How to amplify a 5V square wave to 28V square wave???

Thread Starter

Cole2014

Joined Jun 26, 2014
5
Hi ALL,:)
I am a new in electronic.

Do anyone know how to amplify a 5V square wave to 28V square wave???
Should I use an amplifier or transformer? Or other IC?? Please recommend.

Please kindly help!!
Many many thanks!!
Cole
 

GopherT

Joined Nov 23, 2012
8,009
It might help to have a description of what/how it will be used (28volt side), a schematic of what you are connecting to And a nice schematic for the 5 volt source.

Some info on frequency is also helpful and spec for your definition of just how square it must be.
 

wayneh

Joined Sep 9, 2010
18,133
You're assuming the OP has a 28V power supply, which we don't know. Until he answers Gopher's questions, there's nothing more to say that isn't guessing. Maybe Scott has guessed right - we'll see.
 

Alberto

Joined Nov 7, 2008
169
Scottwang, please check the value of R4 & R5, it seems that PNP will never conduct with the values given in the schematic you posted.

Cheers

Alberto
 

crutschow

Joined Mar 14, 2008
38,624
................... please check the value of R4 & R5, it seems that PNP will never conduct with the values given in the schematic you posted.
Sure it will. With Q1 saturated there will be about 1mA going through R4 and about 0.7V / 2.7K = .26mA going through R5. That leaves .74mA coming from the base of Q2. That's way more than enough to saturate Q2 with its half mA collector current.
 

ian field

Joined Oct 27, 2012
6,536
Hi ALL,:)
I am a new in electronic.

Do anyone know how to amplify a 5V square wave to 28V square wave???
Should I use an amplifier or transformer? Or other IC?? Please recommend.

Please kindly help!!
Many many thanks!!
Cole
A similar question came up recently - you can use a c_a_s_c_o_d_e (have to do it like that or the spellcheck sneakily changes it to cascade!).

Basically you add a common gate MOSFET with gate to +5V and source to the 5V output, It has current gain of about unity, so if you also need more current as well, add an emitter follower stage.

By now, it starts to look vaguely like the video O/P stage in a CRT monitor - you can use a complementary pair of emitter followers - if crossover distortion might be a problem, put the 2 bias diodes between the CE bases in series between the common base transistor's drain and its drain load resistor.
 

Thread Starter

Cole2014

Joined Jun 26, 2014
5
It just needs a voltage level conversion.
What's the load of 28V square wave and the current of load?
What's the frequency of 5V square wave?

The below is a 5V to 28V voltage level conversion..
5V frequency is 30Hz. Can we not change the frequency? I mean the input frequency = output frequency??

Many thanks!
 

Thread Starter

Cole2014

Joined Jun 26, 2014
5
It might help to have a description of what/how it will be used (28volt side), a schematic of what you are connecting to And a nice schematic for the 5 volt source.

Some info on frequency is also helpful and spec for your definition of just how square it must be.
I got 28V DC supply on the other side. I dont want to change the input frequency 30Hz. Is it possible?:)
 

ronv

Joined Nov 12, 2008
3,770
The circuit posted will do what you want, but it cannot sink much current. How much current does you 28 volt circuit take?
 

GopherT

Joined Nov 23, 2012
8,009
I got 28V DC supply on the other side. I dont want to change the input frequency 30Hz. Is it possible?:)
Yes, it is possible - to a limit. Flip-flops can easily cut frequency by half. Frequency doubling circuits also exist. Using counters you can get many, many options - (you may have to run back through a flip-flop to re square you re-square your wave.
 

ScottWang

Joined Aug 23, 2012
7,504
5V frequency is 30Hz. Can we not change the frequency? I mean the input frequency = output frequency??

Many thanks!
Please go back to see the new posted circuit on #8, the output frequency will be the same with the input frequency.

The problem is that how much output current do you need?
 

anhnha

Joined Apr 19, 2012
904



With 5V input voltage and assuming that transistor Q1 2N3904 has Vbe = 0.6V, the current flowing through resistor 1K is 0.6mA and through 4.7K is 1mA. So the base current of Q1 is 1mA - 0.6mA = 0.4mA.
Could you tell me how can I find the beta of Q1 from its datasheet corresponding to that base current?
The datasheet shows DC current gain with different Ic but here I don't know Ic, how can I do that???
 
Last edited:

ScottWang

Joined Aug 23, 2012
7,504


With 5V input voltage and assuming that transistor Q1 2N3904 has Vbe = 0.6V, the current flowing through resistor 1K is 0.6mA and through 4.7K is 1mA. So the base current of Q1 is 1mA - 0.6mA = 0.4mA.
Could you tell me how can I find the beta of Q1 from its datasheet corresponding to that base current?
The datasheet shows DC current gain with different Ic but here I don't know Ic, how can I do that???
There are two ways to solve the problem as what you said about the Ib and Ic, the one is to take the 1k Rbe away, and reducing the values of Rb from 4.7k to 4.3k, that is to match the official info of datasheet, when Ib=1mA then the Ic will be equal to 10mA and the Vce will get into the saturation region.

The another way is to reducing the Rb from 4.7k to 2.4k, how did the 2.4k came out, that is :
Rb=(Vin-Vbe1)/(Ib+(Vbe1/Rbe1))
=(5v-0.7v)/(1mA+(0.7v/1k))
=4.3v/0.0017mA
=2.529k, it can use 2.5k or 2.4k.

The official description that the Ib=1mA, Ic=10mA, hfe=10, the Vce will be in saturation, but i like to use it according to my experiment to adjust the values.

The Rb and Rbe will be different, when the Vin and frequency are changing.
 
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