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This question came to me while connecting a lower voltage battery to a higher voltage battery by connecting their grounds using an NPN transistor switch.

I know it works, but I don't know how to calculate the limits.

Why isn't the lower battery destroyed from the higher voltage and current and how can we know the limits ?

(I am using "lower voltage" and "higher voltage" instead of specific values because I want the answer to be a general rule.)

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Are you just connecting the grounds? What's going on at the other end of the battery? \$\endgroup\$ – pjc50 Apr 7 at 10:49
  • \$\begingroup\$ There is a tool on this site to draw circuits. Can you show us what you mean? \$\endgroup\$ – Warren Hill Apr 7 at 10:49
  • \$\begingroup\$ Are you only connecting the negative terminals of each battery and leaving the positive terminals unconnected? \$\endgroup\$ – Andy aka Apr 7 at 10:49
  • \$\begingroup\$ @Andyaka I edited in a sample circuit \$\endgroup\$ – soundslikefiziks Apr 7 at 10:55
  • \$\begingroup\$ @WarrenHill I edited in a sample circuit \$\endgroup\$ – soundslikefiziks Apr 7 at 10:55
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enter image description here

Figure 1. There are two current-loops in the circuit: control and power.

Notice that provided the transistor's ratings are not exceeded that no current flows from the power circuit back into the control circuit.

The control circuit turns on the transistor and allows current to flow from V2 through the R3 - Q1 loop.

In this circuit the base of Q1 will be held at 0.7 V no matter what voltage V1 is (provided it's above a volt or two) and independent of V2.

... because of the high electric fields from the 200v or higher, seemed they could affect a small battery just like it can affect a small component , and i thought that just like any other smaller components, batteries would also have some limit as to which voltage they can be connected to in this configuration.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. A transistor consists of two P-N diode junctions.

From this it should be a bit clearer that V2's potential is dropped across D1 so V1 isn't affected by V2. It's just like connecting two different voltage batteries with a diode in reverse-biased configuration.

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  • \$\begingroup\$ so the only limitations come from the circuit components, resistors , transistors,diodes, wires, etc, and in theory even a 200V battery can be connected with a 1.5 battery as the switch voltage ? \$\endgroup\$ – soundslikefiziks Apr 7 at 11:28
  • \$\begingroup\$ @soundslikefiziks Yes, we can use 200V and 1.5V. But why this configuration worries you? \$\endgroup\$ – G36 Apr 7 at 11:49
  • \$\begingroup\$ @soundslikefiziks: That's correct. There will be practical considerations in building a 200 V device such as ensuring adequate separation between the pads on the transistor as well as the transistor's rating (mentioned in the answer). I didn't mention it in the answer but the emitter of Q1 will have the combined collector current and base current but that doesn't affect the answer. \$\endgroup\$ – Transistor Apr 7 at 11:50
  • \$\begingroup\$ @G36 because of the high electric fields from the 200v or higher, seemed they could affect a small battery just like it can affect a small component , and i thought that just like any other smaller components, batteries would also have some limit as to which voltage they can be connected to in this configuration \$\endgroup\$ – soundslikefiziks Apr 7 at 11:52
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    \$\begingroup\$ @soundslikefiziks yes. In the case of the transistor, the number you're looking for is the "base-collector breakdown voltage", usually listed as "V(br)CBO" on datasheets. \$\endgroup\$ – pjc50 Apr 7 at 12:07

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