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Please help transistor amplifier

by michael1978
Tags: amplifier, transistor
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michael1978
#181
Dec18-12, 03:03 AM
P: 133
Quote Quote by Jony130 View Post
I don't know why. Maybe you made a mistake in simulation program.

yes Jony, i do it one more time and is now 0,68 Vac now smaller, of i have to change the emitter capacitor , because the value of capacitor i make 1000u, of transistor , wich transistor you think to use do you know some type?
because input is 2mv*50gain=100MV
thnx for reply
NascentOxygen
#182
Dec19-12, 04:54 AM
HW Helper
Thanks
P: 5,149
The BC109C is a popular high gain general purpose transistor.
michael1978
#183
Dec19-12, 04:56 AM
P: 133
Quote Quote by NascentOxygen View Post
The BC109C is a popular high gain general purpose transistor.
i will change now

i dont have that type of transistor
michael1978
#184
Dec19-12, 05:53 AM
P: 133
look my circuit, i dont know where i make mistake, normal gain have to be 50*2Mv=100mV Peak
Attached Thumbnails
AMPLIFIER 1.png  
NascentOxygen
#185
Dec19-12, 07:09 AM
HW Helper
Thanks
P: 5,149
It's not a very good design. For bias point stability (against VB and variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across RC as VCE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?
Jony130
#186
Dec19-12, 07:37 AM
P: 406
No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420
michael1978
#187
Dec19-12, 09:41 AM
P: 133
Quote Quote by NascentOxygen View Post
It's not a very good design. For bias point stability (against VB and variations) you should allow a couple of volts across the emitter resistor; you have just 0.4V.

For maximum symmetrical output swing, you should allow approx as much voltage across RC as VCE. But you have ~1.8V and 7.8V, resp.

At what frequency are you measuring the ac gain?
at 20kh
thnx for reply
michael1978
#188
Dec19-12, 09:50 AM
P: 133
Quote Quote by Jony130 View Post
No the design is good. Simply the BJT that michael1978 use in his simulation has a very low current gain.
Ie = 0.4V/220Ω = 1.8mA

And Ib = IR4 - IR5

IR4 = 1V/30K = 33.4μA

IR5 = (10V - 1V)/150kΩ = 9V/150KΩ = 60μA

So Ib = 60μA - 33.4μA = 26.6μA

And Hfe = β = (Ie/Ib - 1) = 66.6

And this circuit was design for Hfe > 420
so result is correct 66.6, for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply
michael1978
#189
Dec20-12, 05:36 AM
P: 133
Quote Quote by michael1978 View Post
so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply
thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply[/QUOTE]
thnx for reply
michael1978
#190
Dec22-12, 05:41 PM
P: 133
Quote Quote by michael1978 View Post
thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply
thnx for reply[/QUOTE]

hi Jony, can you answer me please
Jony130
#191
Dec23-12, 07:14 AM
P: 406
Quote Quote by michael1978 View Post
thnx for reply
]so result is correct 66.6, or voltage gain is 66.6, if so 66.6*2Mv=133mv again the result is not corrcect
for ib like this is formula for current Ir5-Ir4? is not the same for ir4 and ir5 they are in serie ir4 and ir5
but Jony you told Ve to be 1v why is now Ve 0.4V?
thnx for reply
66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.
michael1978
#192
Dec23-12, 08:26 AM
P: 133
Quote Quote by Jony130 View Post
66 is a BJT current gain, not a amplifier voltage gain.
So in order to make this circuit work properly. You need change the BJT that you use in your simulation program. You need to find the BJT with Hfe > 300 in your simulation.
ah 66 is curent gain, so you mean to change type of transistor, but is possible for you to tell me any type of transistor wich work, because there are a lot of type of transistor, and somebody he tell me one type of transistor but that transistor dont exist in my list of transistors
and yes Jony if you can tell me, i try to find self soms but i cant find it, that is the problem
thnx for reply
Jony130
#193
Dec23-12, 08:35 AM
P: 406
BC548C
What simulation program do yo use ?
michael1978
#194
Dec23-12, 08:44 AM
P: 133
Quote Quote by Jony130 View Post
BC548C
What simulation program do yo use ?
o joney i dont have, i have bc548A but not BC548C, i use B2 Spice A/V
michael1978
#195
Dec23-12, 09:00 AM
P: 133
hey Jony i try with bc548A, and i get voltage gain of 84mv is this correct?
Jony130
#196
Dec23-12, 09:03 AM
P: 406
Use BC547C
michael1978
#197
Dec23-12, 09:10 AM
P: 133
Quote Quote by Jony130 View Post
Use BC547C
yes Joney i have that type i change, and now voltage gain is 94mv is correct now?
Jony130
#198
Dec23-12, 09:56 AM
P: 406
Well first you need to learn how to use the simulation program.
I use B2 Spice V5 and as a BJT use BC547C. And AC sweep analysis show that voltage gain is equal to 50.1V/V
For 20mV at input I get 1V at output.
Attached Thumbnails
aa.PNG   aa2.PNG  


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