current change accordingly?

Extra question: can I in some way destroy/burn the aurdriono pin with this circuit?

\$\endgroup\$
3
  • 2
    \$\begingroup\$ The inductive load needs a flyback diode - back-emf kills the transistor otherwise. \$\endgroup\$
    – Turbo J
    Sep 5, 2017 at 12:51
  • \$\begingroup\$ Using a BC337 Vbe will not be 1.2 V but 0.6 V. But I consider the TIP120 a better choice (higher Ic,max more robust), it does have a Vbe of 1.2 V. Since the TIP120 is a Darlington transistor, it will have a very high beta. You can just use that 1 kohm series resistor or even 10 kohm, it does not matter much. Not much use in calculating it as beta is so high, it is a better strategy to just supply enough base current, 1 k ohm or 10 kohm will do just that. \$\endgroup\$
    – Bimpelrekkie
    Sep 5, 2017 at 13:05
  • \$\begingroup\$ You could use a FET and not worry about how much current you need. \$\endgroup\$
    – George Herold
    Sep 5, 2017 at 20:05

2 Answers 2

Reset to default
1
\$\begingroup\$

The formula \$I_c = \beta I_b\$ applies in forward active operation of the BJT. But to minimize heat dissipated by the transistor in this circuit, you don't want to operate the BJT in forward active mode. You want to operate it in saturation mode. Do this by increasing the base current (by reducing the base resistor value) until the collector current required to maintain forward-active operation is more than the 9-12 V source can supply.

What would happen if I choose 1.5v, 5v, or 12v etc ?

The BJT isn't going to produce a negative voltage on its collector. So the voltage applied to the load (solenoid valve in your case) will be equal to the source voltage or less. If you use a 1.5 V supply, then only 1.5 V can be applied to the load.

So choose a supply voltage equal to what is required to operate your load, plus a small allowance for the saturated collector-emitter voltage of your BJT.

\$\endgroup\$
0
\$\begingroup\$

In bipolar devices, once a minimum Vce is reached, almost all the charges injected by the emitter end up at the collector. Increasing the Vce further just moves the device operation into a region described by "Early Effect", where the base-collector region becomes slightly narrower and the beta slightly increases.

enter image description here

\$\endgroup\$

Your Answer

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

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