Diode transistor logic

[kishore680]

The diode transistor logic uses negative power supply in order to remove base charge from the transistor. IF we want to use only positive power supply and not negative power supply ,one solution i heard is to connect two diodes in place of R3. and R4 and V- will be eliminated in that design. how? And other solutions also are welcome

 

[hakeen]

Hi Kishore680,

yes it is possible, but the turn off speed of transistor will be much lower. A simple circuit is shown in below link. R4 should stay there, because it supplies Vbe and discharging path for different timings (i.e. ON/OFF). Juts R3 is rplaced by diodes. It helps to adjust different threshold levels. Otherwise would NPN always stay ON.

https://www.google.de/imgres?imgurl...HnT8yBH-jP4QTasIG9AQ&ved=0CGQQ9QEwBQ&dur=1295

Cheers

 

[erikl]

Here's a DTL circuit schematic from Siemens LSL ("slow, fail-safe logic") from 1979. Intended for 12..15V operation, however: View attachment Siemens_1979-80_Digitale_Schaltungen_p69.pdf

 

[kishore680]

The circuit with negative power supply has lesser propagation delay compared to other i.e without negative power supply right? but the solution of two diodes in place of R3 just addresses the problem of NPN (on/off) ?