Forum: Analog Circuits How to make a Attenuation (Channel Out) without it acting as a Voltage Divider?

von R098 (Guest)

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Hi, I want to rebuild a amplifier that Amplifies a 2.8Vpp DC-50MHZ 
Signal. The Output should be 12dB (11,3Vpp) as in the technical 
Datasheet mentioned. So far so good, it works with the THS3491 (right on 
the edge, but it works in the simulation.

Now I want to design the Attenuation on CH1 and CH2. It attenuate 33dB. 
Problem I have with this is, that the Attenuation build a Voltage 
Divider with a T/Pi Attenuator and the 50 Ohm&also the 50 Ohm Load. It 
changes my Output Voltage..

How can I make that the CH1&CH2 Path (for measuring internal and output 
Voltage) don't change my Voltage? I thought that I could use a Buffer, 
than the Input is high impedance. But I couldn't find any Buffer that 
works with 0.1dB flatness at 50MHz and also the Input would be 22.5Vpp 
on the one at CH1..

How would you tackle this?

Datasheet of Reference Design I want to build: 


: Moved by Moderator
von Der_Zahn_der_Zeit (Guest)

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I suspect the TO speaks mainly German. So my answer written in German 
further back, first the translation made by DeepL:

I assume that the following happens in the B-Amp 12: The two attenuators 
are resistor dividers with 50 ohm output impedance. They don't need to 
be T or Pi dividers. However, the load and the change in output 
impedance must be compensated by making the series resistance and gain 
slightly larger. The circuit shows estimated values, an exact 
calculation is more complex.

There is of course a feedback from the Att_Out to the VOut. The high 
attenuation is chosen so that it can be neglected. Otherwise another 
buffer would be necessary.


Ich vermute, der TO spricht hauptsächlich Deutsch. Also meine in Deutsch 
geschriebene Antwort weiter hinten, zuerst die von DeepL erstellte 

Ich nehme an, dass im B-Amp 12 folgendes passiert: Die beiden 
Abschwächer sind Widerstands-Teiler mit 50 Ohm Ausgangsimpedanz. Das 
brauchen keine T- oder Pi-Teiler zu sein. Allerdings muss die Belastung 
und die Änderung der Ausgangsimpedanz dadurch kompensiert werden, dass 
der Längswiderstand und die Verstärkung etwas größer werden. Die 
Schaltung zeigt Schätzwerte, eine genaue Berechnung ist aufwändiger.

Es gibt natürlich eine Rückwirkung vom Att_Out auf den VOut. Die hohe 
Dämpfung ist gewählt, damit sie vernachlässigt werden kann. Andernfalls 
wäre ein weiterer Buffer erforderlich.


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