Saturated NPN base to emitter voltage drop?

Thread Starter

morphy_richards

Joined Jan 3, 2013
1
Hi,
I'm checking for possible problem with a simple H bridge motor control circuit I'm designing using only NPN transistors. The motor supply is 4.5 V and the base will be 'switched' using a 5 V supply.

I understand that as long as the transistor is in saturation there is no appreciable voltage drop from the collector to the emitter but what about from the base to the emitter?

Something I have read elsewhere suggests there is a 0.7 v drop here which would put the base voltage at 4.3 V and below the value of the (4.5 v) emitter voltage which means the transistor wont work as expected? Thanks for any help and advice.
 

GopherT

Joined Nov 23, 2012
8,009
In general, you are right. Emitter will be 0.5 to 0.7 volts below base voltage (it can be more). If you have a darlington NPN, it will be double. Check your DATASHEET.
 
A schematic would definitely be helpful.

If you tie the emitter of your drive transistor, npn, to 4.5 volts, the base of the transistor would have to reach approximately 5.2 volts in order to turn the transistor ON. The 5.2 volts is above both your drive and logic supplies. It requires additional "high side drive" circuitry to use only npn transistors.
 

tinamishra

Joined Dec 1, 2012
39
A bipolar junction transistor could be a design of transistor that relies on the get in touch with of 2 types of semiconductor for its function. BJTs are often used as built-in amps already built in, changes, or in oscillators. BJTs are often discovered either as personal individual areas, or in large figures as components of incorporated tour.

Bipolar transistors place system thus known as due to their function includes each electrons and gaps. These 2 types of cost providers place system attribute of the 2 types of doped semiconductor content. In difference, unipolar transistors like the field-effect transistors have just one quite cost service provider.
 
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