1
\$\begingroup\$

I'm in the process of designing this small circuit using an SS8050 NPN transistor.

I have an input base voltage of 1.28V and I'd like it to be working as a switch to turn on a little buzzer.

According to the datasheet, the base-emitter on voltage is 0,7 V ish when the collector current is 30 mA, so I picked a 4000 ohm base-resistance to get the right voltage drop.

Unfortunately, I think the circuit doesn't work well and I hope you help me to sort it right.

Can you help me to check the calculations? Is the base-emitter-on voltage the voltage I must apply to turn on the transistor and allow that Ic current? What's the meaning of VCe=1V on the diagram I mentioned before?

  • I2= I1/hfe= 30 10-3/200=0,15 mA
  • ΔV= 1,28 V -0,68V=0,6 V
  • ΔV= R2*I2 ---> R2=4000 Ω

Datasheet

enter image description here

enter image description here

enter image description here

\$\endgroup\$
7
  • 3
    \$\begingroup\$ You've drawn the transistor upside down, with the emitter labelled "C". How is it actually connected? \$\endgroup\$
    – John Doty
    Jan 14, 2023 at 17:56
  • 1
    \$\begingroup\$ To use it as a switch, drive the base with more current. A 1:10 ratio is often used, so use 3 mA, not 0.15. \$\endgroup\$
    – John Doty
    Jan 14, 2023 at 17:59
  • 1
    \$\begingroup\$ You've drawn a PNP transistor, but the SS8050 is an NPN transistor. \$\endgroup\$
    – Hearth
    Jan 14, 2023 at 18:00
  • 2
    \$\begingroup\$ Your schematic wrongly shows a PNP transistor with its E and C terminals and voltages like it is an NPN transistor. An NPN transistor should turn on when the base current is 1/10th the collector current because hFE is used only when the transistor is an amplifier with plenty of collector voltage and here it is not an amplifier, it is a switch. \$\endgroup\$
    – Audioguru
    Jan 14, 2023 at 18:04
  • \$\begingroup\$ Thank you very much for yours detailed answers. You are right, the transistor is an NPN. I'm really sorry for the misanderstanding. \$\endgroup\$
    – Andico
    Jan 14, 2023 at 18:24

2 Answers 2

2
\$\begingroup\$

Assuming an NTC thermistor in an unbuffered resistor divider, the circuit is more like:

schematic

simulate this circuit – Schematic created using CircuitLab

(Using a PTC thermistor would swap the variable resistor and \$R_1\$.)

The above circuit would cause the voltage at the divider node to climb as the temperature increased.

Looking at a Murata datasheet:

enter image description here

The resistance at \$100^\circ\:\text{C}\$ (circled in blue) is about \$\frac1{10}\text{th}\$ of its nominal value at room temperature (circled in green.)

At nominal temperature, the NTC thermistors from Murata come in three ranges:

\$\quad\quad\quad\$enter image description here

If the thermistor is a common \$10\:\text{k}\Omega\$ variety then its resistance at \$100^\circ\:\text{C}\$ will be about \$1\:\text{k}\Omega\$ at the point where the buzzer should turn on. Assuming that \$+5\:\text{V}\$ also supplies the thermistor+resistor divider, I'd guess \$R_1=2.7\:\text{k}\Omega\$. This suggests a Thevenin impedance at the trigger point of about \$700\:\Omega\$.

Even if \$R_2=0\:\Omega\$, the design isn't well-managed. Setting \$\beta=10\$ and therefore \$I_{_\text{B}}=3\:\text{mA}\$ means that the managed plan is for a \$2.1\:\text{V}\$ drop across the thevenin resistance relative to its unloaded value. And that just won't work.

Worse, there's no designed hysteresis. Instead, a gradually increasing collector current occurs as the thermistor resistance gradually changes with temperature.

Is an unbuffered resistor divider arrangement being used?

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

I think I spotted my mistakes and I want to share them with you.

  1. The transistor is actually a BJT NPN
  2. I didn't know that "Ib is 1/10th the collector current because hFE is used only when the transistor is an amplifier with plenty of collector voltage and here it is not an amplifier, it is a switch" so I used the wrong plot from the datasheet.
  3. Taking Ib= Ic/10 allows the transistor to be working in saturation (switch fully on). That's because the current base is higher than the one you would obtain referring to the plot for hfe on the datasheet. While working in saturation, the Vce is smaller for the same Ib. That entails less power dissipation
  4. The correct circut should be the one attached below:

enter image description here

  1. I think the purpose of this plot is giving you the voltage drop between base and emittor dependign on Ic, isn't it?

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ You did not notice that the graph of base-emitter voltage is when it is an amplifier with 1V collector to emitter. The graph for SATURATED base-emitter ON voltage has a voltage a little higher than when it is a saturated switch because the base-collector is forward biased and drawing a little current. \$\endgroup\$
    – Audioguru
    Jan 14, 2023 at 23:09
  • 1
    \$\begingroup\$ The thermistor and circuit are not a switch. The circuit will audibly make a noise in when the temperature is a normal room temperature and get louder as the temperature rises. \$\endgroup\$
    – Audioguru
    Jan 15, 2023 at 0:46

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.