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Increase the amperage fo PNP transistor

koklimabc

Sep 7, 2013
20
I'm current using 2N2907 type of PNP but the limitation of "collector pin" only able to produce 50 mAh of current, Emitter pin take the 500 mAh as current for input.

My Question:
How can i be able to produce in around 250 to 350 mAh at collector pin in order to drive my 6 volt small motor. Appreciate Thanks to any of help.

Bluejets

Oct 5, 2014
6,862
Show a circuit diagram of what you have there.

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,648
limitation of "collector pin" only able to produce 50 mAh of current,
Where do you take this data from? The datasheet states a max. collector current of 600 mA:

How can i be able to produce in around 250 to 350 mAh at collector pin in order to drive my 6 volt small motor.
Have a look at our resources (1 and 2).

koklimabc

Sep 7, 2013
20
Where do you take this data from? The datasheet states a max. collector current of 600 mA:
View attachment 54278

Have a look at our resources (1 and 2).
I'd measure the circuit current from collector leg where it got 50 mAh only.
ok, I'll go through from your resource as given by you. Thanks a million.

Tha fios agaibh

Aug 11, 2014
2,251
X2 On "Show a circuit diagram"

You may not have enough current driving the base to fully turn it on.

crutschow

May 7, 2021
834
For 350mA (not mAh which is battery capacity) through the collector, the base-emitter current should be at least 35mA.
If your circuit cannot provide that much base current, then you will need to add another transistor (for example to make a Darlington stage).

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,648
I'd measure the circuit current from collector leg where it got 50 mAh only.
Show a circuit diagram of what you have there.
Then it is obviously a problem of your circuit, not the transistor.

koklimabc

Sep 7, 2013
20
X2 On "Show a circuit diagram"

You may not have enough current driving the base to fully turn it on.
yes correct, my base was measured in around 6 mA only

[mod edit ... mA not mAh ]

Last edited by a moderator:

koklimabc

Sep 7, 2013
20
For 350mA (not mAh which is battery capacity) through the collector, the base-emitter current should be at least 35mA.
If your circuit cannot provide that much base current, then you will need to add another transistor (for example to make a Darlington stage).
ok, I'll refers from your suggestion. TQ so much.

Sep 7, 2013
20

Attachments

• circuit.jpg
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Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,648
my base was measured in around 6 mAh only
mAh is a measurement of capacity, as @crutschow mentioned. Your measurement surely is 6 mA. It is a bit on the low side, I would have expected something around 8 mA (given the 1 kΩ base resistor and the Vbe of the transistor), but 6 mA is not unreasonable. A 2N2907 has a minimum DC current gain of 50 which allows for up to 300mA collector current from 6 mA base current.
How much current actually flows depends on the load. What are the specs of the motor and the LEDs?

koklimabc

Sep 7, 2013
20
mAh is a measurement of capacity, as @crutschow mentioned. Your measurement surely is 6 mA. It is a bit on the low side, I would have expected something around 8 mA (given the 1 kΩ base resistor and the Vbe of the transistor), but 6 mA is not unreasonable. A 2N2907 has a minimum DC current gain of 50 which allows for up to 300mA collector current from 6 mA base current.
How much current actually flows depends on the load. What are the specs of the motor and the LEDs?
FYI, I'm using 6 - 9 volt volt 200 rpm DC motor & 5mm LED superbright.

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,648
These are not specs.
What current does the motor draw at 6 V, at 9V? Or what power ( current is power / voltage).
What is the voltage drop of the LEDs? The color of the LEDs would at least give a coarse indication. LED voltage can vary from 1.5 V to 4 V or more.

crutschow

May 7, 2021
834
A 2N2907 has a minimum DC current gain of 50 which allows for up to 300mA collector current from 6 mA base current.
But that's not with the transistor in saturation.
BJT's are typically specified for saturation with a base current 1/10th of the collector current.

koklimabc

Sep 7, 2013
20
These are not specs.
What current does the motor draw at 6 V, at 9V? Or what power ( current is power / voltage).
What is the voltage drop of the LEDs? The color of the LEDs would at least give a coarse indication. LED voltage can vary from 1.5 V to 4 V or more.
FYI, the current draw from motor is around 200 mA and the voltage drop from series of these three LED is around 0.69 volt.

koklimabc

Sep 7, 2013
20
For 350mA (not mAh which is battery capacity) through the collector, the base-emitter current should be at least 35mA.
If your circuit cannot provide that much base current, then you will need to add another transistor (for example to make a Darlington stage).
I'd currently succeeded built to a Darlington circuit with three series of PNP transistor => produce 280 mA and able to drive the motor (comsumed 200mA and 3.8 volt) but unable to power on to my three LED built in series (consumed 6 volt and 60 mA) under parallel circuit.

Can I use any boost converter (adjustable 5v to 24volt , 2 Amp) to fix this issues in order to let my LED to light on and motor to work out at the same time?

Harald Kapp

Moderator
Moderator
Nov 17, 2011
13,648
the voltage drop from series of these three LED is around 0.69 volt.
That is not plausible. A red LED drops ~ 1.6 V, other colors have higher voltage drops, up to ~ 4 V for blue ones.
Assuming red LEDs, the total voltage drop would be ~ 4.8 V. Add 0.2 V for the transistor's C-E drop then the voltage across the 100 Ω series resistor to the LEDs will be around 4 V (with a good 9 V power source).The current through the LEDs will then be around 5 V / 100 Ω = 50 mA.
With a motor current of 200 mA the total is 250 mA. The transistor should be able to drive this current.

What is the 9 V power source? A battery block? Check the voltage of the battery when the motor is on. I suspect your 9 V aren't 9 V but much less due to the battery not being able to deliver the current required.

koklimabc

Sep 7, 2013
20
correct, I'd wrongly assume that 0.69 is the result deducted from main source.

FYI, I'd need to cascade with three pnp transistor to form as darlington stage to be able to produce the 340 mA (as ready measured) before it could be able to drive the motor.

Yes, you're right, It is 8 volt lesser than 9 volt.

crutschow

May 7, 2021
834
I'd need to cascade with three pnp transistor to form as darlington stage
You only need two transistor to form a Darlington, why are you using three?

Moderator
Nov 8, 2019
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