I have a S8050 with an Ib/Ic current gain of about 100. The diagram for the current gain in the datasheet shows Ic being 330mA at Ib 2mA for this transistor. In my circuit I drive a 3.3-5V motor with the transistor, both are fed from the same 4V source. The motor takes about 65mA when directly connected to a 4V battery pack. I measured the following values:

Provided current for base (with 2.2kOhm resistor): 1.8mA, measured current flowing to base: 0.2 mA, Ie = ~50mA

Provided current for base (with 1.0kOhm resistor): 4mA, measured current flowing to base: 0.22 mA, Ie = ~56mA

While I see a relation between the measured current flowing through Ib and Ie, I don't see it with the provided current for the base. In my understanding the motor should receive the full 65mA from the emitter as soon as I provide a minimum of around 0.5 mA to the base, which would lead to around 80mA for Ie (at least that's what the datasheet claims). But instead the measured current flow into Ib seems to stay static at around 0.2mA and Ie is smaller than expected (but still rising on higher input current values for Ib). Can anybody explain, why? Does transistors have a non-linear resistance on the base?

  • 1
    \$\begingroup\$ 1) I measured the following values Show the schematic of how you measured that. 2) beta = Ic/Ib is never an accurate value, the manufacturers cannot control beta very well. 3) You will only be able to measure the beta if the transistor is in linear mode meaning Vce is high enough (about 1 V should do). 4) it should be possible for the expected Ic or Ie to flow, maybe the way you're using the transistor prevents a higher current to flow. \$\endgroup\$ Commented Mar 12, 2018 at 13:10
  • 2
    \$\begingroup\$ Show your circuit. There is some ambiguity in your words. \$\endgroup\$
    – Andy aka
    Commented Mar 12, 2018 at 13:10
  • \$\begingroup\$ The non-linearity is in the Ib/Ic current ratio when Vce is less than a couple of volts (i.e. when most of the supply volts are across the motor). This effect is called saturation. \$\endgroup\$
    – user16324
    Commented Mar 12, 2018 at 14:25

3 Answers 3


You are doing this:


simulate this circuit – Schematic created using CircuitLab

You should be doing this:


simulate this circuit

  • \$\begingroup\$ Ouch, I indeed put the motor on the emitter side and wasn't aware of any difference by doing so. I will try out at home and report back. \$\endgroup\$ Commented Mar 12, 2018 at 17:34
  • \$\begingroup\$ @needfulthing the difference is that in the first one (called common collector) the voltage across the resistor is Vce-Vbe. In the second one (called common emitter), the voltage across the resistor is the battery voltage minus vbe. \$\endgroup\$
    – τεκ
    Commented Mar 12, 2018 at 19:12
  • \$\begingroup\$ First tries with placing the motor on the collector side leads to measurements very close to the datasheet values. So i.e. I can use a 4.8kOhm resistor now and still get sufficient current for the motor. But just as I told Tony below, I really have to learn more about voltage behaviors around a transistor :p \$\endgroup\$ Commented Mar 13, 2018 at 10:57

Your current gain will be about 10 when you try and drive your transistor into decent saturation. Look at figure 4 in the data sheet: -

enter image description here

I've added a vertical red line at a collector current of 60 mA and this corresponds to a VCE saturation voltage of 200 mV i.e. your BJT does not short out to zero ohms when you drive it hard! There is an effective ohmic value of about 3 ohms (200 mV/60 mA) and this is with a base current of about 6 mA.

When you drive a BJT into the collector-emitter saturation region, the hFE of the device drops big-time.

Provided current for base (with 2.2kOhm resistor): 1.8mA, measured current flowing to base: 0.2 mA, Ie = ~50mA

This informs me that you have your motor load connected to the emitter. This form of amplifier is called an emitter follower and it's not ideal for turning a motor on because the base voltage must be about 0.8 volts higher than the emitter voltage to turn the transistor on into saturation.

  • \$\begingroup\$ You are right. Please don't mind that I chose τεκ's answer for guessing my circuit and providing a graphical solution. \$\endgroup\$ Commented Mar 13, 2018 at 10:52

In addition to what Andy said ....

As an emitter follower at low current you lose ~0.6V out of 4V 15%, so it runs at lower RPM and current is less. But hFE was 50/0.2=250 hFE has a very wide range with load current then saturates by reducing Hfe to 10% of max but Vce(sat) is certified at Ic/Ib=10

When used as grounded NPN emitter with motor between collector and + to Vcc with a pullup to Vcc, the base current will now be (4V-0.7)/Rb=Ib and you can expect a no load motor to be 100x this when stalled or just starting so you will need to measure DCR of coil to compute this exactly . Vcc/DCR=Isurge(t=0)=10*Imax at rated RPM full load.

At high currents Saturation or the rapid reduction of hFE begins at ~ Vce=2V and reduces down to 10% of HFE at the rated Vce(sat) @ I

Once you understand this, you may opt for low Rce transistors from Diodes Inc, etc or low RdsOn MOSFETs using Vce(sat)/Isat as your indicator of voltage drop effective switch ON resistance.

  • \$\begingroup\$ lamers -2 , with no comments \$\endgroup\$ Commented Mar 13, 2018 at 2:30
  • \$\begingroup\$ Not from me, thanks for your effort. The various answers here made clear that I have to learn a lot more about transistors besides the simple put current on base and get some more current on CE :) \$\endgroup\$ Commented Mar 13, 2018 at 10:48
  • \$\begingroup\$ You must show your schematic in future. \$\endgroup\$ Commented Mar 13, 2018 at 15:20
  • \$\begingroup\$ Will do, will do. \$\endgroup\$ Commented Mar 14, 2018 at 16:57

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.