Author Topic: the art of electronics : switching the high side of a load  (Read 261 times)

0 Members and 1 Guest are viewing this topic.

Offline Camelkos

  • Newbie
  • Posts: 4
  • Country: fr
the art of electronics : switching the high side of a load
« on: June 01, 2020, 04:26:54 pm »
hi everybody,
i'm reading the art of electronics 2 and i'm stuck at the circuit B to switch the high side of a load returned to ground in the chapter of bipolar transistors,
can someone explain it to me please, thank you all !!
 

Offline bob91343

  • Frequent Contributor
  • **
  • Posts: 759
  • Country: us
Re: the art of electronics : switching the high side of a load
« Reply #1 on: June 01, 2020, 08:59:44 pm »
We don't all have that book so give us the source and perhaps we can help.
 

Offline MarkF

  • Super Contributor
  • ***
  • Posts: 1660
  • Country: us
Re: the art of electronics : switching the high side of a load
« Reply #2 on: June 01, 2020, 09:25:48 pm »
Is this the circuit?

999164-0

 

Offline bob91343

  • Frequent Contributor
  • **
  • Posts: 759
  • Country: us
Re: the art of electronics : switching the high side of a load
« Reply #3 on: June 02, 2020, 04:06:56 pm »
If B is the circuit, then driving Q2 with a pulse will turn it on, which in turn provides base current for Q3 and thus energizing the load.
 

Offline MarkF

  • Super Contributor
  • ***
  • Posts: 1660
  • Country: us
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf