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

Discussion in 'General Electronics Chat' started by Cole2014, Jun 26, 2014.

1. ### Cole2014 Thread Starter New Member

Jun 26, 2014
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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.

Many many thanks!!
Cole

2. ### GopherT AAC Fanatic!

Nov 23, 2012
6,303
4,026
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.

3. ### ScottWang Moderator

Aug 23, 2012
4,925
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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..

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4. ### wayneh Expert

Sep 9, 2010
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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.

5. ### Brownout Well-Known Member

Jan 10, 2012
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He better have a 28V supply if he wants to amplify to that level

6. ### Ramussons Active Member

May 3, 2013
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Hmmmmmm... a Little more than 28 V ? Say 30 V ?

Ramesh

7. ### Alberto Active Member

Nov 7, 2008
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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

8. ### ScottWang Moderator

Aug 23, 2012
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Thank you for your mentioned, but the problem is not there, I had attached a new circuit.
Because the npn stage was copied from my another circuit, and i forgot to recheck it.

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9. ### crutschow Expert

Mar 14, 2008
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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.

10. ### Alec_t AAC Fanatic!

Sep 17, 2013
5,968
1,135
..... but if the OP doesn't have a 28V supply already then a switch-mode DC-DC converter is called for.

11. ### wayneh Expert

Sep 9, 2010
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Or a transformer. We're still guessing. Although he did say square wave.

12. ### ian field Distinguished Member

Oct 27, 2012
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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.

13. ### Cole2014 Thread Starter New Member

Jun 26, 2014
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5V frequency is 30Hz. Can we not change the frequency? I mean the input frequency = output frequency??

Many thanks!

14. ### Cole2014 Thread Starter New Member

Jun 26, 2014
5
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I got 28V DC supply on the other side. I dont want to change the input frequency 30Hz. Is it possible?

15. ### ronv AAC Fanatic!

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

16. ### GopherT AAC Fanatic!

Nov 23, 2012
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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.

17. ### ScottWang Moderator

Aug 23, 2012
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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?

18. ### anhnha Active Member

Apr 19, 2012
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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: Jun 30, 2014
19. ### anhnha Active Member

Apr 19, 2012
776
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Well, now I see that the transistor is in saturation region. I will try to calculate the output voltage now!

20. ### ScottWang Moderator

Aug 23, 2012
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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.

anhnha likes this.