please can anyone show me how to calculate the base resistor for an NPN transistor used as a switch. the supply voltage is 12 volts, the load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

" wrote:

please can anyone show me how to calculate the base resistor for an NPN transistor used as a switch. the supply voltage is 12 volts, the load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

When the NPN transistor is saturated (fully on), the emitter-collector
voltage will be about 0.2-0.3 volts. Using the other values you have,
you can calculate Ic; the current flowing through the collector
(and the relay).

To saturate the transistor, the base current must be greater than
Ic / beta
where "beta" is the minimum current gain of the transistor.

The last piece of information you need is that the base-emitter
voltage is one "diode-drop"; about 0.7 volts.

On 10/8/2017 1:12 PM, natp wrote:
" wrote:

please can anyone show me how to calculate the base resistor for an NPN transistor used as a switch. the supply voltage is 12 volts, the load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

When the NPN transistor is saturated (fully on), the emitter-collector
voltage will be about 0.2-0.3 volts. Using the other values you have,
you can calculate Ic; the current flowing through the collector
(and the relay).

To saturate the transistor, the base current must be greater than
Ic / beta
where "beta" is the minimum current gain of the transistor.

The last piece of information you need is that the base-emitter
voltage is one "diode-drop"; about 0.7 volts.

It's customary to overdrive the base. I normally use a factor of
10.
It's also customary to put a reverse diode across the coil
so that the turn-off transient doesn't overvoltage the collector.

On Sunday, 8 October 2017 19:53:50 UTC+1, wrote:

please can anyone show me how to calculate the base resistor for an NPN transistor used as a switch. the supply voltage is 12 volts, the load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

12v 300ohms is 40mA.
Assume beta of at least 30, so we need 40/30mA into the base = 1.3mA
V across base R is 12v - 0.7v = 11.3v
R=V/I so R = 11.3/1.3m = 8.5k. Preferred values below that to give a bit more i are 8k2 or 6k8 or 4k7.


NT

On Sunday, October 8, 2017 at 6:53:50 PM UTC, wrote:
please can anyone show me how to calculate the base resistor for an NPN transistor used as a switch. the supply voltage is 12 volts, the load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

Thanks all your suggestion are well taken

On 09/10/17 07:33, mike wrote:
On 10/8/2017 1:12 PM, natp wrote:
" wrote:

please can anyone show me how to calculate the base resistor for an
NPN transistor used as a switch. the supply voltage is 12 volts, the
load is a 12 volts relay with a resistance of 300 Ohms.
thank you.

When the NPN transistor is saturated (fully on), the emitter-collector
voltage will be about 0.2-0.3 volts. Using the other values you have,
you can calculate Ic; the current flowing through the collector
(and the relay).

To saturate the transistor, the base current must be greater than
Â*Â*Â*Â*Ic / beta
where "beta" is the minimum current gain of the transistor.

The last piece of information you need is that the base-emitter
voltage is one "diode-drop"; about 0.7 volts.

It's customary to overdrive the base.Â* I normally use a factor of
10.

Beta is reduced near saturation. 10x is an attempt to compensate.