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I have a very basic inverter circuit with a 2n3053. I basically need around 5V at the collector terminal when the base is at 0V. I am able to achieve this provided the load is not connected. When the load is connected at the collector, the 5V drops to around 0.6 V or so (depending on the pull up resistor). After going through the load's data sheet I realized that there is an input impedance of 1.5K ohm. I could reduce the pull up resistor value to 100 ohms, but don't want to do so because of the high current that will flow through it.

I don't know much about impedance matching but I think that's the way to sort this problem out. Could someone help me? Can I make an emitter follower to match the impedance?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Switch to PNP and use the load itself as the pull-down resistor? \$\endgroup\$ – The Photon Feb 2 '15 at 19:18
  • \$\begingroup\$ Could it be that you need to get rid of the collector resistor, connect your load's CTRL input directly to the collector, and the 5V Reference to +5V? What is your load, BTW? \$\endgroup\$ – EM Fields Feb 2 '15 at 19:21
  • \$\begingroup\$ I guess I could do that just remove the resistor altogether. My load is an LED driver. I'll probably just remove the resistor altogther. Thanks for the suggestion :) \$\endgroup\$ – acexa616 Feb 2 '15 at 19:51
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I you need rail-to-rail output, you should use a class D output (replace R1 by a PNP - this may need adaptation when you attack the base to avoid excessive current - or better replace both bipolar by an NMOS and a PMOS).

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