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schematic

simulate this circuit – Schematic created using CircuitLab

I am building a so called protection section for my mini project which involves a 3.9V zener diode parallel to a battery as a shunt voltage regulator.

Assuming the voltage at the 1N4002 diode is 4.1 V and that the battery has reach 3.9 V to cause the zener diode to conduct, will the voltage at base of the transistor be 0.2 V?

The transistor I intend to use is a BC549C low noise transistor instead of the general 2N2222.

Assume that the circuit is current limiting with a 20 mA into the battery, will this be enough to turn on the base of the transistor (without a sufficient 0.7 V across Vbe to turn it on)?

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closed as unclear what you're asking by Marcus Müller, brhans, Daniel Grillo, ThreePhaseEel, jonk Nov 30 '16 at 23:51

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Please add a scematic. There is a built in function for this on this site when you edit your question. \$\endgroup\$ – winny Nov 30 '16 at 18:17
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    \$\begingroup\$ Voltage flowing into the base? Current flows into the base of an NPN, voltage is a potential with respect to ground or some other specified point. Please add a schematic and be a little more precise with the wording of your question. \$\endgroup\$ – John D Nov 30 '16 at 18:30
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    \$\begingroup\$ It's been nearly half an hour without any attempt by OP to clarify. Voting to close as unclear. \$\endgroup\$ – Marcus Müller Nov 30 '16 at 18:43
  • \$\begingroup\$ Just look up "Shunt Voltage Regulator with Zener Diode" and plenty of results come up. Make your own calculations from there and we will help you if you have any other issues. But we also need a schematic of your circuit. \$\endgroup\$ – 12Lappie Nov 30 '16 at 18:45
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It appears that you want the LED to be on when the battery voltage is above the limit set by the zener.

Unfortunately, the way you have it will not work as you expect.

You are using the transistor as a high side switch. To make it conduct, you must make the voltage on the base higher than the sum of the Vbe and the LED forward voltage. Then there's the 3.9V for the zener. So, 1.5V for the LED plus 0.7V for Vbe and 3.9V for the zener means the LED will only light when the battery voltage is above 6.1V.

The usual way would be to use the transistor as a low side switch (between the LED and ground.). Then the battery only has to reach 0.7V plus the zener voltage to make the LED light.

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  • \$\begingroup\$ What I read from similar circuits of this design is that they just say the zener conducts and the remaining current will stop flowing into the battery but into the transistor to ground (They did not specify why, been trying to figure this out). Does this mean if I were to use it as a low side switch, it will not function as intended due to the total 0.7 Vbe + 3.9 V zener while the battery only supplies the 3.9 V? \$\endgroup\$ – Jack Daniels Nov 30 '16 at 19:55
  • \$\begingroup\$ A diode conducts when the voltage on its anode is higher than the cathode. For a normal diode, there has to be a difference of about 0.7 volts. The base-emitter junction of a transistor is like a normal diode. For LEDs, it varies by color, with red LEDs being somewhere a bit below 1.5V. Zeners work differently. The forward voltage is about like a regular diode, but they have a specific, higher voltage at which they conduct when connected backwards. \$\endgroup\$ – JRE Nov 30 '16 at 20:01
  • \$\begingroup\$ This means that the battery/zener protection part will not work unless I were to increase the voltage at the 1N4002 diode above the extra +0.7 Vbe needed to cause a forward bias in the diode. Rather troublesome for the strict 4.2 V limit to charge the lithium ion battery. Need to rethink my design. \$\endgroup\$ – Jack Daniels Nov 30 '16 at 20:12
  • \$\begingroup\$ Urk. Lithium battery. You may want to reconsider building your own charger, especially using such a primitive regulator. Lithium battteries tend to express themselves violently if charged improperly. Does the expression "venting with flames" mean anything to you? The recent excitement in the news about exploding smart phones was related to improper charging of lithium batteries. \$\endgroup\$ – JRE Nov 30 '16 at 20:22
  • \$\begingroup\$ Tried to charge it up to 3.9 V, that was why I was thinking of utilising the zener diode to bypass the current as a 'safety threshold'. But the design does not follow through. \$\endgroup\$ – Jack Daniels Nov 30 '16 at 20:28

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