{"id":7839,"date":"2022-05-05T18:18:33","date_gmt":"2022-05-05T12:48:33","guid":{"rendered":"https:\/\/www.goseeko.com\/blog\/?p=7839"},"modified":"2025-06-02T20:47:03","modified_gmt":"2025-06-02T15:17:03","slug":"what-is-the-cb-hybrid-model","status":"publish","type":"post","link":"https:\/\/www.goseeko.com\/blog\/what-is-the-cb-hybrid-model\/","title":{"rendered":"What is the CB Hybrid Model?<\/strong>"},"content":{"rendered":"\n

We use h-parameter model to analyze the Small signal response. Before moving on to the CB (common base) hybrid model we first understand about the hybrid model. We then first derive the general expression of various transistor parameters in terms of hybrid parameters. Later in this article we will derive expressions of the CB (common base) hybrid model.<\/p>\n\n\n\n

The hybrid two port model is shown below.<\/p>\n\n\n\n

<\/strong><\/p>\n\n\n\n

Vi<\/sub> = h11<\/sub>Ii<\/sub> + h12<\/sub>Vo<\/sub><\/p>\n\n\n\n

Io<\/sub> = h21<\/sub>Ii<\/sub> + h22<\/sub>Vo<\/sub><\/p>\n\n\n\n

h11<\/sub> = hi<\/sub> = Vi<\/sub>\/Io<\/sub> for Vo<\/sub> =0  [Input resistance]<\/p>\n\n\n\n

Reverse voltage gain <\/p>\n\n\n\n

h12<\/sub> = hr<\/sub> = Vi<\/sub>\/Vo<\/sub> for Ii<\/sub> =0 <\/p>\n\n\n\n

Then Forward current gain <\/p>\n\n\n\n

h21<\/sub> = hf<\/sub> = Io<\/sub>\/Ii<\/sub> for Vo<\/sub> =0  <\/p>\n\n\n\n

Hence,<\/p>\n\n\n\n

h22<\/sub> = ho<\/sub> = Io<\/sub>\/Vo<\/sub> for Ii<\/sub> =0   [output admittance]<\/p>\n\n\n\n

The hybrid equivalent model is shown below<\/p>\n\n\n\n

\"\"\/
hybrid Equivalent Model<\/figcaption><\/figure>\n\n\n\n

The transistor mode<\/a>l has three terminals with two ports.<\/p>\n\n\n\n

Input resistance = hi<\/sub> <\/p>\n\n\n\n

Output conductance= ho<\/sub><\/p>\n\n\n\n

The reverse transfer voltage <\/p>\n\n\n\n

hr<\/sub>= Vi<\/sub>\/Vo<\/sub> <\/p>\n\n\n\n

hf<\/sub>= Io<\/sub>\/Ii<\/sub> = forward transfer current ratio<\/p>\n\n\n\n

The simplified model is shown below<\/p>\n\n\n\n

\"\"\/
Simplified hybrid Model<\/figcaption><\/figure>\n\n\n\n

Simplified Hybrid Model parameters<\/h2>\n\n\n\n

Finding current gain Ai<\/sub><\/strong><\/p>\n\n\n\n

Applying KCL at the output of above circuit<\/p>\n\n\n\n

Io<\/sub> =I + hf<\/sub> Ib<\/sub> = Vo<\/sub> ho<\/sub> + hf<\/sub> Ib<\/sub><\/p>\n\n\n\n

Vo<\/sub> = -Io<\/sub> RL<\/sub><\/p>\n\n\n\n

Then <\/p>\n\n\n\n

Io<\/sub> = -Io<\/sub> RL<\/sub> ho<\/sub> + hf<\/sub> Ib<\/sub><\/p>\n\n\n\n

Io<\/sub> (1 + RL<\/sub> ho<\/sub>) = hf<\/sub> Ib<\/sub><\/p>\n\n\n\n

Ai<\/sub> = Io<\/sub>\/ Ib<\/sub> = hf<\/sub>\/(1 + RL<\/sub> ho<\/sub>)<\/p>\n\n\n\n

Finding Voltage Gain<\/a> AV<\/sub><\/strong><\/p>\n\n\n\n

Applying KVL at input of the above h-model<\/p>\n\n\n\n

Vi<\/sub> = hi<\/sub> Ib<\/sub> + hr<\/sub> Vo<\/sub><\/p>\n\n\n\n

But Ib<\/sub> = (1 + ho<\/sub> RL<\/sub>)\/hf<\/sub><\/p>\n\n\n\n

Io<\/sub> = – Vo<\/sub>\/RL<\/sub><\/p>\n\n\n\n

Substituting in above equation and solving for Vo<\/sub>\/Vi\u00ad\u00ad<\/sub> we get<\/p>\n\n\n\n

Av<\/sub> = – hf<\/sub> RL\/hi<\/sub> + (hi<\/sub> ho<\/sub>– hf<\/sub> hr<\/sub>) RL<\/sub><\/p>\n\n\n\n

Finding Input Impedance Z<\/strong>i<\/sub><\/strong><\/p>\n\n\n\n

Vi<\/sub> = hi<\/sub> Ib<\/sub> + hr<\/sub> Vo<\/sub><\/p>\n\n\n\n

Then<\/p>\n\n\n\n

Io<\/sub> = – Vo<\/sub>\/RL<\/sub><\/p>\n\n\n\n

Ai<\/sub> = Io<\/sub>\/ Ib<\/sub><\/p>\n\n\n\n

Since<\/p>\n\n\n\n

Vi<\/sub> = hi<\/sub> Ib<\/sub> – hr<\/sub>RL<\/sub>Ai<\/sub>Ib<\/sub><\/p>\n\n\n\n

Hence the input impedance is<\/p>\n\n\n\n

Zi<\/sub> = Vi<\/sub>\/Ib<\/sub> = hi<\/sub> – hr<\/sub>RL<\/sub>Ai<\/sub><\/p>\n\n\n\n

Finding Output impedance Z<\/strong>o<\/sub><\/strong><\/p>\n\n\n\n

It is ratio of output voltage to output current with Vs<\/sub> =0. Then the value of output current becomes<\/p>\n\n\n\n

Io<\/sub> = Vo<\/sub>ho<\/sub> + hf<\/sub>Ib<\/sub><\/p>\n\n\n\n

Ii<\/sub> = -hr<\/sub>Vo<\/sub>\/Rs<\/sub> + hi<\/sub><\/p>\n\n\n\n

Zo<\/sub> = Vo<\/sub>\/Io<\/sub> = 1\/[ho<\/sub> \u2013 (hf<\/sub> h r<\/sub> \/hi<\/sub>+ Rs<\/sub>)]<\/p>\n\n\n\n

CB hybrid Model Parameters<\/h2>\n\n\n\n

The common base configuration hybrid model is also shown below. It is the hybrid equivalent of a CB transistor. As we also know input in common base transistors is applied between base emitter terminals. Then the output is collected from the collector emitter junction. Then value of input voltage Vbe<\/sub> and output current IC<\/sub> are as <\/p>\n\n\n\n

Vbe<\/sub> =hib<\/sub> ib<\/sub> +hrb<\/sub> VC<\/sub> <\/p>\n\n\n\n

ie<\/sub> = hfb<\/sub>ib<\/sub> + hob<\/sub>VC<\/sub> <\/p>\n\n\n\n

\"\"\/
CB hybrid model<\/figcaption><\/figure><\/div>\n\n\n\n

The hybrid expression can also be obtained from the general hybrid formula discussed above by adding a second subscript letter b which commonly stands for base with the h-parameters.<\/p>\n\n\n\n

Current Gain<\/h3>\n\n\n\n

As derived earlier the value of current gain for CB configuration will be<\/p>\n\n\n\n

Ai<\/sub> = -(hfb<\/sub> \/ (1+ hob<\/sub> rL<\/sub>)<\/p>\n\n\n\n

rL<\/sub>\u2192 AC load resistance. <\/p>\n\n\n\n

But rL <\/sub>= Rc<\/sub>|| RL<\/sub><\/p>\n\n\n\n

Since hfb<\/sub> is a positive number therefore Ai<\/sub> of a CE amplifier is negative.<\/p>\n\n\n\n

Input Resistance<\/h3>\n\n\n\n

Then value of resistance at the input terminal in the figure is <\/p>\n\n\n\n

Ri <\/sub>= hib<\/sub>+ hrb<\/sub> Ai <\/sub>rL<\/sub>= hib <\/sub>– ((hrb<\/sub>hfb <\/sub>)\/ (hob<\/sub> + (1\/rL<\/sub>)))<\/p>\n\n\n\n

The input resistance of the amplifier stage also depends upon the biasing arrangement. Then a fixed biased the stage input resistance is,<\/p>\n\n\n\n

Ris<\/sub>= Ri <\/sub><\/p>\n\n\n\n

Voltage Gain<\/h3>\n\n\n\n

Av<\/sub> = Ai <\/sub> r1 <\/sub>\/Ri<\/sub><\/p>\n\n\n\n

We also know that common base amplifiers have positive current gain. Hence, the voltage gain is also positive. Therefore, the output and input are in phase. Then h-parameter is<\/p>\n\n\n\n

Av<\/sub> = hfb <\/sub> rL <\/sub>\/(hib<\/sub> + \u0394h rL<\/sub>)<\/p>\n\n\n\n

\u0394h = hib <\/sub> hob <\/sub>– hrb <\/sub> hfb<\/sub><\/p>\n\n\n\n

Output Resistance<\/h3>\n\n\n\n

Ro<\/sub> = Rs <\/sub> + hib  <\/sub>\/(Rs <\/sub>hob<\/sub> + \u0394h)<\/p>\n\n\n\n

\u0394h = hib <\/sub> hob <\/sub>– hrb <\/sub> hfb<\/sub><\/p>\n\n\n\n

Overall Voltage Gain<\/strong><\/p>\n\n\n\n

Avs<\/sub> = Av <\/sub> Ris <\/sub>\/(Rs<\/sub> + Ris<\/sub>)<\/p>\n\n\n\n

Overall Current Gain<\/strong><\/p>\n\n\n\n

Ais<\/sub> = Avi<\/sub> Rs <\/sub>\/(Rs<\/sub> + Ris<\/sub>)<\/p>\n","protected":false},"excerpt":{"rendered":"

We use small signal transistor model to analyze h-parameters. A transistor can be treated as a two port terminal.<\/p>\n","protected":false},"author":3,"featured_media":7841,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[37],"tags":[],"class_list":["post-7839","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-electronics"],"yoast_head":"\nWhat is the CB Hybrid Model? - Goseeko blog<\/title>\n<meta name=\"description\" content=\"We use small signal transistor model to analyze h-parameters. A transistor can be treated as a two port terminal. 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