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need a advice choosing resistor value

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btminzon

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Hi, I need a advice in choose the better value os resistor, as shown in figure below. I have a digital signal (3.3V) with maximum 100Khz in transistor base, and the signal output is in colector. The input signal contains maximum 25mA and 3.3V, and I need to amplifi to about 80mA more or less, and 5V is desirable. Can somebody help me? tnkx!!!
 

you may use this circuit
 

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    btminzon

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For equal output source and sink currents:
 

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    btminzon

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ok, but wich transistor can i use, in this case? if i'm not wrong, this is a comom-emmiter with a cascode.... I liked very much!!!
 

you circuit is analog inverting amplifier and not suitable for digital signal amplification.
for you purpose, you can parallel 4 buffer gate such as 74HC244

Regards
Davood Amerion
 

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    btminzon

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Davood Amerion what you said is simply not true. Btminzon intention is to use inverter. You should look at his first post carefully.
 

Davood Amerion said:
you circuit is analog inverting amplifier and not suitable for digital signal amplification.
for you purpose, you can parallel 4 buffer gate such as 74HC244
Click to expand...

hey davood Amerion, i did this, and was a better solution for me. I tried to use the transistor amplifier, but for my application, i had better results with this buffer!!!

best regards

Breno
 

Hi btminzon,
R3 in your “amp.JPG” has to be low for driving 80mA and when the out put is low the BC337 has to sink the current through R3 – which you may not like.
Borber’s “cscs,gif” is definitely a better solution with minimum wastage of power.

You may remove Q3 and connect a diode parallel to the base-emitter of Q2 [base = anode & emitter = cathode] so that Q1 sinks the current through D1 when the out put is low, else Q2 sources the current when the out put is high.
 

ok suvendubose, tkns the help. so you mean, i may put a diode with the anode in q2 emitter and conected to gnd, or just conected to it?
 

Anode goes to emitter of Q2 and cathode to base of Q2, Q3 is removed.
 

Yes, I agree with Borber. In fact I am new in the group. I am yet to learn the methods of editing the circuit drawing while answering, and I had to describe the connection.

One 10K - 22K resistor may be connected from emitter of Q2 [anode of diode] to gnd.
 

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    btminzon

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So, this is the final circuit, as figure above. If i am wrong, please correct me. Tnks suvendubose, Borber, Davood Amerion and leotim.

Regards

Breno
 

Sorry, the circuit needs one modification. Just reverse the diode, connect cathode of diode [D1] to the base of Q2 [collector of Q1], and connect anode of D1 to output [emitter of Q2].
 

ok, now i got it. but it seams to me that this is an inverter amplifier. If i don't want to invert,can i simply put a inverter befor, or there is a way to modify his amplifier to be a non-inverter one?
 

Yes, you can put an inverter before, and also can get a much larger swing with an increased supply voltage in place of +5V.

In other applications where you do not need voltage amplification but need more current driving capability, you can also use the ckt posted by Borber [cscs.gif] with little changes:
Remove R2 and Q1, connect other end of the input resistor R1 directly to the shorted base junction of Q2 & Q3.
But in this case the out put swing is lower and governed by input voltage swing, loading at out put, beta and Vbe drops of transistors [you will get an out put swing of decreased by ~1.6V].
In you present case since you have 3.3 Volt input swing, the out put will be about +0.8 V for 0V input and about +2.5V for 3.3V input.
 

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